When studying the motion of a body in a vertical circle we have to consider the effect of gravity. Due to the influence of the earth’s gravitational field, the magnitudes of the velocity of the body and tension in the string change continuously. It is maximum at the lowest point and minimum at the highest point. Hence the motion in vertical circle is not uniform circular motion.

The Expression for Velocity of Body Moving in a Vertical Circle:

Consider a small body of mass ‘m’ attached to one end of a string and whirled in a vertical circle of radius ‘r’. In this case, the acceleration of the body increases as it goes down the vertical circle and decreases when goes up the vertical circle. Hence the speed of the body changes continuously. It is maximum at the bottommost position and minimum at the uppermost position of the vertical circle. Hence the motion of the body is not uniform circular motion. Irrespective of the position of the particle on the circle, the weight ‘mg’ always acts vertically downward.

Let ‘v’ be the velocity of the body at any point P on the vertical circle. Let L be the lowest point of the vertical circle. Let ‘h’ be the height of point P above point L. let ‘u’ be the velocity of the body at L. By the law of conservation of energy

Energy at point P = Energy at point L

This is an expression for the velocity of a particle at any point performing a circular motion in a vertical circle.

The Expression for Tension in the String in Motion of Body in Vertical Circle:

Consider the centripetal force at point P

Substituting in equation (2)

This is the expression for the tension in the string.

Special Cases:

Case – I: When the body is at the lowermost position i.e. body is at L) (h = 0)

Case – II: When the body is at the uppermost position i.e. body is at H (h = 2r)

Case – III: When the string is horizontal i.e. body is at M (h = r)

Relation Between Tension at the Highest point and at the Lowest Point:

Thus the difference in tensions at the two positions

Thus the tension in the string at the lowest point L is greater than the tension at the highest point H by six times the weight of the body.

Minimum Velocity of Body at Different Positions When Looping a Loop:

Lowest Point L (h = 0):

This is the minimum velocity of the body required so that the body looping a loop i.e. to go round the circle once completely.

This is the minimum velocity at the lowest point of the vertical circle required for a body looping a loop.

Highest Point H (h = 2r):

This is the minimum velocity at the highest point of the vertical circle required for a body looping a loop.

When String is Horizontal  (h = r):

This is the minimum velocity when the string is horizontal at the point M of the vertical circle required for a body looping a loop.

If the velocity ‘v’ of the body is such that, v ≤ √2gr, then the body oscillates about point L, the lowest point of the vertical circle.

If the velocity ‘v’ of the body is such that √2gr < v < √5gr , then the body leaves the circular path and acts like a projectile. It will leave circular motion between M and H.

Energy of Body Moving in a Vertical Circle:

The energy of the body has two components a) kinetic energy (E_K) and b) potential energy (E_P). Sum of the two energies is total energy (E_T)

At Lowest Point L:

At Highest Point H:

When String is Horizontal:

Study Above Proof in Video Lecture

Notes:

Kinetic energy is maximum at the lowest point of the circular path and minimum at the highest point of the circular path. Total energy is conserved.

Uniform Circular Motion in a Vertical Circle

Let us consider a body performing U.C.M. in a vertical circle with constant speed ‘v’. Then,

Tension in String and Acceleration of Particle:

At the Highest Point:

a=v²/r

The direction of tension acceleration is vertically downward

At the Lowest Point:

a=v²/r

The direction of tension acceleration is vertically upward

When String is Horizontal:

a=v²/r

The direction of tension acceleration is horizontal and towards the centre.

At Any Position:

Note:

The difference in tension at the lowest point and the highest point is = 2mg

Numerical Problems on Motion in Vertical Circle:

Example – 01:

A stone weighing 1 kg is whirled in a vertical circle at the end of a rope of length 0.5 m. Find the velocity of a stone at a) lowest position b) midway when the string is horizontal c) topmost position to just complete the circle.

Given: Radius of circular path = r = 0.5 m, Mass of the body = m = 1 kg, g = 9.8 m/s²,

To find: velocity at lowest point = v_L =? velocity when string is horizontal = v_M = ?, velocity at topmost point = v_H = ?.

Solution:  

Ans: Velocity of stone at the lowermost point = 4.95 m/s, The velocity of stone when the string is horizontal = 3.83 m/s, The velocity of stone at the topmost point = 2.21 m/s

Example – 02:

A stone weighing 1 kg is whirled in a vertical circle at the end of a rope of length 0.5 m. Find the tension in the string at a) lowest position b) midway when the string is horizontal c) topmost position to just complete the circle.

Given: Radius of circle = r = 0.5 m, mass of the body = m = 1 kg, g = 9.8 m/s².

To find: Tension at lowest point = T_L =? Tension when string is horizontal = T_M = ?, Tension at topmost point = T_H = ?.

Solution:

Ans: Tension at the lowermost point = 58.8 N, Tension when string is horizontal  = 29.4 N, Tension at topmost point = 0 N

Example – 03:

A stone weighing 50 g is whirled in a vertical circle at the end of a rope of length 1.8 m. Find the tension in the string at a) lowest position b) midway when the string is horizontal c) topmost position to just complete the circle.

Given: Radius of circle = r =1.8 m, mass of the body = m = 50 g =0.050 kg, g = 9.8 m/s²,

To find: Tension at lowest point = T_L =? Tension when string is horizontal = T_M = ?, Tension at topmost point = T_H = ?,

Solution:

Ans: Tension at lowermost point = 2.94 N, Tension when string is horizontal  = 0 N, Tension at topmost point = 1.47 N

Example – 04:

A stone weighing 1 kg is whirled in a vertical circle at the end of a rope of length 1 m. Find the tension in the string and velocity of the stone at a) lowest position b) midway when the string is horizontal c) topmost position to just complete the circle.

Given: Radius of circle = r =1 m, mass of the body = m = 1 kg, g = 9.8 m/s²,

To find: Tension at lowest point = T_L = ?, Tension when string is horizontal = T_M = ?, Tension at topmost point = T_H = ?, velocity at lowest point = v_L = ?, velocity when string is horizontal = v_M = ?, velocity at topmost point = v_H = ?

Solution:

Ans: Velocity of stone at the lowermost point = 7 m/s, Velocity of stone when string is horizontal = 5.42 m/s, Velocity of stone at topmost point = 3.13 m/s, Tension at lowermost point = 58.8 N, Tension when string is horizontal = 0 N, Tension at topmost point = 29.4 N

Example – 05:

An object of mass 0.5 kg attached to a string of length 0.5 m is whirled in a vertical circle at constant angular speed if the maximum tension in the string is 5 kg wt. Calculate the speed of the object and maximum number of revolutions it can complete in one minute

Given: Mass of object = m = 0.5 kg, Radius of circle = r = 0.5 m, Tension in the string = T = 5 kg wt = 5 x 9.8 N.

To Find: Speed of the object = ?, the maximum number of revolutions per minute = N =?

Solution:

In vertical circle maximum tension is at the lowermost point

Ans: Speed of object = 6.64 m/s and maximum number of revolutions = 126.7 r.p.m.

Example – 06:

A pilot of mass 75 kg in a jet aircraft while executing a loop the loop with a constant speed of 360 km/h. If the radius of a circle is 200, compute the force exerted by seat on the pilot a) at the top of loop b) at the bottom loop.

Given: mass of piolot = m = 75 kg, radius of circle = r = 200 m, velocity of plane = v =360 km/h = 360 x 5/18 = 100 m/s  ,

To find: force at the top of loop = F_T =? Force at bottom of loop = F_B = ?,.

Solution:

At the topmost point, the centrifugal force acts vertically upward and the weight of the body acts vertically downward. Thus the net force on the pilot by the seat is given by

At the bottom-most point, both the centrifugal force and the weight of the body act vertically downward. Thus the net force on the pilot by the seat is given by

Ans: The force exerted by the seat on the pilot at the topmost point is 3015 N, The force exerted by the seat on the pilot at the bottom-most point is 4485 N.

Example – 07:

A pilot of mass 50 kg in a jet aircraft while executing a loop the loop with a constant speed of 250 m/s. If the radius of circle is 5 km, compute the force exerted by sent on the pilot a) at the top of loop b) at the bottom loop.

Given: mass of piolot = m = 50kg, radius of circle = r = 5 km = 5000 m, velocity of plane = v = 250 m/s,

To find: force at the top of loop = F_T =?, Force at bottom of loop = F_B = ?,.

Solution:

At the topmost point, the centrifugal force acts vertically upward and the weight of the body acts vertically downward. Thus the net force on the pilot by the seat is given by

At the bottom-most point, both the centrifugal force and the weight of the body act vertically downward. Thus the net force on the pilot by the seat is given by

Ans: The force exerted by the seat on the pilot at the topmost point is 135 N, The force exerted by the seat on the pilot at the bottom-most point is 1115 N.

Example – 08:

A ball is released from a height h along the slope and at the end of a slope moves along a circular track of radius R without falling vertically downwards. Determine the height h in terms of R.

Solution:

As the ball is released from a height along the slope and moves along a circular track thus it is looping a loop or it has sufficient velocity at the lowest point to loop a loop.

At lowest point, v = √5gr

By the law of conservation of energy

Potential energy at P = Kinetic energy at Q

Example – 09:

• A block of mass 1 kg is released from point P on a frictionless track which ends in a quarter-circular track of radius 2m at the bottom. What is the magnitude of radial acceleration and total acceleration of the block when it arrives at Q?

• Given: Height of point P above datum = 6 m, Radius of circular track = 2 m
• To find: the magnitude of radial acceleration = a_R = ? and magnitude of total acceleration = a =?
• Solution:

By the principle of conservation of energy

Total Energy at P = Total energy at Q

The magnitude of the radial acceleration is given by

At point Q the tangential acceleration is the acceleration due to gravity

Ans: Magnitude of radial acceleration = 39.2 m/s² and the magnitude of total acceleration = 40.4 m/s².

Example – 10:

A 3 kg ball is swung in a vertical circle at the end of an inextensible string 3 m long. What is the maximum and minimum tension in the string if the ball moves 90/π revolutions per minute.

Given: Mass of body = m = 3 kg, radius of circle = r = 3 m, rpm = N = 90/π r.p.m.  ,

To find: Maximum tension = T_max =? minimum tension = T_min = ?,

Solution:

At the bottom-most point, the centrifugal force acts vertically downward and the weight of the body acts vertically downward. Thus the tension in the string is maximum.

T_max = mrω² + mg = 3 x 3 x (3)² + 3 x 9.8 = 110.4 N

At the topmost point, the centrifugal force acts vertically upward and the weight of the body acts vertically downward. Thus the tension in the string is minimum.

T_min = mrω² – mg = 3 x 3 x (3)² – 3 x 9.8 = 51.6 N

Ans: Maximum tension at the bottom-most point is 110.4 N, Minimum tension at the topmost point is 51.6 N.

Example – 11:

A 2 kg ball is swung in a vertical circle at the end of an inextensible string 2 m long. What are the speed and angular speed of the ball if the string can sustain maximum tension of 119.6 N.

Given: Mass of body = m = 2 kg, radius of circle = r = 2 m, Maximum tension = T = 119.6 N,

To find: Linear speed = v =? angular speed = ω = ?

Solution:

At the bottom-most point, the centrifugal force acts vertically downward and the weight of the body acts vertically downward. Thus the tension in the string is maximum.

Ans: Maximum linear speed = 10 m/s, Maximum angular speed = 5 rad/s.

Previous Topic: The Concept of Conical Pendulum

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Science > Physics > Circular Motion > Motion in Vertical Circle

The post Motion in a Vertical Circle appeared first on The Fact Factor.

A conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’. In this case, the string makes a constant angle with the vertical. The bob of pendulum describes a horizontal circle and the string describes a cone.

Expression for Period of Conical Pendulum:

Let us consider a conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’. Let the string makes a constant angle ‘θ’ with the vertical. let ‘h’ be the depth of the bob below the support.

The tension ‘F’ in the string can be resolved into two components. Horizontal ‘Fsin θ’ and vertical ‘Fcos θ’.

The vertical component (F cos θ) balances the weight mg of the vehicle.

F cosθ  = mg  ………….. (1)

The horizontal component (F sin θ) provides the necessary centripetal force.

F sin θ = mv²/r  ………… (2)

Dividing equation (2) by (1) we get,

But v = rω
Where ω is angular speed and T is the period of the pendulum.

From figure tan θ = r/h

Substituting in equation (4)

This is an expression for the time period of a conical pendulum.

The time taken by the bob of a conical pendulum to complete one horizontal circle is called the time period of the conical pendulum

Notes:

The semi-vertical angle θ or angle made by the string with vertical depends on the length and period of the conical pendulum. If the time period decreases, then the quantity cosθ decreases. in turn θ increases. θ can never be 90° because for this period T = 0.

Expression for Tension in the String of Conical Pendulum:

Let us consider a conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’. Let the string makes a constant angle ‘θ’ with the vertical. let ‘h’ be the depth of the bob below the support.

The tension ‘F’ in the string can be resolved into two components. Horizontal ‘Fsin θ’ and vertical ‘Fcos θ’.

The vertical component (F cos θ) balances the weight mg of the vehicle.

F cosθ = mg ………….. (1)

The horizontal component (F sin θ) provides the necessary centripetal force.

F sin θ = mv²/r ………… (2)

Dividing equation (2) by (1) we get,

Squaring equations (1) and (2) and adding

Substituting equation (5)

This is an expression for the tension in the string of a conical pendulum.

Notes:

A simple pendulum is a special case of a conical pendulum in which angle made by the string with vertical is zero i.e. θ = 0°.

Then the period of the simple pendulum is given by

For conical pendulum θ < 10° time period obtained is almost the same as the time period for simple pendulum having the same length as that of the conical pendulum. Hence when performing an experiment on a simple pendulum, the teacher advises not to increase θ beyond 10°.

Characteristics of Semi-vertical Angle of Conical Pendulum:

• The semi-vertical angle is the angle made by the string of conical pendulum with the vertical.
• It is given by the expression

• It is independent of the mass of the bob of the conical pendulum.
• The semi-vertical angle θ depends on the length and period of the conical pendulum. If the time period decreases, then θ increases.
• θ can never be 90° because for this period T = 0.

Factors Affecting Time Period of Conical Pendulum:

• The time period of a conical pendulum is given by

• The time period of a conical pendulum is directly proportional to the square root of its length.
• The time period of a conical pendulum is directly proportional to the square root of the cosine ratio of the semi-vertical angle that is the angle made by the string of conical pendulum with the vertical.
• The time period of a conical pendulum increases with the increase in the value of the semi-vertical angle.
• The time period of a conical pendulum is inversely proportional to the square root of the acceleration due to gravity at that place.
• The time period of a conical pendulum is independent of the mass of the bob of the conical pendulum.

Numerical Problems:

Example – 01:

A cord 5.0 m long is fixed at one end and to its other end is attached a weight which describes a horizontal circle of radius 1.2 m. Compute the speed of the weight in the circular path. Find its time period.

Given: Length of conical pendulum = l = 5.0 m, radius of circular path of bob = r = 1.2 m, g = 9.8 m/s²,

To find: velocity of weight = v =? Period = T = ?

Solution:

By Pythagoras theorem

Ans: The speed of the weight is 1.7 m/s, the period of the motion of the weight is 4.41 s.

Example – 02:

A stone of mass 1 kg is whirled in a horizontal circle attached at the end of 1 m long string making an angle of 30° with the vertical. Find the period and centripetal force if g = 9.8m/s².

Given: Length of pendulum = l = 1 m, angle with vertical = θ = 30°, g = 9.8 m/s²,

To find: Period = T =? Centripetal force = F = ?

Solution:

Ans: Period of motion = 1.867 s, Centripetal force = 5.657 N

Example – 03:

A string of length 0.5 m carries a bob of mass 0.1 kg with a period of 1.41 s. Calculate the angle of the inclination of the string with vertical and tension in the string.

Given:  Length of pendulum = l = 0.5 m, mass of bob = m = 0.1 kg, Period = T = 1.41 s, g = 9.8 m/s²,

To find:  The angle with vertical = θ =? Tension = F = ?

Solution:

For equilibrium, Total upward force = Total downward force

Ans: Angle of the string with vertical = 8°52’, The tension in string = 0.992 N

Example – 04:

A conical pendulum has a length of 50 cm. Its bob of mass 100 g performs a uniform circular motion in a horizontal plane in a circle of radius 30 cm. Find a) the angle made by the string with the vertical b) the tension in the string c) the period d) the speed of the bob e) centripetal acceleration of the bob f) centripetal and centrifugal force acting on the bob.

Solution:

Given: length of pendulum = l = 50 cm = 0.5 m, mass of bob = m = 100 g = 0.1 kg, radius of circle = r = 30 cm = 0.3 m, g = 9.8 m/s²,

To find:  Angle made with vertical = θ =? Tension = T =  ?, Period, Centripetal force = F = ?, centrifugal force = F = ?

Calculation of tension in string (In diagram F = T)

For equilibrium, Total upward force = Total downward force

Calculation of time period

Calculation of centrifugal force:

Ans: The angle of the string with vertical = 36°52’, The tension in the string = 1.225 N, Period = 1.27 s, The velocity of a bob = 1.48 m/s, Centripetal force = 0.73 N radially inward, Centrifugal force = 0.73 N radially outward

Note: The above explanations and problems are based on the assumption that the reference frame is inertial.

Period of a simple pendulum in Non-Inertial Reference Frame:

Reference frames which are at rest or moving with constant velocity with respect to the earth are called inertial reference frames.

Reference frames which are moving with acceleration with respect to the earth are called non-inertial reference frames. In the case of the non-inertial reference frame, we have to consider the pseudo force acting on the bob of the pendulum and corresponding changes should be done in the formula.

• For a simple pendulum oscillating in a vehicle moving horizontally with acceleration (a) the time period  is

The pendulum will make an angle θ = tan^-1(g/a)  with the vertical

• For a simple pendulum oscillating in a lift moving upward with acceleration (a) the time period of oscillation is

• For a simple pendulum oscillating in a lift moving downward with acceleration (a) the time period of oscillation is

• For a conical pendulum oscillating in a vehicle moving horizontally with acceleration (a) the time period is

The pendulum will make an angle θ + tan-1(g/a)  with the vertical

• For a conical pendulum oscillating in a lift moving upward with acceleration (a) the time period of oscillation is

• For  a conical pendulum oscillating in a lift moving downward with acceleration (a) the time period of oscillation is

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Science > Physics > Circular Motion > Conical Pendulum

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Example – 01:

To what angle must a racing track of radius of curvature 600 m be banked so as to be suitable for a maximum speed of 180 km/h?

Given: maximum speed v = 180 km/hr = 180 x 5/18 = 50 m/s, Radius of curvature = r = 600 m,  g = 9.8 m/s²

To Find: Angle of banking = θ =?

Solution:

Ans: Angle of banking = 23° 18’

Example – 02:

A curve in the road is in the form of an arc of a circle of radius 400 m. At what angle should the surface of the road be laid inclined to the horizontal so that the resultant reaction of the surface acting on a car running at 120 km/h is normal to the surface of the road?

Given: speed of the car = v = 120 km/hr = 120 x 5/18 = 33.33 m/s, radius of curve = r = 400m, g = 9.8 m/s²

To Find:  Angle of banking = θ = ?

Solution:

Ans: Angle of banking = 15° 49’

Example – 03:

A train of mass 10⁵ kg rounds a curve of radius 150 m at a speed of 20m/s. Find the horizontal thrust on the outer rail if the track is not banked. At what angle must the track be banked in order that there is no thrust on the rail? g= 9.8 m/s².

Given: mass of train = m = 10⁵ kg, velocity of train = v = 20 m/s, radius of curve = r = 400m, g = 9.8 m/s²

To Find: Horizontal thrust = F =?  Angle of banking = θ = ?

Solution:

The horizontal thrust is equal to the centripetal force in magnitude.

Ans: Horizontal thrust = 2.67 x 105 N, the angle of banking = 15° 13’

Example – 04:

The radius of curvature of a metre gauge railway line at a place where the train is moving at 36 km/h is 50m. If there is no side thrust on the rails find the elevation of the outer rail above the inner rail.

Given: speed of train = v = 36 km/hr = 36 x 5/18 = 10 m/s, radius of curve = r = 50m,   g = 9.8 m/s², For metre gauge, distance between rail = l   = 1 m.

To Find: elevation = h =?

Solution:

Ans: The elevation of outer rail over inner rail = 0.2 m

Example – 05:

Find the angle of banking of the railway track of radius of curvature 3200 m if there is no side thrust on the rails for a train running at 144 km/h. Find the elevation of the outer rail above the inner one if the distance between the rails is 1.6 m.

Given: Speed of train = v = 144 km/hr = 144 x 5/18 = 40 m/s, radius of curve = r = 3200m, g = 9.8 m/s², distance between rail = l = 1.6 m.

To find: Angle of banking = θ =? Elevation = h =?

Solution:

Ans: The angle of banking = 2°55’, the elevation of outer rail over inner rail = 81 mm

Example – 06:

A metre gauge train is moving at 60 km/hr along a curved rod of a radius of curvature 500m at a certain place. Find the elevation of outer rail over the inner rail, so that there is no side pressure on the rail. (g = 9.8 m/s²)

Given: Speed of train = 60 kmph = 60 x 5/18 =16.67 m/s, radius of curve = r = 500 m, distance between rail = l = 1 m. g = 9.8 m/s²

To find: elevation of outer rail over the inner rail = h =?

Solution: 

The angle of banking is given by

Elevation h = l sinθ =  1 x sin (3^o15’) = 1x 0.0567 =0.0567 m = 5.67 cm

Ans: The angle of banking is 3^o15’ and the elevation of outer rail over the inner rail is 5.67 cm

Example – 07:

A vehicle enters a circular bend of radius 200 m at 72 km/h. The road surface at the bend is banked at 10°. Is it safe? At what angle should the road surface be ideally banked for safe driving at this speed? If the road is 5m wide, what should be the elevation of the outer edge of the road surface above the inner edge?

Given: Speed of vehicle = v = 72 km/hr = 72 x 5/18 = 20 m/s, Angle of banking = θ = 10°, radius of curve = r = 200 m, g = 9.8 m/s²

To Find:  Part – I: To check safety, part – II: nagle of banking = θ =? for given speed, elevation = h =?

Solution:

Part – I:

Now, the velocity of the vehicle (20 m/s) is greater than the safe velocity (18.59 m/s). Hence it is unsafe to drive at a speed 72 km/hr.

Part – II:

Ans: It is unsafe to drive at the speed 72 km/hr. The angle of banking for speed 72 km/hr = 11°32’, Elevation of the outer edge over inner edge = 1m

Example – 07:

Find the minimum radius of an arc of a circle that can be negotiated by a motorcycle, riding at 21 m/s if the coefficient of friction between the tyres and the ground is 0.3. What is the angle made with the vertical by the motorcyclist? g = 9.8 m/s².

Given: Velocity of motor cycle = v = 21 m/s, Coefficient of friction = μ  = 0.3,  g = 9.8 m/s²

To Find: radius of curvature = r =?,  Angle with vertical = θ= ?

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: Radius of the circular path = 150 m, Angle made by the motorcyclist with vertical = 16°42’

Example – 08:

What is the angle of banking necessary for a curved road of 50 m radius for safe driving at 54 km/h? If the road is not banked, what is the coefficient of friction necessary between the road surface and tyres for safe driving at this speed?

Given: velocity of vehicle = v = 54 km/hr = 54 x 5/18 = 15 m/s, radius of curve = r = 50m, g = 9.8 m/s²

To Find: Angle of banking = θ =? coefficient of friction = μ = ?

Solution:

When the road is not banked, the necessary centripetal force is provided by the friction between the road and tyres.

Ans: The angle of banking = 24°42’, the coefficient of friction = 0.4592

Example – 09:

A motorcyclist at a speed of 5 m/s is describing a circle of radius 25 m. Find his inclination with the vertical. What is the value of the coefficient of friction between the road and tyres?

Given: Speed of motor cycle = v = 5 m/s, radius of circle = r = 25 m, g = 9.8 m/s²,

To find:  Angle of inclination = θ=? Coefficient of friction = μ = ?

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: Angle made with the vertical = 5°49’, The coefficient of friction = 0.1020

Example – 10:

A motor van weighing 4400 kg rounds a level curve of radius 200 m on the unbanked road at 60 km/hr. What should be the minimum value of the coefficient of friction to prevent skidding? At what angle the road should be banked for this velocity?

Given: Mass of vehicle = m = 4400 kg, velocity of vehicle = v = 60 km/hr = 60 x 5/18 = 16.67 m/s, r = 200m,  g = 9.8 m/s²

To Find:  Coefficient of friction = μ =?, Angle of banking = θ = ?,

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The coefficient of friction = 0.1418, The angle of banking = 8°4’

Example – 11:

A circular road course track has a radius of 500 m and is banked to 10°. If the coefficient of friction between the road and tyre is 0.25. Compute (i) the maximum speed to avoid slipping (ii) optimum speed to avoid wear and tear of the tyres.

Solution:

Given: radius of curve = r = 500m, Angle of banking = θ = 10°, coefficient of friction = μ = 0.25, g = 9.8 m/s²,

To find: safe velocity = v =? to avoid wear and tear v = ?

Maximum sped to avoid slipping.

Maximum speed to avoid wear and tear

Ans: The maximum velocity to avoid skidding = 46.74 m/s, The maximum velocity to avoid wear and tear of tyres. = 29.39 m/s.

Example – 06:

An aircraft in level flight completes a circular turn in 100 seconds. What is the radius of the circular turn? What is the angle of banking, if the velocity of aircraft is 40 m/s?

Given: Time taken = T = 100 s. velocity of aircraft = 40 m/s.

To Find: radius of circular turn = ?, angle of banking = θ = ?

Solution:

100 seconds are taken to complete a circular turn (circumference)

Ans: The angle of banking of aircraft is 14°25′

Previous Topic: Theory of Banking of Road

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Science > Physics > Circular Motion > Numerical Problems on Banking of Road

The post Numerical Problems on Banking of Road appeared first on The Fact Factor.

In this article, we shall study the necessity of the banking of road and the factors affecting it.

Safe Velocity of a Vehicle on an Unbanked Road:

The necessary centripetal force required to negotiate a turn by a vehicle moving along a horizontal unbanked curved road is provided by the force of friction between the wheels (tyres) and the surface of the road.

Let us consider a vehicle of a mass ‘m’ is moving along a horizontal curved road of radius ‘r’ with speed ‘v’.

Let μ be the coefficient of friction between the road surface and the wheels then

This expression gives the maximum speed with which a vehicle can be moved safely along a horizontal curved road. If speed is more than this velocity, then there is a danger that the vehicle will get thrown (skid) off the road.

Thus the safe velocity on the unbanked road depends on the coefficient of friction between the road and tyres, the radius of the circular path, and acceleration due to gravity at that place.

Banking of Road:

To make the turning of a vehicle on a curved road safer, the outer edge of the road is raised above the inner edge making some inclination with the horizontal. This is known as banking of road.

When the road is banked then, the inclination of the surface of the road with the horizontal is known as the angle of banking.

Necessity of Banking of Road:

• As the speed of vehicle increases, the centripetal force needed for the circular motion of vehicle also increases. In the case of unbanked road necessary centripetal force is provided by the friction between the tyres and the surface of the road. But there is a maximum limit for frictional force, which depends on the coefficient of friction between the wheels and road.
• When the centripetal force needed exceeds the maximum limit of frictional force, the vehicle skids and tries to go off the curved path resulting in an accident.
• Without proper friction, a vehicle will not be able to move on the curved road with a large speed. To avoid this we may increase the force of friction making the road rough. However, this results in the wear and tear of the tyres of the vehicle. Also, the force of friction is not always reliable because it changes when roads are oily or wet due to rains etc. To eliminate this difficulty, the curved roads are generally banked.
• Due to banking of the road, the necessary centripetal force is provided by the component of the normal reaction.

The Expression for the Angle of Banking of Road:

Consider a vehicle of mass ‘m’ is moving with speed ‘v’ on a banked road of radius ‘r’ as shown in the figure. Let ‘θ’ be the angle of banking. The weight “mg” of the vehicle acts vertically downwards through its centre of gravity G, and N is the normal reaction exerted on the vehicle by banked road AC. N is perpendicular to road AC. Let ‘f’ be the frictional force between the road and the tyres of the vehicle.

Now, the normal reaction is resolved into two components (N cosθ) along the vertical (acting vertically upward) and (N sinθ) along the horizontal (towards left) as shown in the figure. Similarly, the frictional force ‘f’ can be resolved into two components (f sinθ) along the vertical (acting vertically downward) and( f cosθ) along the horizontal (towards left) as shown in the figure.

The free-body diagram of a car is as follows

Considering equilibrium

Total upward force = Total downward force

∴   N cosθ  = mg + f sinθ

∴ mg = N cosθ – f sinθ ….. (1)

The horizontal components N sinθ  and f cosθ  provides the necessary centripetal force

mv²/r = N sinθ + f cosθ …… (2)

Dividing equation (2) by (1)

Now, frictional force f  = μ_sN
Where μ_s = coefficient of friction between road and tyres

This is an expression for the velocity of a vehicle on a curved banked road.

Note:

For the given pair of road and tyre μ_s = tan λ, where λ = angle of friction

Case – I: When the road is horizontal θ = 0°

This is an expression for safe velocity on the unbanked road

Case – II:

When the frictional force between the road and tyres of the vehicle is negligible μ_s = 0.

This is an expression for the angle of banking of a road.

When friction is not mentioned in the problem, use this expression to solve the problems.

Safe Velocity on Banked Road:

The expression for safe velocity on the banked road is

The speed will be maximum when tan θ = 1 i.e. θ = 45°. It means the vehicle can be driven with maximum safe speed only when the angle of banking = 45°.

Factors affecting the angle of banking:

• The angle of banking depends on the speed of the vehicle, the radius of the curved road and the acceleration due to gravity g at that place.
• The expression does not contain the term m representing mass, thus the angle of banking is independent of mass ‘m’ of the vehicle. Thus the angle of banking is the same for heavy and light vehicles.
• the angle of banking depends on the radius (r) of the curved road. The angle of banking is inversely proportional to the radius of curvature.
• For the given radius of the curve, the angle of banking increases with the speed.
• For the given radius of the curve and speed of the vehicle angle of banking is inversely proportional to the acceleration due to gravity at that place. The acceleration due to gravity is more on the pole than that at the equator. Thus for the given radius of the curve and speed of the vehicle angle of banking is less at the pole than that at the equator.

If l is the width of the banked road, then the elevation of an outer edge of the road over the inner edge (provided angle of banking is small) is given by h = l sinθ

In actual practice, some frictional forces are always present even on the banked road. So that the actual safe velocity is always greater than the calculated safe velocity on the banked road.

Characteristics of Angle of Banking:

• The angle of banking is given by

• The angle of banking depends on the speed of the vehicle, the radius of the curved road and the acceleration due to gravity g at that place.
• The angle of banking is independent of mass ‘m’ of the vehicle. Thus the angle of banking is the same for heavy and light vehicles.
• the angle of banking is directly proportional to the square of the velocity (v) of the vehicle, inversely proportional to the radius (r) of curvature of the path and inversely proportional to the acceleration due to gravity (g) at that place.
• The acceleration due to gravity is more on the pole than that at the equator. Thus for the given radius of the curve and speed of the vehicle angle of banking is less at the pole than that at the equator.
• The elevation outer edge of a banked road above the inner edge is given by

h = l sinθ

Characteristics of Safe Velocity on Banked Road:

• The safe velocity on the banked road is given by

• The safe velocity of a vehicle on the banked road depends on the radius of the curved road, the angle of banking, and the acceleration due to gravity g at that place.
• Safe velocity is independent of mass ‘m’ of the vehicle. Thus it is the same for heavy and light vehicles.
• For the given radius of the curve and angle of banking, the safe velocity is more at the pole than that at the equator.

High Speed on Unbanked Road:

• During motorcycle race, the riders negotiate the curve on a flat road in that case they have to tilt their motorcycle and themselves to compensate the angle of banking at that place.

• A cycling race track is called velodrome which has a saucer-shaped track. The rider has to take a position on a track as per his/her speed.

Overturning of Vehicle:

Unbanked Road:

When a vehicle moves along a curved path with very high speed, then there is a chance of overturning of the vehicle. Inner wheel leaves ground first. Let a vehicle of a  mass ‘m’ is negotiating a turn along a curved path of radius ‘r’ with speed ‘v’. Let ‘2a’ be the length of the axle i.e. the distance between the two wheels. Let h be the height of the centre of gravity of the vehicle above the ground. Let R₁ and R₂ be the reactions on the inner wheel and outer wheel of the vehicle exerted by the road surface.

The frictional force ‘F’ provides the necessary centripetal force.

Taking the moment of forces about the centre of gravity ‘G’

Fh  + R₁a = R₂a  ………… (2)

Reactions are balanced by the weight of the vehicle

R₁ + R₂ = mg …….. (3)

Solving equations |(1), (2) and (3)

From these equations, we can see that as the speed increases reaction R₁ decreases while reaction R₂ increases. When reaction R1 is zero overturnings of the vehicle takes place.

This is an expression for maximum velocity of the vehicle by which it can be driven beyond which overturning of the vehicle takes place.

Banked Road:

If μ < d/h, vehicle skids and if μ >d/h, vehicle overturns

Previous Topic: Problems on Centripetal and Centrifugal Force

Next Topic: Numerical Problems on Banking of Road

Science > Physics > Circular Motion > Banking of Road

The post Banking of Road appeared first on The Fact Factor.

Example – 19:

The vertical section of a road over a bridge in the direction of its length is in the form of an arc of a circle of radius 19.5 m. Find the greatest velocity at which a car can cross the bridge without losing contact with the road at the highest point if the c.g. of the car is 0.5 m from the ground.

Given: Radius of circle = r = 19.5 + 0.5 = 20 m,

To find: velocity of car = v =?

Solution:

At the highest point, the centrifugal force and weight of the car are equal in magnitude.

Ans: The velocity of car = 14 m/s

Example – 20:

A flyover bridge is in the form of a circular arc of radius 30 m. Find the limiting speed at which a car can cross the bridge without losing contact with the road at the highest point.  Assume the centre of gravity of the car to be 0.5 m above the road.

Given: Radius of circle = r = 30 + 0.5 = 30.5 m,

To find: velocity of car = v =?

Solution:

At the highest point, the centrifugal force and weight of the car are equal in magnitude.

Ans: The velocity of car = 17.3 m/s

Example – 21:

A motorcyclist rides in a vertical circle in a hollow sphere of radius 3 m. Find the minimum speed required so that he does not lose contact with the sphere at the highest point. Also, find its angular speed.

Given: Radius of hollow sphere = r = 3 m,

To find: Minimum speed required = v = ?, Angular speed = ω = ?

Solution:

At the highest point, the centrifugal force and weight of the motorcyclist and motorcycle are equal in magnitude.

Ans: The velocity of car = 5.42 m/s, Angular speed = 1.807 rad/s.

Example – 22:

A motorcyclist rides in a vertical circle in a hollow sphere of radius 12.8 m. Find the minimum speed required so that he does not lose contact with the sphere at the highest point.

Given: Radius of sphere = r = 12.8 m,

To find: Minimum speed = v =?

Solution:

At the highest point, the centrifugal force and weight of the motorcyclist and motorcycle are equal in magnitude.

Ans: Minimum speed of motorcycle = 11.2 m/s

Example – 23:

An object of mass 100 g moves around the circumference of a circle of radius 2m with a constant angular speed of 7.5 rad/s. Compute its linear speed and force directed towards centre.

Given: mass of the body = m = 100 g = 0.1 kg, Radius of circular path = r = 2 m, Angular speed = ω = 7.5 rad/s,

To find: Linear speed = v =? Centripetal force = F = ?

Solution:

v = r ω = 2 x 7.5 = 15 m/s

F = m r ω²

∴ F = 0.1 x 2 x (7.5)² = 11.25 N

Ans: Linear speed = 15 m/s, Centripetal force = 11.25 N radially inward.

Example – 24:

A car of mass 2000 kg rounds a curve of radius 250m at 90 km/hr. Compute its angular speed, centripetal acceleration and centrifugal force.

Given: Mass of car = m = 2000 kg, Radius of circular path = r = 250 m, Linear velocity of car = v = 90 km/hr = 90 x 5/18 = 25 m/s,

To find:  Angular speed = ω =? centripetal acceleration = a = ?, centripetal force = F = ?,

Solution:

v = r ω

∴ ω = v/r = 25/250 = 0.1 rad/s

a = r ω²

∴ a = 250 x (0.1)² = 2.5 m/s²

F = m a = 2000 x 2.5 = 5000 N radially inward

Ans: Angular speed = 0.1 rad/s, Centripetal acceleration = 2.5 m/s² radially inward

The centripetal force and centrifugal force are equal in magnitude but opposite in direction.

Centrifugal force = 5000 N

Ans: Centrifugal force = 5000 N radially outward

Example – 25:

A 0.5 kg mass is tied to one end of a string and rotated in a horizontal circle of 1.25 m radius about the other end. What is the tension in the string if the period of revolution is 5 s. What is the maximum speed of rotation and the corresponding period if the string can withstand a maximum tension of 150 N.

Given: Mass of body = m = 0.5 kg, radius of curvature = r = 1.25 m, Period of revolution = T = 5 s,

To Find: Maximum speed of rotation = v =? F =? Part – II: Linear speed v =? when maximum tension = F = 150 N

Solution:

Part – I:

The necessary centripetal force acting on a body is given by tension in the string

Part – II:

The necessary centripetal force acting on a body is given by tension in the string

Ans: Tension in string = 0.986 N Maximum speed = 19.36 m/s. Period of revolution at maximum speed = 0.41 s

Example – 26:

A coin kept on a horizontal rotating disc has its centre at a distance of 0.1 m from the axis of rotation of the disc. If the coefficient of friction between the coin and the disc is 0.25, find the speed of the disc at which the coin would be about to slip off.

Given: radius of circular path = r = 0.1 m, coefficient of friction = μ = 0.25, g = 9.8 m/s²

To find: Speed of disc = v =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The speed of the disc = 0.495 m/s.

Example – 27:

A coin kept on a horizontal rotating disc has its centre at a distance of 0.25 m from the axis of rotation of the disc. If μ = 0.2, find the angular velocity of the disc at which the coin is about to slip off.   (g = 9.8 m/s²)

Given: radius of circular path = r = 0.25 m, coefficient of friction = μ = 0.2, g = 9.8 m/s²

To find: Angular speed of disc = ω =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The angular speed of the disc = 7.84 rad/s.

Example – 28:

A coin just remains on a disc rotating at 120 r.p.m. when kept at a distance of 1.5 cm from the axis of rotation. Find the coefficient of friction between the coin and the disc.

Given: Radius of circular path = r = 1.5 cm = 0.015 m, rpm = N = 120 r.p.m., g = 9.8 m/s²

To find: coefficient of friction = μ =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: Coefficient of friction = 0.2415

Example – 29:

A coin just remains on a disc rotating at a steady rate of 180 rev/min if kept at a distance of 2 cm from the axis of rotation. Find the coefficient of friction between the coin and the disc.

Given: Radius of circle = r = 2 cm = 0.02 m, rpm = N = 180 r.p.m., g = 9.8 m/s²

To find: coefficient of friction = μ =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The coefficient of friction = 0.724

Example – 30:

A coin kept with its centre at a distance of 9 cm from the axis of rotation of a disc starts slipping off when the disc speed reaches 60 r.p.m. Up to what speed the coin will remain on the disc if its centre is 16 cm from the axis of rotation of the disc?

Given: Initial radius of circle = r₁ = 9 cm, initial rpm = N₁= 60 r.p.m., final radius of circle = r₂ = 16 cm

To find: Final rpm = N₂ =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The required angular speed = 45 r.p.m.

Example – 31:

A coin placed on a turntable rotating at 30 r.p.m. revolves with the table without slipping provided it is not more than 12 cm away from the axis. How far from the axis can the coin be placed so that it revolves with the turntable without slipping if the speed of rotation is 45 r.p.m.?

Given: Initial radius of circle = r₁ = 12 cm, initial rpm = N₁= 45 r.p.m., Final rpm = N₂ = 45 r.p.m.

To find: final radius of circle = r₂ =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The maximum displacement from centre  = 5.33 cm.

Example – 32:

A coin is placed at a distance of 10 cm from the centre of a turntable of radius 1m just begins to slip, when the turntable rotates at a speed of 90 r.p.m. Calculate the coefficient of static friction between the coin and the turntable.

Given: Radius of circle = r = 10 cm = 0.1 m, speed of rotation = N = 90 r.p.m.

To Find: Coefficient of Friction = m =?

Solution:

The necessary centripetal force is provided by the friction between the turntable and coin

Ans: The coefficient of friction is 0.9066

Example – 32:

With what maximum speed a car be safely driven along a curve of radius 40 m on a horizontal road if the coefficient of friction between the car tyres and the road surface is 0.3? (g = 9.8 m/s²)

Given: Radius of circular path = r = 40 m, coefficient of friction = μ = 0.3, g = 9.8 m/s²

To find: Maximum speed = v =?

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The speed of the car = 10.84 m/s.

Example – 33:

Find the maximum speed at which a car can be safely driven along a curve of 100m radius. The coefficient of friction between tyres and the surface of the road is 0.2.

Given: radius of curve = r = 100 m, coefficient of friction = μ = 0.2, g = 9.8 m/s²

To find: maximum speed = v =?

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The speed of the car = 14 m/s.

Example – 34:

A car travelling at 18 km/h just rounds a curve without skidding. If the road is plain and the coefficient of friction between the road surface and the tyres is 0.25, find the radius of the curve.

Given: Speed of car = v = 18 km/hr = 18 x 5/18 = 5 m/s, coefficient of friction = μ = 0.25, g = 9.8 m/s²

To find: radius of curve = r =?

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The radius of road = 10.2 m.

Example – 35:

A car can just go around a curve of 20 m radius without skidding when travelling at 36 km/h. If the road is plain, find the coefficient of friction between the road surface and tyres.

Given: Speed of the car = v = 36 km/hr = 36 x 5/18 = 10 m/s, r = 20m,  g = 9.8 m/s²

To find: coefficient of friction = μ =?

Solution:  

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The coefficient of friction = 0.5102

Example – 04:

A flat curve highway has a radius of curvature 400 m. A car rounds the curve at a speed of 32 m/s. What is the minimum value of the coefficient of friction that will prevent the car from sliding?

Given: Radius of curvature = r = 400 m, speed of car = 32 m/s, g = 9.8 m/s².

To Find: Coefficient of Friction = m =?

Solution:

The necessary centripetal force is provided by the friction between the road and tyres of the car

Ans: The coefficient of friction is 0.2612

Example – 36:

A rotor has a diameter of 4.0 m. The rotor is rotated about the central vertical axis. The occupant remains pinned against the wall. When the linear velocity of the drum is 8 m/s. Compute the coefficient of static friction between the wall of the rotor and the clothing of occupant. Also, calculate the angular velocity of the drum. How many revolutions will it make in a minute?

Given: diameter = d = 4 m, radius = r = 4/2 = 2 m, linear speed = v =.8 m/s , g = 9.8 m/s² ,

To find:  Coefficient of friction = μ = ?, Angular velocity = ω = ?, rpm =  N = ?

Solution:

As the occupant remains pinned against the wall, his weight is equal to the frictional force.

Frictional force = Weight of a body

Ans: Coefficient of friction = 0.3063, Angular velocity = 4 rad/s, Number of revolutions per minute = 38.21.

Previous Topic: More Problems on Centripetal Force

Next Topic: Concept of Banking of Road

Science > Physics > Circular Motion > Numerical Problems on Centripetal Force – 02

The post Numerical Problems on Centripetal Force – 02 appeared first on The Fact Factor.

Example – 01:

A 0.5 kg mass is rotated in a horizontal circle of radius 20 cm.  Calculate the centripetal force acting on it, if its angular speed of revolution is 0.8 rad /s.

Given: Mass of a body = m = 0.5 kg, radius of circular path = r = 20cm = 0.2 m, Angular speed = ω = 0.8 rad/s,

To find: Centripetal force = F =?

Solution:

F = m r ω²

∴ F = 0.5 x 0.2 x (0.8)²

∴ F = 0.5 x 0.2 x (0.64) = 0.064 N

Ans: Centripetal force = 0.064 N acting radially inward.

Example – 02:

An object of mass 0.5 kg is whirled in a horizontal circle of radius 20 cm. Calculate centripetal force acting on it, if its angular speed of revolution is 0.6 rad/s.

Given: Mass of a body = m = 0.5 kg, radius of circular path = r = 20cm = 0.2 m, Angular speed = ω = 0.6 rad/s,

To find: Centripetal force = F =?

Solution:

F = m r ω²

∴ F = 0.5 x 0.2 x (0.6)²

∴ F = 0.5 x 0.2 x (0.36) = 0.036 N

Ans: Centripetal force = 0.036 N acting radially inward.

Example – 03:

A one kg mass tied at the end of the string 0. 5 m long is whirled in a horizontal circle, the other end of the string being fixed. The breaking tension in the string is 50 N. Find the greatest speed that can be given to the mass.

Given: Mass of a body = m = 1 kg, radius of circular path = r = 0.5 m, Centripetal force  = F = 50 N.

To find: Greatest speed = v =?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

Ans: The greatest speed can be given to mass = 5 m/s

Example – 04:

2 kg mass is tied to a string at one end and rotated in a horizontal circle of radius 0.8 m about the other end. If the breaking tension in the string is 250 N, find the maximum speed at which mass can be rotated.

Given: Mass of a body = m = 2 kg, radius of circular path = r = 0.8 m, Centripetal force  = F = 250 N.

To find: Maximum speed = v =?

Solution:

Ans: The greatest speed can be given to mass =10 m/s

Example – 05:

An object of mass 0.5 kg is whirled in a horizontal circle of radius 20 cm. Calculate the maximum number of revolutions per minute, so that the string does not break. The breaking tension of the string is 9.86 N.

Given: Mass of a body = m = 0.5 kg, radius of circular path = r = 20 cm = 0.2 m, Centripetal force  = F = 9.86 N.

To find: Maximum rpm = N =?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

Ans: Max. No. of revolutions = 94.87 r.p.m.

Example – 06:

A mass of 5 kg is tied at the end of the string 1.2 m long rotates in a horizontal circle. If the breaking tension in the string is 300 N, find the maximum number of rotations per minute the mass can make.

Given: Mass of a body = m = 5 kg, radius of circular path = r = 1.2 m, Centripetal force  = F = 300 N.

To find: Maximum rpm = N =?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

Ans: Max. No. of revolutions per minute = 67.56

Example – 07:

A stone is tied to a string 50 cm long and rotated uniformly in a horizontal circle about the other end. If the string can support a maximum tension ten times the weight of the stone, find the maximum number of revolutions per second the string can make before it breaks.

Given: Mass of a body = m, radius of circular path = r = 50 cm = 0.5 m, Centripetal force  = 10 mg.

To find: Maximum number of revolutions per second = n =?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

Ans: Max. No. of revolutions per second = 2.23

Example – 08:

A certain string breaks under a tension of 45 kg-wt. A mass of 100 g is attached to one end of a piece of this string 500 cm long and rotated in a horizontal circle. Neglecting the effect of gravity, find the greatest number of revolutions which, the sting can make without breaking.

Solution:

Given: Mass of a body = m = 100g = 0.1 kg, radius of circular path = r = 500 cm = 5 m, Centripetal force  = F = 9.86 N.

To find: Maximum rps = n =?,

The necessary centripetal force acting on a body is given by tension in the string

Ans: Max. No. of revolutions per second = 4.73

Example – 09:

The breaking tension of a string is 80 kg.-wt. A mass of 1 kg is attached to the string and rotated in a horizontal circle on a horizontal surface of radius 2 m. Find the maximum number of revolutions made without breaking.

Given: Mass of a body = m = 1 kg, radius of circular path = r = 2 m, Centripetal force = F = 80 kg wt = 80 x 9.8 N.

To find: Maximum rps = n =?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

Ans: Max. No. of revolutions per second = 3.15

Example – 10:

A string breaks under a tension of 10 kg-wt. If a string is used to revolve a body of mass 1.2 g in a horizontal circle of radius 50 cm, what is the maximum speed with which a body can be resolved? When a body is revolving at maximum speed, what is its period? (g = 9.8 m/s²)

Given: Mass of a body = m = 1.2 g = 1.2 x 10^-3 kg, radius of circular path = r = 50 cm = 0.5 m, Centripetal force = F = 10 kg wt = 10 x 9.8 N.

To find: Maximum speed = v =? Period = T = ?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

Ans: Max. Speed = 202.1 m/s, Period of revolution at maximum speed = 0.016 s

Example – 11:

A spherical body of mass 1 kg and diameter 2 cm rotates in a horizontal circle at the end of a string 1.99 m. long. What is the tension in the string when the speed of rotation is 6 revolutions in 1.5 s?

Given: Mass of the body = m = 1 kg, diameter of sphere = d = 2cm = 0.02 m. Radius of sphere = 0.01 m, radius of circular path = r = 1.99 + 0.01 = 2m, No. of revolutions = 6, time taken t = 1.5 s.

To find: Tension in string = F =?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

n = 6/1.5 = 4 rad/s

ω = 2πn = 2 x 3.14 x 4 = 25.12 rad/s

F = m r ω²

∴ F = 1 x 2 x (25.12)² = 1262 N

Ans: The tension in the string = 1262 N radially inward.

Example – 12:

A spherical bob of diameter 3 cm having a mass 100 g is attached to the end of a string of length 48.5 cm. Find the angular velocity and the tension in the string, if the bob is rotated at a speed of 600 r.p.m. in a horizontal circle. If the same bob is now whirled in a vertical circle of the same radius, what will be the difference in the tensions at the lowest point and the highest point?

Given: Mass of bob = m = 100 g = 0.1 kg, Radius of circular path = r = 48.5 cm + 1.5 cm = 50 cm = 0.5 m, rpm = N = 600 r.p.m.,

To Find: Angular velocity = ω=? Tension in the string = F = ?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

The difference in the tensions at the lowest point and the highest point

= 6mg = 6 x 0.1 x 9.8 = 5.88 N

Ans: Angular speed = 62.84 rad/s, The tension in string =.197.2 N, The difference in the tensions at the lowest point and the highest point is 5.88 N

Example – 13:

A body of mass 2 Kg is tied to the end of a string of length 1.5 m and revolved about the other end (kept fixed) in a horizontal circle. If it makes 300 rev/min, calculate the linear velocity, the acceleration and the force acting upon the body.

Given: Mass of the body = m = 2 kg, radius of circular path = r = 1.5 m, Revolutions per minute = N = 300 r.p.m.,

To find: Linear speed = v =? Acceleration = a = ?, Force = F = ?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

Ans: Linear speed of body = 47.13 m/s; acceleration of body = 1480 m/s2;  Force acting on body = 2958N radially inward

Example – 14:

A body of mass 20 g rests on a smooth horizontal plane. The body is tied by a light inextensible string 80 cm long to a fixed point in the plane. Find the tension in the string if the body is rotated in a circular path at 30 rev/min. What is the force experienced by a fixed point?

Given:  Mass of the body = m = 20 g = 0.020 kg, radius of circular path = r = 80 cm = 0.8 m, rpm = N = 30 r.p.m.,

To find: Tension in string = F =?

Solution:

The necessary centripetal force acting on a body is given by tension in the string

Ans: Tension in the string = 0158 N radially inward. The fixed point experiences a force of 0.158 N Radially outward.

Example – 15: 

How fast should the earth rotate about it axis so that the apparent weight of a body at the equator be zero? How long would a day be then? Take the radius of the earth = 6400 km.

Given:  Radius of earth = R = 6400 km = = 6.4 x 10⁶ m.,

To find: Angular speed of earth = ω =?, period of earth = T = ?

Solution:

As the apparent weight of the body is zero. The centrifugal force and the weight of the body are equal in magnitude. Let m be the mass of the body.

Ans: The angular speed of earth then = 1.24 x 10^-3 rad/s

T = 2π/ ω = (2 x 3.142)/( 1.24 x 10^-3)  = 5077 s

Ans: The duration of the day then = 5077 s

Example – 02:

An object of mass 100 g moves around the circumference of a circle of radius 2m with a constant angular speed of 7.5 rad/s. Compute its linear speed and force directed towards centre.

Solution:

Given: mass of thre body = m = 100 g = 0.1  kg, Radius of circular path = r = 2 m,  Angulae speed = ω = 7.5 rad/s,

To find: Linear speed = v  = ?, Centripetal force = F = ?

v = r ω = 2 x 7.5 = 15 m/s

F = m r ω²

∴  F = 0.1 x 2 x (7.5)² = 11.25 N

Ans: Linear speed = 15 m/s, Centripetal force = 11.25 N radially inward.

Example – 16:

An electron of mass 9 x 10^-31 kg is revolving in a stable orbit of radius 5.37 x 10^-11 m.  If the electrostatic force of attraction between electron and proton is 8 x 10^-8 N. Find the velocity of the electron.

Given: Mass of electron = m = 9 x 10^-31 kg, r = 5.37 x 10^-11 m, F = 8 x 10^-8 N

To find: velocity of electron = v =?

Solution:

The necessary centripetal force is given by electrostatic attraction between an electron and a proton.

Ans: The velocity of the electron is 2.185 x 10⁶ m/s

Example – 17:

A bucket containing water is tied to one end of a rope 8 m long and rotated about the other end in a vertical circle. Find the minimum number of rotations per minute in order that water in the bucket may not spill? (g = 9.8 m/s²)

Given:  Radius of circular path = r = 8 m, g = 9.8 m/s²,

To find: rpm = N =?

Solution:

At the highest point, the centrifugal force and the weight of water and bucket are equal in magnitude. Let m be the mass of the water and bucket.

mg = m r ω²

∴   g = r ω²

∴   ω² =   g /r = 9.8/8 = 1.225

∴   ω=   1.107 rad/s

ω= 2πN / 60

∴   N = 60ω/2π = (60 x 1.107) / (2 x 3.142) = 10.57

Ans: Max. No. of revolutions per minute = 10.57

Example – 18:

A bucket containing water is tied to one end of a rope 0.75 m long and rotated about the other end in a vertical circle. Find the speed in order that water in the bucket may not spill? Also, find the angular speed. (g = 9.8 m/s²)

Given: Radius of circle = r = 0.75 m, g = 9.8 m/s²

To find: linear speed = v =? angular speed = ω= ?

Solution:

At the highest point, the centrifugal force and the weight of water and bucket are equal in magnitude.

Let m be the mass of the water and bucket.

Ans: The linear velocity of bucket = 2.711 m/s, Angular speed of bucket = 3.615 rad/s

Previous Topic: Theory of Centripetal Force and Centrifugal Force

Next Topic: More Problems on Centripetal Force

Science > Physics > Circular Motion > Numerical Problems on Centripetal Force – 01

The post Numerical Problems on Centripetal Force – 01 appeared first on The Fact Factor.

In this article, we shall study the concept of centripetal force and centrifugal force, their expressions, and examples.

Centripetal Force:

A force which acts on a body performing circular motion and acts along the radius of the circular path and directed towards the centre of the path is called centripetal force.

Example: When a stone tied to one end of a string is whirled horizontally in a circle, there is an inward force exerted by the string on the stone called tension. The tension in the string provides necessary centripetal force for circular motion.

Without centripetal force, circular motion is not possible. If centripetal force vanishes at some instant, then the body ceases to move in the circular path and flies along the tangent at that point to the circular path.

On the surface of the earth, the centripetal force is maximum at the equator and zero on the poles.

Expression For Centripetal Force:

If ‘m’ is the mass of the particle performing circular motion then the magnitude of the centripetal force is given by

Centripetal force = mass  x  centripetal acceleration.

It is a vector quantity and always directed towards the centre of the circle. In vector form

Characteristics of Centripetal Force:

• It is a real force because it is provided by gravitational, electromagnetic or nuclear interaction.
• It arises in an inertial frame of reference. The inertial reference frame is that frame of reference which is moving with uniform velocity w.r.t. earth.
• It is always directed towards the centre of the circular path.
• Without it, the circular motion is not possible.
• The work done by the centripetal force is zero because the displacement of the particle (tangential) is perpendicular to the direction of the centripetal force (radial) i.e. there is no displacement in the direction of the centripetal force.
• The torque produced by the centripetal force at the center of the circular path is zero.
• The direction of the centripetal force does not depend on the sense of rotation of the body.

Notes:

The kinetic energy of a body performing a circular motion in terms of centripetal force is

K.E. = F.r/2

The linear momentum of a body performing a circular motion in terms of centripetal force is

p = F.r / v

Change in the velocity of a body performing U.C.M. and when it moves through angle θ is

dv = 2 sin(θ/2)

Examples of Centripetal Force:

• When a stone tied to one end of a string is whirled horizontally, there is an inward force exerted by the string on the stone called tension. This force provides necessary centripetal force for circular motion.

• Satellite revolves around the earth in a circular orbit. Necessary centripetal force is provided by the gravitational force of attraction between satellite and earth.

• When a vehicle moves around a horizontal circular road, the centripetal force for circular motion is provided by the frictional force between the road and the wheels.

• In an atom, an electron moves around the nucleus in an orbit. The centripetal force required for the motion of the electron is provided by the electrostatic force of attraction between the negatively charged electron and positively protons.

• When a coin is kept on rotating disc necessary centripetal force is provided by friction between the coin and the surface of the rotating disc.

• In a circus, when the motorcyclist moves in a vertical circle in the sphere of death, at the highest point of the vertical circle, the necessary centripetal force is provided by the weight of the motorcycle and motorcyclist.

• When a body is performing looping a loop, at the highest point of the vertical circle, the necessary centripetal force is provided by the weight of the body. For looping a loop the speed of motorcyclist at the lowest point should be greater than √5gr  and at the highest point, the speed should be greater than √gr.

Centrifugal Force:

The imaginary force which acts on the particle performing a circular motion in the direction away from the centre along the radius of the circular path having the same magnitude as that of centripetal force is called centrifugal force.

Example: When moving a car takes a turn along a horizontal curved road, persons in the car experience a force in the outward direction. This force is centrifugal force

It has the same magnitude and opposite direction to that of centripetal force.

It should be noted that the centrifugal force is not real force’ since it does not arise due to either gravitational or electrostatic or nuclear interaction.

Centrifugal Force is a Pseudo Force:

• Centrifugal force is not real force since it does not arise due to either gravitational or electrostatic or nuclear interaction.
• Centrifugal force has no independent existence. It comes to play with the action of the centripetal force. It does not arise due to either gravitational or electrostatic or nuclear interaction.
• Centrifugal force acts in the non-inertial reference frame
  Hence centripetal force is a pseudo force.

Theoretical Explanation:

Case – I:

Let us consider body P is tied to a string and lying on a frictionless turntable and performing uniform circular motion along with turntable in a clockwise sense as shown. The necessary centripetal force is provided by the tension in the string. Let A be an observer standing between the centre of the table and the particle at P. Let P₁, P₂, P₃ be different positions of the particle on the circular path. Let A₁, A₂, A₃ be corresponding positions of observer A. Observer B is standing near table B outside the turntable. As observer B is stationary w.r.t. the earth he is an inertial frame of reference, while observer A is acted upon by centripetal acceleration. Thus observer A is in acceleration w.r.t. earth, observer A is a non-inertial frame of reference.

Let us check the opinion of observer B about the motion of particle P. His observation is that he is stationary and the particle P is in a circular motion and a centripetal force acts on it. The necessary centripetal force is provided by the tension in the string. There is a centripetal acceleration and is provided by centripetal force. Thus his observations by B are in accordance with Newton’s laws of motion.

Now let us check observation of observer A. He says that he is stationary and every instant particle is in front of me at a constant distance, hence particle is stationary. There is tension in the string. Thus force is acting on particle directed towards centre but particle is stationary. By Newton’s second law is F = ma. Thus if there is a force, the state of motion of the body should change. But as per observer A, the state of motion of the body is not changing in spite of the action of the force. Thus Newton’s second law of motion fails. To correct A we have to assume some imaginary force (centrifugal force) which is equal to the tension in the string acts radially outward i.e. opposite and equal to centripetal force. Under the action of these equal and opposite forces acting on the body, the body does not change its change of state of motion.

We can see that the concept of centrifugal force (imaginary) is not required by B but is required by observer A who is the noninertial frame of reference.

Case- II:

Now let us assume the string breaks when the particle is at position P₁. The particle is thrown off with uniform velocity along the tangent to the circular path at position P₁.

Let us check the new opinion of observer B about the motion of particle P. His observation is that I am he is stationary and the body is moving along a straight line with uniform velocity and no force acts on it. It is in accordance with Newton’s first law of motion.

Now let us check new observation of observer A. He says that he is stationary and the body P moves outward with constant acceleration but no force acts on it. Thus Newton’s second law of motion fails. To correct A we have to assume some imaginary force (centrifugal force) which is equal to the tension in the string acts radially outward. Under the action of this force, the body moves radially outward with acceleration.

Thus in both the case we have to assume the existence of imaginary or pseudo force to correct observer A, which is non-inertial reference frame..

Centripetal Force and Centrifugal Force Do Not Constitute Action-Reaction Pair:

A statement ” A particle moving uniformly along a circle experiences a force directed towards the centre (centripetal force) and an equal and opposite force directed away from the centre (centrifugal force). The two forces keep the particle in equilibrium” is completely wrong.

Because the particle in a circular motion is not in equilibrium because a net force (centripetal force) acts towards the centre which is required to change the direction of particle continuously. Centripetal force is a real force.

The centrifugal force is an imaginary force is required by an observer moving with the particle. The observer is a non-inertial reference frame and for the observer, the particle is at rest.

Centripetal force is required for circular motion. Centrifugal force does not have an independent existence. It comes into play when the centripetal force starts acting. Thus the centripetal force and centrifugal force do not constitute the action-reaction pair.

Characteristics of Centrifugal Force:

• It is an imaginary force or pseudo force.
• It is experienced in non – inertial frame of reference.
• The magnitude of the centripetal force is equal to that of the centripetal force.
• It is always directed away from the centre of the circle along the radius
• it is directed opposite to the centripetal force.
• Centrifugal force doesn’t have an independent existence.

Examples of Centrifugal Force:

• When a moving car along a horizontal curved road takes a turn, persons in the car experience a force in an outward direction. This force is centrifugal force.

• When the horizontal merry go round rotates about the vertical axis the chairs are pulled out due to centrifugal force.

• When a stone is whirled in a circle, we feel that stone is pulling our hand because of centrifugal force.

• The earth is flattened at the poles and bulged at the equator because the centrifugal force acting on the particles on the equator is maximum.

• The drier of a washing machine acts on the principle of centrifugal force. Water particles from wet clothes are thrown outward due to centrifugal force acting on them. Drier in the washing machine consists of a cylindrical vessel with perforated walls. As the cylindrical vessel is rotated fast, centrifugal force acts on the water particles of wet clothes. Under the action of this centrifugal force, water particles are forced out of the perforations, thereby drying of the clothes.

• When a pilot moves the plane in looping a loop, he does not fall down because his weight is balanced by the centrifugal force acting on it.

• A coin kept slightly away from the center of rotating gramophone disc slips off towards the edge of the disc at a particular speed. This is due to centrifugal force acting on the coin.

• A bucket full of water is rotated in a vertical circle at a particular speed, so that water does not fall. This is because the weight of the water is balanced by centrifugal force.

• The centrifuge is a device which is used for separating heavier particles and light particles and works on the principle of centrifugal force. In the centrifuge, the tube containing liquid along with the suspended particles is whirled in a horizontal circle. Dense particles are acted upon by a centrifugal force. Hence they get accumulated at the bottom, which is outside while rotating. This is because buoyant force towards the center is greater for the lighter particle.

• Centrifugal governor works on the principle of centrifugal force. When the speed of the vehicle increases beyond a certain set limit, the fly balls under the action of centrifugal force, fly away from the axis of rotation. A bell and crank arrangement attached to it reduces the flow of fuel to the engine. Thus the speed of the vehicle is governed.

• Centrifugal pump works on this principle. It sucks water through a pipe from the reservoir and throws outward under the action of centrifugal action in the involute casing and thus produces a draft and water is lifted up.

• When wheels are rotating mud stuck to the wheels and then under the action of centrifugal force are thrown tangentially away from the wheels. These mud particles tarnish the vehicle and spray the mud on the following vehicles. At the same time, these mud article may degrade the performance of the brake. When mudguards are used the mud particles are stopped dead and fall down under gravity.

Distinguishing Between Centripetal Force and Centrifugal Force:

Previous Topic: Concept of Centripetal Acceleration

Next Topic: Numerical Problems on Centripetal Force

Science > Physics > Circular Motion > Centripetal and Centrifugal Force

The post Centripetal and Centrifugal Force appeared first on The Fact Factor.

In this article, we shall study the concept of centripetal acceleration and expression for it. Centripetal acceleration is also called radial acceleration.

Accelerated Motion:

Velocity is a vector quantity which has magnitude as well as direction. Either magnitude or direction or both change the motion is nonuniform motion

• Case – I: When direction remains the same, but magnitude changes e.g. Motion under gravity.
• Case – II: When magnitude remains the same, but the direction changes continuously e.g. Uniform circular motion:
• Case – III:  When both the magnitude and direction changes continuously e.g. Projectile motion

The acceleration of the body performing circular motion which is directed towards the centre of the circular path along the radius is called a radial acceleration or centripetal acceleration.

Characteristics of Centripetal Acceleration:

• It is the acceleration of a particle performing the circular motion, which is directed towards the centre of the circular path along the radius.
• It is always directed towards the centre of the circular path along the radius.
• The direction of centripetal acceleration changes continuously.
• For U.C.M. the magnitude of the centripetal acceleration is constant.
• It is denoted by the letter ‘a’. Its S.I. unit is metre per square second (m s^-2). Its dimensions are [MºL¹T ^-2].
• The magnitude of centripetal acceleration is given by

The Expression for Centripetal Acceleration

Geometric Method:

The magnitude of the velocity of a particle performing uniform circular motion is constant but its direction changes constantly. Hence the particle in circular motion has linear acceleration.

Let us consider a particle performing uniform circular motion with a linear velocity of magnitude ‘v’ and angular velocity of magnitude ‘ω’ along a circle of radius ‘r’ with centre O in an anticlockwise sense (moving from initial position A to final position B)as shown in the figure.

Triangle QBR is drawn for the velocities of the particles at A and B.

Now by definition,

The triangles AOB and RBQ are similar. Hence ∠ QBR = δθ
For smaller angular displacement δθ,

Substituting in equation (1), we get

This is the expression for the acceleration of particle performing the uniform circular motion.

This acceleration is directed towards the centre of the circular path along the radius. This acceleration is called radial acceleration or centripetal acceleration. In vector form, centripetal acceleration can be given as

The negative sign indicates that centripetal acceleration is oppositely directed to that of radius vector i.e. directed towards the centre of the circle along the radius.

Note: the alternate formula is

Centripetal acceleration is also called radial acceleration. It always acts along the radius towards the centre of the circular path. The angle between radius vector and centripetal acceleration is π radian or 180°.

Cartesian Method or Calculus Method:

Let us consider a particle performing uniform circular motion with a linear velocity of magnitude v and angular velocity of magnitude ω along a circle of radius r with centre O in an anticlockwise sense. Let the particle moves from A to P, in time ‘t’ Such that ∠ POA = θ.

But for uniform Circular motion θ = ω t. Thus, ∠ POA = ω t

Let us draw seg PM perpendicular to seg OA. Thus, ∠ POM = ω t

The radius vector at time t at P is given by

Substituting these values in equation (1)

 The linear velocity of a particle can be obtained by differentiating equation (2) w.r.t. time t.

The linear acceleration of a particle can be obtained by differentiating equation (3) w.r.t. time t.

From equations (1) and (4) we have

Thus the magnitude of the acceleration is v²/r and its direction is along the radius and the negative sign indicates that it is opposite to the radius vector i.e. the acceleration is directed towards the centre of the circular path. This acceleration is called the centripetal acceleration.

Relation between linear velocity (v) and angular velocity (ω) by calculus method:

From the equation (3) we get the instantaneous velocity as

Thus the linear velocity of a particle performing U.C.M. is radius times its angular velocity.

Angle between linear velocity (v) and radius vector by calculus method:

From the equations (2) and (3) we have

Thus the scalar or dot product of the velocity of the particle performing U.C.M. and the radius vector is zero. Hence the angle between the velocity of the particle performing U.C.M. and the radius vector.

Acceleration of a Body Performing Circular Motion has Two Components:

The relation between linear velocity and angular velocity In vector form is written as

Thus the acceleration of the body performing circular motion has two components. one along the radius of the circular path towards the centre and is called centripetal acceleration and another tangential component. The net acceleration of the body is given by

Numerical Problems:

Example – 1:

The length of an hour hand of a wristwatch is 1.5 cm. Find the magnitudes of following w.r.t. tip of the hour hand a) angular velocity b) linear velocity c) angular acceleration d) radial acceleration e) tangential acceleration f) linear acceleration

Given: r = 1.5 cm = 1.5 x 10^-2 m, For hour hand T_H = 12 hr = 12 x 60 x 60 sec

To Find: Angular velocity = ω = ?, linear velocity = v = ?, angular acceleration = α = ?, radial acceleration = a_r = ?, tangential acceleration = a_T =?, linear acceleration a = ?

Solution:

Now v = r ω = 1.5 x 10^-2 x 1.454 x 10^-4 = 2.181 x 10^-6 m/s

Tip of hour hand performs uniform circular motion

∴  α = 0 and a_T = 0

Radial acceleration is given by

Now linear acceleration

Ans: Angular speed = 1.454x 10^-4 rad/s, Linear speed =2.181 x 10^-6 m/s,
Angular acceleration = 0, Radial acceleration = 3.171 x 10^-10 m/s².
Tangential acceleration = 0, Linear acceleartion = 3.171 x 10^-10 m/s².

Example – 2:

To simulate the acceleration of high-speed fighter plane, astronauts are spun at the end of a long rotating beam of radius 5 m. Find the angular velocity required to generate a centripetal acceleration 3 times the acceleration due to gravity.

Given:  r = 5 m, a= 3g, g = 9.8 m/s².

To find:  ω =?

Solution:

Ans: Required angular velocity = 2.425 rad/s

Example – 3:

To simulate the acceleration of large rockets, astronauts are spun at the end of a long rotating beam of radius 9.8 m. Find the angular velocity required to generate a centripetal acceleration 8 times the acceleration due to gravity.

Given:  r = 9.8 m, a= 8g, g = 9.8 m/s².

To find:  ω = ?

Solution:

Ans: Required angular velocity = 2.828 rad/s

Example – 4:

A body of mass 2 Kg is tied to the end of a string of length 1.5 m and revolved around the other end (kept fixed) in a horizontal circle. If it makes 300 rev/min, calculate the linear velocity, the acceleration and the force acting upon the body.

Given:  m = 2 kg,  r = 1.5 m,  N = 300 r.p.m.,

To find: v = ?, a = ?, F = ?

Solution:

Centripetal force F = ma = 2 x 1479 = 2958 N

Ans: Linear speed of body = 47.13 m/s, Acceleration of body = 1480 m/s²,

Force acting on body = 2958 N radially inward.

Example – 5:

The tangential acceleration of a body performing circular motion is 29.48 m/s² and its linear acceleration is 52.3 m/s². Find its radial acceleration

Given: a_T = 29.48 m/s² and a = 52.3 m/s².

To Find: a_r =?

Solution:

Ans: The radial acceleration is 43.1 m/s²

Example – 6:

A particle is revolving in a circle. Its angular speed increases from 2 rad/s to 40 rad/s in 19 s. The radius of a circle is 20 cm. Compare the ratios of centripetal acceleration to tangential acceleration, at end of 19s.

Given: ω₁ = 2 rad/s, ω₂ = 40 rad/s, t = 38 s, r = 20 cm = 0.2 m.

To Find: a_r : a_t =?

Solution:

Now a_t = r a = 0.2 x 2 = 0.4 m/s²

Centripetal acceleration at end of 19 s

Ar = r ω ² = 0.2 x (40)² = 320 m/s²

Ans: the required ratio a_r : a_t =320: 0.4 = 800:1

Previous Topic: Numerical Problems on Circular Motion

Next Topic: Concept of Centripetal and Centrifugal Force

Science > Physics > Circular Motion > Centripetal Acceleration

The post Centripetal Acceleration appeared first on The Fact Factor.

For uniform circular motion Angular speed ω, Linear speed t, kinetic energy, Angular momentum (L) are constant. The angular acceleration α and the tangential acceleration a_T are zero.

Problems Based on Hands of Clock:

1. The second hand of a clock takes 60 seconds to complete one rotation. Its angular speed is 0.105 rad /s.
2. The minute hand of a clock takes 60 minutes = 60 x 60 seconds to complete one rotation. Its angular speed is 1.746 x 10^-3 rad /s.
3. The hour hand of a clock takes 12 hours = 12 x 60 x 60 seconds to complete one rotation. Its angular speed is 1.455 x 10^-4 rad /s.
4. The ratio of angular speeds of the second hand of a clock and the minute hand of a clock is 60:1.
5. The ratio of angular speeds of the minute hand of a clock and an hour hand of a clock is 12:1.
6. The ratio of angular speeds of the second hand of a clock and the hour hand of a clock is 720:1.

Example – 01:

Calculate the angular speed of the second hand, minute hand and hour hand of a clock.

Given: For second Hand T_S = 60 sec, For minute hand T_M = 60 min = 60 x 60 sec, For hour hand T_H = 12 hr = 12 x 60 x 60 sec.

To Find: Angular speed ω_S=? ω_M=? ω_H=?

Solution: 

The tips of the second hand, the minute hand, and the hour hand perform the uniform circular motion.

Ans: Angular speed of second hand = 0.105 rad/s,

Angular speed of minute hand =1.746 x 10^-3 rad/s,

Angular speed of hour hand =1.454 x 10^-4 rad/s.

Example – 02:

What is the angular velocity of the minute hand of a clock? What is the angular displacement of the minute hand in 20 minutes? If the minute hand is 5 cm long, what is the linear velocity of its tip?

Given: For minute hand  T_M = 60 min = 60 x 60 sec, t = 20 min = 20 x 60 sec, r = 5 cm = 5 x 10^-2 m

To Find: Angular speed ω_M=?, Angular displacement  θ =?, Linear velocity = v =?

Solution: 

The tip of the minute hand performs the uniform circular motion.

Ans: The angular speed of minute hand =1.746 x 10^-3 rad/s,

The angular displacement of minute hand in 20 minutes =2.095 rad,

The linear speed of the tip of minute hand = 8.73 x 10^-5 m/s.

Example – 03:

What is the angular displacement of the minute hand of a clock in 25 minutes?

Given: For minute hand  T_M = 60 min = 60 x 60 sec, t = 25 min = 25 x 60 sec,

To Find:  Angular displacement  θ =?

Solution: 

The tip of the minute hand performs the uniform circular motion.

Ans: Angular displacement of minute hand = 2.618 rad

Example – 04:

What is the angular velocity of the second hand of a clock? If the second hand is 10 cm long find the linear velocity of its tip.

Given: For second hand  T_s = 60  sec, r = 10 cm = 10 x 10^-2 m,

To Find:  Linear speed = v =?

Solution: 

The tip of the second hand performs the uniform circular motion.

Now v = r ω = 10 x 10^-2 x 0.105 = 1.05 x 10^-2 m/s = 1.05 cm/s

Ans: Angular velocity of second hand = 0.105 rad/s, Linear speed of tip of second hand = 1.05 cm/s

Example – 05:

The second hand of a watch is 1.5 cm long. Find the linear speed of a point on the second hand at a distance of 0.5 cm from the tip.

Given: For second hand  T_s = 60  sec, r = 1.5 cm – 0.5 cm = 1 cm = 1 x 10^-2 m,

To Find:  Linear speed = v =?

Solution:

The tip of the second hand performs the uniform circular motion.

Now v = r ω = 1 x 10^-2 x 0.105 = 1.05 x 10^-3 m/s = 1.05 mm/s

Ans: Linear speed of the point on the second hand = 1.05 mm/s

Example – 06:

The extremity of the hour hand of a clock moves 1/20th as fast as the minute hand. What is the length of the hour hand if the minute hand is 10 cm long?

Given: v_H = 1/20 v_M , r_M = 10 cm

To Find: r_H =?

Solution:

Ans: Length of hour hand = 6 cm

Example – 07:

The length of an hour hand of a wristwatch is 1.5 cm. Find the magnitudes of following w.r.t. tip of the hour hand a) angular velocity b) linear velocity c) angular acceleration d) radial acceleration e) tangential acceleration f) linear acceleration

Given: r = 1.5 cm = 1.5 x 10^-2 m, For hour hand T_H = 12 hr = 12 x 60 x 60 sec

To Find: Angular velocity = ω = ?, linear velocity = v = ?, angular acceleration = α = ?, radial acceleration = a_r = ?, tangential acceleration = a_T =?, linear acceleration a = ?

Solution:

Now v = r ω = 1.5 x 10^-2 x 1.454 x 10^-4 = 2.181 x 10^-6 m/s

Tip of hour hand performs uniform circular motion

∴  Angular acceleration = α = 0 and tangential acceleartion = a_T = 0

Radial acceleration is given by

a_r = v² /r = ( 2.181 x 10^-6 ) ² / (1.5 x 10^-2) = 3.171 x 10^-10 m/s²

a² = a_r ² + a_T ²

as a_T = 0, a = a_r = 3.171 x 10^-10 m/s²

Ans: Angular speed = 1.454x 10^-4 rad/s, Linear speed =2.181 x 10^-6 m/s,
Angular acceleration = 0, Radial acceleration = 3.171 x 10^-10 m/s².
Tangential acceleration = 0, Linear acceleartion = 3.171 x 10^-10 m/s².

Problems Based on Earth:

The earth takes 24 hours to complete one rotation about its axis. The angular speed of the earth of its rotation about its axis is 7.273 x 10^-5 rad /s.

Example – 08:

Calculate the angular velocity of the earth due to its spin motion.

Given: For the earth = T = 24 hr = 24 x 60 x 60 sec

To Find: Angular velocity = ω = ?

Solution:

Ans: Angular speed of the earth due to its spin motion is 7.273 x 10^-5 rad/s.

Problems Based on Rotating Discs:

Example – 09:

A turntable rotates at 100 rev/sec. Calculate its angular speed in rad/s and degrees/s.

Given: n = 100 r.p.s.

To Find: Angular speed =?

Solution:

Ans: Angular speed = 10.47 rad/s, Angular speed = 600 degrees/s

Example – 10:

Propeller blades of an aeroplane are 2 m long. When the propeller is rotating at 1800 rev/min, compute the tangential velocity of the tip of the blade. Also, find the tangential velocity at a point on the blade midway between tips and axis.

Given: r = 2 m, Angular speed = N = 1800 r.p.m.

To Find: v_Tip =? v_Mid =?

Solution:

Ans: The angular speed of the tip of blade = 376.8 rad/s, AThe angular speed of point midway = 188.4 rad/s

Example – 11:

A turntable has a constant angular speed of 45 r.p.m. Express this in rad per second and degrees per second. If the radius of the turntable is 0.5 m, what is the linear speed of a point on the rim?

Given: N = 45 r.p.m., r = 0.5 m

To Find: angular speed in rad/s and degrees/s, linear velocity = v = ?

Solution:

Ans: Angular speed = 2.355 rad/s = 270 degrees/s Linear speed on point on rim = 2.355 m/s.

Example – 12:

The linear velocity of a point on the rotating disc is 3 times greater than at a point on the at a distance of 8 cm from it. What is the diameter of the disc?

Given: v_T = 3 v_P, Let r_T = r, r_P = (r – 8) cm

To Find: Diameter of the disc =?

Solution:

Angular velocity for both the points is the same.
Ans: Diameter of disc = 24 cm.

Example – 13:

A disc has a diameter of one metre and rotates about an axis passing through its centre and at right angles to its plane at the rate of 120 rev/min. What is the angular and linear speed of a point on the rim and at a point halfway to the centre.

Given: d = 1m , r_T = r = 0.5 m, N = 120 r.p.m., for r_P = r/2 = 0.25 m

To Find: v_T = ?, V_P = ?

Solution:

Angular velocity for both the points is the same.

The linear speed of point on the rim

v_T =  r_Tω  = 0.5 x 12.56 = 6.28 m/s

v_P =  r_Pω  = 0.25 x 12.56 = 3.14 m/s

Ans: Angular speed of point on rim = 12.57 rad/s, Linear peed of point on rim = 6.28 m/s,
Angular speed of point on halfway =12.57m/s, Linear speed of point on halfway = 3.14 m/s.

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Angular Displacement:

For a particle performing a circular motion the angle, traced by the radius vector at the centre of the circular path in a given time is called the angular displacement of the particle at that time. It is denoted by ‘θ’. Its S.I. unit is radian (rad). It is a dimensionless quantity. [MºLºTº]

The direction of angular displacement: 

For smaller magnitude (infinitesimal) angular displacement is a vector quantity and its direction is given by the right-hand thumb rule.

Right-Hand Thumb Rule:

If we curl the fingers of our right hand and hold the axis of rotation with fingers pointing in the direction of motion then the outstretched thumb gives the direction of the angular displacement vector.

Sign Convention: 

An angular displacement in counter clock-wise direction is considered positive and that in the clockwise direction is considered as negative.

Vector relation between linear and angular displacement is

Characteristics of Angular Displacement:

• It is the angle, traced by the radius vector at the centre of the circular path in a given time is called the angular displacement of the particle at that time.
• For smaller magnitude (infinitesimal) angular displacement is a vector quantity and its direction is given by the right-hand thumb rule.
• Finite angular displacement is a vector quantity.
• Instantaneous angular velocity is a vector quantity.
• The direction of angular displacement in an anticlockwise sense is considered as positive, while the direction of angular displacement in a clockwise sense is considered as negative.
• The angular displacement of the particle performing uniform circular motion in equal time is equal.
• It is denoted by ‘θ’. Its S.I. unit is radian (rad). It is a dimensionless quantity. [MºLºTº].

Larger angular displacement is not treated as a vector quantity.

If a quantity has both the direction and magnitude then it seems to be vector quantity but it can only be treated as vector quantity if its satisfies laws of vector addition. Consider the following two cases of angular displacement

The commutative law of vector addition which states that if we add two vectors, the order in which we add them does not matter.

We can see that if the order is interchanged the final outcome is different. Thus the angular displacement fails to obey the law of vector addition. Hence larger angular displacement is not a vector quantity.

Angular Velocity:

The rate of change of angular displacement with respect to time is called the angular velocity of the particle. It is denoted by the letter ‘ω’. Its S.I. unit is radians per second (rad s^-1). Its dimensions are [MºLºT ^-1].

Mathematically,

For uniform circular motion, the magnitude of angular velocity is given by

Where ω = Angular speed, T = Period
N = Angular speed in r.p.m., n = Angular speed in r.ps. or Hz.
θ = Angular displacement, t = time taken

The Direction of Angular Velocity: 

For smaller magnitude (infinitesimal) the angular velocity is the vector quantity. Its direction is given by the right-hand thumb rule. It states that “If we curl the fingers of our right hand and hold the axis of rotation with fingers pointing in the direction of motion then the outstretched thumb gives the direction of the angular velocity vector”. Thus, the direction of angular velocity is the same as that of angular displacement.

By this rule, the direction of the angular velocity of the second hand, the minute hand, and the hour hand is perpendicular to the dial and directed inwards.

Angular Speed:

The angle traced by radius vector in unit time is called the angular speed or The magnitude of angular velocity is known an angular speed.

Uniform motion is that motion in which both the magnitude and direction of velocity remain constant. In UCM the magnitude of velocity is constant but its direction changes continuously. Hence UCM is not uniform motion. For uniform circular motion, the angular velocity is constant.

For uniform circular motion, the magnitude of velocity at P = magnitude of velocity at Q = magnitude of velocity at R and the direction of velocity at P ≠ direction of velocity at Q ≠ direction of velocity at R. In uniform circular motion a body moves in a circle describes equal angles in equal interval of time. Thus for a body performing UCM has uniform speed.

For non-uniform circular motion, The magnitude of velocity at P ≠ magnitude of velocity at Q ≠ magnitude of velocity at R and the direction of velocity at P ≠ direction of velocity at Q ≠ direction of velocity at R.  In non-uniform circular motion a body moves in a circle describes unequal angles in equal interval of time.

Characteristics of Angular Velocity:

• The rate of change of angular displacement with respect to time is called the angular velocity of the particle.
• Its direction is given by the right-hand thumb rule.
• The direction of angular velocity is the same as that of angular displacement.
• For uniform circular motion, the magnitude of angular velocity is constant.
• The magnitude of angular velocity (ω) is related to the magnitude of linear velocity (v) by the relation v = rω.
• It is denoted by the letter ‘ω’. Its S.I. unit is radians per second (rad s-1). Its dimensions are [MºLºT ^-1].

Example – 1:

The graph shows angular positions of a rotating disc at different instants. What is the sign of angular displacement and angular acceleration?

The angular velocity at any instant is given by ω = dθ/dt,

At t = 1 second the graph is rising up, thus the slope (dθ/dt) of the tangent at t = 1 second is positive. Hence angular velocity is positive.

At t = 2 seconds the graph reaches the topmost point, thus the slope (dθ/dt) of the tangent at t = 2 seconds is zero. Hence angular velocity is zero.

At t = 3 seconds the graph is going down, thus the slope (dθ/dt) of the tangent at t = 3 seconds is negative. Hence angular velocity is negative.

We can see the change in angular velocity as positive → zero → negative. Thus angular velocity is decreasing. Hence angular acceleration is negative.

Angular Acceleration: 

The average angular acceleration is defined as the time rate of change of angular velocity. It is denoted by the letter ‘α’. Its S.I. unit is radians per second square (rad /s²). Its dimensions are [MºLºT ^-2]. Mathematically,

If the initial angular velocity of the particle changes from initial angular velocity ω₁  to final ω₂ angular velocity in time ‘t’ then

The direction of Angular Acceleration:

The direction of angular acceleration is given by right-hand thumb rule. If the angular velocity is increasing then the angular acceleration has the same direction as that of the angular velocity.

If the angular velocity is decreasing then the angular acceleration has the opposite direction as that of the angular velocity.

For uniform circular motion angular acceleration is zero.

Characteristics of Angular Acceleration:

• The average angular acceleration is defined as the time rate of change of angular velocity.
• If the angular velocity is increasing angular acceleration is positive (e.g. the angular acceleration of the tip of a fan just switched on). If the angular velocity is decreasing angular acceleration is negative (e.g. the angular acceleration of the tip of a fan just switched off)
• If the angular velocity is increasing then the angular acceleration has the same direction as that of the angular velocity. If the angular velocity is decreasing then the angular acceleration has the opposite direction as that of the angular velocity.
• For uniform circular motion angular acceleration is zero.
• The magnitude of angular acceleration (α) is related to the magnitude of linear acceleration (a) by the relation a = rα.
• It is denoted by the letter ‘α’. Its S.I. unit is radians per second square (rad /s²). Its dimensions are [M⁰L⁰T^-2].

Right Handed Screw Rule:

When a right-handed screw is rotated in the sense of revolution of the particle, then the direction of the advance of the screw gives the direction of the angular displacement vector.

Relation Between Linear Velocity and Angular Velocity:

Consider a particle performing uniform circular motion, along the circumference of the circle of radius ‘r’ with constant linear velocity ‘v’ and constant angular speed ‘ω’ moving in the anticlockwise sense as shown in the figure.

Suppose the particle moves from point P to point Q through a distance ‘δx’along the circumference of the circular path and subtends the angle ‘δθ’ at the centre O of the circle in a small interval of time ‘δt’. By geometry

δx = r . δθ

If the time interval is very very small then arc PQ can be considered to be almost a straight line. Therefore the magnitude of linear velocity is given by

Thus the linear velocity of a particle performing uniform circular motion is radius times its angular velocity. In vector form above equation can be written as

The linear velocity can be expressed as the vector product of angular velocity and radius vector.

The following figure shows relative positions of the linear velocity vector, angular velocity vector, and radius or position vector.

For smaller magnitudes angular displacement, angular velocity are vector quantities. Let ( r) be the position vector of the particle at some instant. Let the angular displacement in small time δt be ( δθ). Let the corresponding linear displacement (arc length) be ( δs). By geometry

Dividing both sides of the equation by δt and taking the limit

Period of Revolution:

Let us consider particle performing a uniform circular motion. Let ‘T’ be its period of revolution. During the periodic time (T), particle covers a distance equal to the circumference 2pr of the circle with linear velocity v.

This is an expression for the period of revolution for particle performing the uniform circular motion.

The Expression for Angular Acceleration:

When a body is performing a non-uniform circular motion, its angular velocity changes. Hence the body possesses angular acceleration.
The rate of change of angular velocity w.r.t. time is called as the angular acceleration. We know that acceleration is the rate of change of velocity with respect to time.

r = radius of circular path = constant.

ω = angular velocity of the particle performing a circular motion

Where ‘α’ is angular acceleration. Hence,

linear acceleration = radius x angular acceleration.

If speed is increasing linear acceleration is in the same direction as that of linear velocity. If speed is decreasing linear acceleration is in the opposite direction to that of linear velocity. It is also referred as tangential acceleration. For uniform circular motion α = 0.

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