In this article, we shall study problems based on the change in kinetic energy and the change in angular momentum of a rotating body.

Example – 01:

A disc begins to rotate from rest with a constant angular acceleration of 0.5 rad/s² and acquires an angular momentum of 73.5 kg m²/s in 15 s after the start. Find the kinetic energy of the disc in 20 s after the start.

Given: Angular acceleration = α = 0.5 rad/s², Change in angular momentum = dL = 73.5 kg m²/s, time taken = t = 15 s, initial angular speed = ω₁ = 0 rad/s

To find: K.E. in 20 s after start.

Solution:

We have,   ω₂ = ω₁ + αt = 0 + 0.5 x 20 = 10 rad/s

Change in angular momentum = dL = L₂ – L₁ = Iω₂ – Iω₁

73.5 = I x 7.5 – I x 0

I = 73.5/7.5 = 9.8 kg m²

Now, Kinetic energy is given by

K.E. = ½ I ω² = ½ x 9.8 x 10² = 490 J

Ans: The kinetic energy of the disc in 20 s after the start is 490 J

Example – 02:

The angular momentum of a body changes by 80 kg m²/s when its angular velocity changes from 20 rad/s to 40 rad/s. Find the changes in its K.E. of rotation.

Given: Change in angular momentum = 80 kg m²/s. Initial angular speed = ω₁ = 20 rad/s, final angular speed = ω₂ = 40 rad/s.

To Find: Change in kinetic energy =?

Solution:

Change in angular momentum = dL = L₂ – L₁ = Iω₂ – Iω₁

80 = I x 40 – I x 20

∴ 80 = 20 I

∴ I = 80/20 = 4 kg m²

Change in kinetic energy = K.E.₂ – K.E.₁ = ½ I ω₂² – ½ I ω₁² = ½ I ( ω₂² – ω₁²)

Change in kinetic energy =   ½ x 4 x (40² – 20²)

Change in kinetic energy =   2 x (1600- 400) = 2400 J

Ans: The change in kinetic energy of rotation is 2400 J

Example – 03:

An energy of 500 J is spent to increase the speed of wheel from 60 r.p.m. to 240 r.pm. Calculate the moment of inertia of the wheel.

Given: Change in kinetic energy = 500 J, Initial angular speed = N₁ = 60 r.pm., final angular speed = N₂ = 240 r.p.m.

To Find: Moment of Inertia of the wheel.

Solution:

ω₁ = 2πN₁/60 = 2π x 60/60 = 2π rad/s

ω₂ = 2πN₂/60 = 2π x 240/60 = 8π rad/s

Change in K.E. = K.E.₂ – K.E.₁

∴ Change in K.E. = ½ Iω₂² – ½ Iω₁² = ½ I (ω₂² – ω₁²)

∴ 500. = ½ I ((8π)² – (2π)²)

∴ 1000. = I (64π² – 4π²)

∴ 1000. = I (60π²)

∴ I = 1000/60π² = 1.688 kgm²

Ans: The moment inertia of wheel is 1.688 kg m²

Example – 04:

A wheel of the moment of inertia 1 kgm² is rotating at a speed of 30 rad/s. Due to friction on the axis, it comes to rest in 10 minutes. Calculate i) total work done by friction, ii) the average torque of the friction iii) angular momentum of the wheel two minutes before it stops rotating.

Given: Moment of inertia of wheel = 1 kg m², Initial angular speed = ω₁ = 30 rad/s, final angular speed = ω₂ = 0 rad/s, time = t = 10 min = 10 x 60 = 600 s

To Find: Work done by friction =? Torque acting = ?, angular momentum of the wheel two minutes before it stops rotating.

Solution:

Work done by friction = Change in K.E. = K.E.₁ – K.E.₂

∴ Work done by friction. = ½ Iω₁² – ½ Iω₂² = ½ I (ω₁² – ω₂²)

∴ Work done by friction. = ½ x 1 ((30)² – (0)²) = 0.5 x 900 = 450 J

Now angular acceleration = α = (ω₂ – ω₁)/t = (0 – 30)/(10 x 60) =  -30/600 = – 0.05 rad/s²

Negative sign indicates it is retardation.

The magnitude of angular acceleration = α = 0.05 rad/s²

Average torque = I α = 1 x 0.05 Nm. It is retarding torque

To find the angular momentum of the wheel two minutes before it stops rotating.

t = 10 min – 2 min = 8 min = 8 x 60 = 480 s.

ω₂ = ω₁+ αt = 30 – 0.05 x 480 = 30 – 24 = 6 rad/s

Angular momentum = Iω₂ = 1 x 6 = 6 kg m/s²

Ans: Work done by friction is 450 J, Retarding torque acting is 0.05 Nm, angular momentum of the wheel two minutes before it stops rotating is 6 kg m/s².

Example – 05:

A ceiling fan of a moment of inertia 30 kgm² rotates about its own axis at the speed of 120 r.p.m. under the action of an electric motor of power 62.8 watts. If electric power is cut off, how many revolutions will it complete before coming to rest?

Given: Moment of inertia of fan = 30 kg m², Initial angular speed = N₁ = 120 r.p.m., final angular speed = ω₂ = 0 rad/s,

To Find: Number of revolutions before coming to rest.

Solution:

ω₁ = 2πN₁/60 = 2π x 120/60 = 4π rad/s

Power = τ ω₁ = I α ω₁

∴ 62.8 = 30 x α x 4π

∴ 62.8 = 120π x α

∴ α = 62.8/(120 x 3.14) = 1/6 rad/s².

ω₂² = ω₁² + 2α θ

∴ (0)² = (4π)² + 2(- 1/6) θ

∴ 0 = 16π² – (1/3) θ

∴ (1/3) θ = 16π²

∴ θ = 48π²

No. of rotations = θ/2π = 48π²/2π = 24π = 24 x 3.14 = 75.36

Ans: Number of revolutions before coming to rest are 24π or 75.35

Example – 06:

When an angular velocity of a body changes from 20 rad/s to 40 rad/s, its angular momentum changes by 80 kgm²/s. Find the change in kinetic energy of rotation.

Given: Initial angular speed = ω₁ = 20 rad/s, final angular speed = ω₂ = 40 rad/s, Change in angular momentum = dL = 80 kgm²/s.

To Find: Change in kinetic energy of rotation =?

Solution:

Change in angular momentum = L₂ – L₁

∴ dL = Iω₂ – Iω₁ = I(ω₂ – ω₂)

∴ 80 = I(40 – 20) = 20 I

∴ I = 80/20 = 4 kg m²

Change in K.E. = K.E.₂ – K.E.₁

∴ Change in K.E. = ½ Iω₂² – ½ Iω₁² = ½ I (ω₂² – ω₁²)

∴ Change in K.E. = ½ x 4 ((40)² – (20)²) = 2 x (1600 – 400) = 2 x 1200 = 2400 J

Ans: the change in kinetic energy of rotation is 2400 J

Example – 07:

A flywheel of mass 2 kg has the radius of gyration of 0.2 m. Calculate the K.E. of rotation when it makes 5 revolutions per second.

Given: Mass of disc = M = 2 kg; radius of gyration = K = 0.2 m, Angular speed = n = 5 r.p.s.

To Find: Kinetic energy =?

Solution:

I = MK² = 2 x (0.2)² = 2 x 0.04 = 0.08 kg m/s

Now, K.E. = ½ Iω² = ½ I (2πn)² = ½ x 0.08 x ( 2 x 3.142 x 5)²

∴ K.E. = ½ Iω² = ½ I(2πn)² = 0.04 x ( 231..42)² = 39.49 J

Ans: Kinetic energy of disc is 39.49 J

Example – 08:

A circular disc of mass 10 kg and radius 0.2 m is set into rotation about an axis passing through its centre and perpendicular to its plane by applying torque 10 Nm. calculate the angular velocity of the disc at the end of 6 s from rest.

Given: Mass of disc = M = 10 kg; radius of disc = R = 0.2 m, Torque acting = τ = 10 Nm, initial angular velocity = ω₁ = 0 rad/s, time = t = 6 s

To Find:  final angular velocity = ω₂ =?

Solution:

Moment of Inertia of a disk is given by

I = ½ MR² = ½ (10)(0.2)² = 5 x 0.04 = 0.2 kg m²

Now, the torque acting on rotating body is given by

τ = I α

∴ 10 = 0.2 x α

∴ α = 10/0.2 = 50 rad/s²

Now, ω₂ = ω₁+ αt = 0 + 50 x 6 = 0 + 300 = 300 rad/s

Ans: The angular velocity of the disc at the end of 6 s from rest is 300 rad/s

Example – 09:

A torque of 400 Nm acting on a body of mass 40 kg produces an angular acceleration of 20 rad/s². Calculate the moment of inertia and radius of gyration of the body.

Given: Mass of body = M = 40 kg; Torque acting = τ = 400 Nm, Angular acceleration = α = 20 rad/s²

To Find:  Moment of inertia = I =? and radius of gyration = K = ?

Solution:

The torque acting on a rotating body is given by

τ = I α

∴ 400 = I x 20

∴ I = 400/20 = 20 kg m²

Now, I = MK²

∴ K² = I/M = 20/40 = 0.5

∴ K = 0.707 m

Ans: Moment of inertia of the body is 20 kg m² and the radius of gyration is 0.707 m.

Example – 10:

If the radius of a solid sphere is doubled by keeping its mass constant, compare the moment of inertia about any diameter.

Given: For sphere R₂ = 2R₁.

To Find:  Ratio of the moment of inertia = I₁/I₂ =?

Solution:

The M.I. of a sphere in two cases about its diameter is given by

Ans: The ratio of the moment of inertia in two cases is 1:4

Example – 11:

A flywheel in the form of a disc is rotating about an axis passing through its centre and perpendicular to its plane loses 100 J of energy, when slowing down from 60 r.p.m. to 30 r.p.m. Find the moment of inertia about the same axis and change in its angular momentum.

Given: lost in energy = 100 J. Initial angular speed = N₁ = 60 r.p.m., final angular speed = N₂ = 30 r.p.m.,

To Find:  Moment of inertia = I =? and change in angular momentum = dL =?

Solution:

ω₁ = 2πN₁/60 = 2π x 60/60 = 2π rad/s

ω₂ = 2πN₂/60 = 2π x 30/60 = π rad/s

Loss in K.E. = K.E.₁ – K.E.₂

∴ Loss in K.E. = ½ Iω₁² – ½ Iω₂² = ½ I (ω₁² – ω₂²)

∴ 100 = ½ I ((2π)² – (π)²)

∴ 200 = I (4π² – π²)

∴ 200 = I (3π²)

∴ I = 200/3π² = 6.753 kgm²

Change in angular momentum = L₂ – L₁

∴ dL = Iω₁ – Iω₂ = I(ω₁ – ω₂)

∴ dL = 6.753(2π – π) = 6.753 π = 6.753 x 3.142 = 21.22 kg m²/s

Ans: Moment of inertia is 6.753 kg m² and change in angular momentum is 21.2 kg m²/s

Example – 12:

A disc of diameter 50 cm and mass 2 kg rotates about an axis passing through its centre and at right angles to its plane with a frequency of 8 revolutions per sec. Find the angular momentum of the disc and the rotational K.E.

Given: Diameter of disc = D = 50 cm, Radius of disc = 50/2 = 25 cm = 0.25 m, Mass of disk = M = 2 kg, Number of revolutions per second = n = 8

To Find:  Angular momentum = L =? rotational kinetic energy = ?

Solution:

ω= 2πn = 2π x 8 = 16π rad/s

Moment of Inertia of a disk is given by

I = ½ MR² = ½ (2)(0.25)² = 1 x 0.0625 = 0.0625 kg m²

L = I ω = 0.0625 x 16π = 0.0625 x 16 x 3.142 = 3.142 kg m²/s

K.E. = ½ Iω² = ½ x 0.0625 x (16π)² =  ½ x 0.0625 x 256 x 3.142²

K.E. = 78.98 J

Ans: Angular momentum is 3.142 kg m²/s and rotational K.E. is 78.98 J.

Example – 13:

A solid sphere of diameter 25 cm and mass 25 kg rotates about an axis through its centre. Calculate its moment of inertia if its angular velocity changes from 2 rad/s to 12 rad/s in 5 seconds. Also, calculate the torque applied.

Given: Diameter of sphere = D = 25 cm, Radius of sphere = 25/2 = 12.5 cm = 0.125 m, Mass of sphere = M = 25 kg, Initial angular speed = ω₁ = 2 rad/s, final angular speed = ω₂ = 12 rad/s,time = t = 5 s

To Find:  Moment of inertia = I =? Torque required = τ = ?

Solution:

Moment of inertia of the sphere is given by

Ans: Moment of inertia is 0.1562 kg m²and torque acting is 0.3124 Nm

Example – 14:

Assuming that the earth is a uniform homogeneous sphere of radius 6400 km and density 5500 kg/m³, calculate its M. I. about its axis of rotation. What is the rotational K. E. of the earth when spinning about its axis?

Given: Radius of earth = 6400 km = 6.4 x 10⁶ m, Density = 5500 kg/m³.

To Find: M.I. of earth =? K.E. of the earth =?

Solution:

Ans: Moment of inertia is 9.896 x 10³⁷ kgm² and kinetic energy is 2.612 x 10²⁹ J

Example – 15:

Assuming the earth to be a sphere, calculate its M. I. about the axis of rotation. Calculate the angular momentum and rotational K. E. of the earth about its axis. Mass of earth = 6 x 10²⁴kg; R = 6400 km.

Given: Mass of earth = M = 6 x 10²⁴ kg; radius of earth = R = 6400 km = 6.4 x 10⁶ m, Time period for earth = 24 hr = 24 x 60 x 60 s.

To Find: Moment of inertia = I =? Angular momentum = L = ?, Kinetic energy = ?

Solution:

Ans : Moment of inertia is 9.3 x 10³⁷ kgm² , angular momentum = 7.149 x 10³³ kg m²/s, and kinetic energy is 2.612 x 10²⁹ J

Example – 16:

Find the ratio of the spin angular momentum of the earth to its orbital angular momentum. Distance of earth from the sun = 1.5 x 10⁸ km; Radius of earth = 6400 km; Mass of earth = 6 x 10²⁴ kg. One year = 365 days.

Given: Distance of earth from the sun = r = 1.5 x 10⁸ km; Radius of earth = R =6400 km = 6.4 x 10⁶ m; Mass of earth = 6 x 10²⁴ kg, Time period for orbital motion = T_o = 365 days = 365 x 24 x 60 x 60 s, spin period = 24 hr = 24 x 60 x 60 s.

To Find: Ratio of the spin angular momentum of the earth to its orbital angular momentum

Solution:

Ans: The ratio of the spin angular momentum of the earth to its orbital angular momentum is 2.658 x 10^-7.

Previous Topic: The Concept of Angular Momentum

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The post Numerical Problems on Kinetic Energy and Angular Momentum appeared first on The Fact Factor.

In this article, we shall study important physical quantity related to the rotational motion called the angular momentum.

The angular momentum of a rigid body about a given axis is defined as the product of its moment of inertia about the given axis and its angular velocity. It is denoted by letter ‘L’. Its S.I. unit is kg m² s^-1 and its dimensions are [M¹L²T^-1]

Relation Between Angular Momentum and Angular Velocity of a Rigid Rotating Body:

Consider a rigid body rotating about an axis passing through point O and perpendicular to the plane of the paper in an anticlockwise sense as shown. Consider infinitesimal element at P of mass dm in the plane of the paper.  Let the distance of point P from the axis of rotation be r.

The moment of inertia of the rigid body is given by

Let  ‘v’ be the linear velocity of the element, then the magnitude of the linear momentum of the element is given by

dP =  v .dm

Then, quantity dL =  r . dp, is the magnitude of the angular momentum of the element. Similarly, we can find the angular moment of each and every element in the body.  As all elements are moving in the same direction, resultant angular momentum can be calculated by integrating the above expression.

This is an expression for the angular momentum of a rotating body.

In vector form,

Principle of Conservation of Angular Momentum:

Statement: 

If the external torque acting on the body or the system is zero, then the total vector angular momentum of a body or of a system remains constant.

Explanation: 

For a rigid body rotating about the given axis, torque is given by

τ = I α

Where τ = torque acting on the rotating body

I =  Moment of Inertia of the body about the given axis of rotation

α =   Angular acceleration

For rigid body moment of inertia about the given axis of rotation is always constant.

As external torque acting on the body or the system is zero, τ =  0

Therefore, there is no change in angular momentum.

∴ L =   constant,

but L = Iω

∴ L =   Iω =   constant

I₁ω₁ = I₂ω₂ = I₃ω₃ = ……. = I_nω_n = constant

Above relation indicates that as the moment of inertia decreases angular velocity increases and vice-versa.  This is known as the principle of conservation of angular momentum.

Moment of Inertia of body changes if the distribution of mass about the axis of rotation changes.

Examples of Conservation of Angular Momentum:

This principle is used by acrobats in the circus, ballet dancers, skaters etc.  By extending or by pulling in the hands, legs, they change the distribution of mass about the axis of rotation and thus their angular velocity changes by keeping angular momentum constant.  If they, extend their hands or legs the moment of Inertia increases thus angular velocity decreases.  If they pull in their hands or legs the moment of inertia decreases thus angular velocity increases. When diver, want to execute somersault, he pulls in his arms and legs together, so that the moment of inertia decreases and his angular velocity increases.

Numerical Problems on Angular Momentum:

Example – 01:

If the earth had its radius suddenly decreased by half when spinning about its axis, what would the length of the day be?

Given: Radius of earth R₂ = ½ R₁, Present period = T₁ = 24 hr

To Find: New period = T₂ =?

Solution:

By the principle of conservation of angular momentum

Ans: New length of the day would be 6 hours

Example – 02:

What will be the duration of the day if the earth suddenly shrinks to 1/27 th of its original volume? The mass being unchanged.

Given: Volume of the earth V₂ = 1/27 V₁, Present period = T₁ = 24 hr

To Find: New period = T₂ =?

Solution:

By the principle of conservation of angular momentum

Ans: New duration of the day would be 2.67 hours

Example – 03:

A disc is rotating in a horizontal plane about a vertical axis at the rate of 5π/3 rad/s. A blob of wax of mass 0.02 kg falls vertically on the disc and adheres to it at a distance of 0.05 m from the axis of rotation. If the speed of rotation thereby becomes 40 rev/min, calculate the M. I. of the disc.

Given: Initial speed of rotation of disc = 5π/3 rad/s, Mass of blob = M = 0.02 kg, Distance from axis of rotation = r = 0.05 m, New angular speed = N = 40 rev/min

To Find: M.I. of disc = I_d =?

Solution:

The angular speed after coupling is

Let M.I. of the disc be I_d and that of bob be I_b.

By the principle of conservation of angular momentum

Ans: M.I. of the disc is 2 x 10^-4 kg m²

Example – 04:

A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 r.p.m. A blob of wax of mass 20 g falls on the disc and sticks to it at a distance of 5 cm from the axis. If the M. I. of the disc about the given axis is 2 x 10^-4 kg m², find the new frequency of rotation of the disc.

Given: Initial speed of rotation of disc = N_d = 100 r.p.m., Mass of blob = M = 20 g = 0.02 kg, Distance from axis of rotation = r = 5 cm = 0.05 m, M.I. of disc = I_d = 2 x 10^-4 kg m²

To Find: New frequency of rotation = N =?

Solution:

Let M.I. of the disc be I_d and that of bob be I_b.

By the principle of conservation of angular momentum

Ans: New angular speed = 50 r.p.m.

Example – 05:

A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 180 r.p.m. A blob of wax of mass 1.9 g falls on the disc and sticks to it at a distance of 25 cm from the axis. If the speed of rotation now is 60 r.p.m. Calculate the moment of inertia of the disc.

Given: Initial speed of rotation of disc = 180 r.p.m., Mass of bob = M = 1.9 g = 0.0019 kg, Distance from axis of rotation = r = 25 cm = 0.25 m, New angular speed = N = 60 r.p.m.

To Find: M.I. of disc = I_d =?

Solution:

Let M.I. of the disc be I_d and that of bob be I_b.

By the principle of conservation of angular momentum

Ans: M.I. of the disc is 5.94 x 10^-4 kg m²

Example – 06:

A ballet dancer spins about a vertical axis at 120 r.p.m. with her arms outstretched. With her arms folded, the M. I. about the axis of rotation decreases by 40%. Calculate the new rate of revolution.

Given: Initial angular speed = 120 r.p.m., M.I. decreases by 40 %, I₂ = I₁ – 40% I₁ = 0.6 I₁,

To Find: New angular speed = N₂ =?

Solution:

By the principle of conservation of angular momentum

Ans: New rate of revolution is 200 r.p.m.

Example – 07:

A ballet dancer spins about a vertical axis at 90 r.p.m. with her arms outstretched. With her arms folded, the M. I. about the axis of rotation changes to 75%. Calculate the new rate of revolution.

Given: Initial angular speed = 90 r.p.m., M.I. decreases to 75 %, I₂ = 0.75 I₁,

To Find: New angular speed = N₂ =?

Solution:

By the principle of conservation of angular momentum

Ans: New rate of revolution is 120 r.p.m.

Example – 08:

A man is standing on a rotating chair with his arms fully outstretched. The chair rotates with a speed of 81 r.p.m. On bringing arms close to his body, the radius of gyration is reduced by 10%. What is the increase in the speed of rotation? (neglect the friction)

Given: Initial angular speed = N₁ = 81 r.p.m., radius of gyration reduces by 10%, K₂ = K₁ – 10% K₁ = 0.90 K₁,

To Find: increase in angular speed = N₂ – N₁ =?

Solution:

By the principle of conservation of angular momentum

Ans: The Increase in angular speed = 100 – 81 = 19 r.p.m.

Example – 09:

A man standing with outstretched arms on a frictionless rotating turntable has a mass 0.2 kg in each hand. The M. I. of the system in this position is 150 kg m². He slowly folds his arms until the M. I. is reduced to 60 kg m². The angular velocity of the system is now 5 rad/s. Find the initial angular velocity and the new K.E. of the system.

Given: Initial M.I. of system = I₁ = 150 kg m². Final M.I. of system = I₂ = 60 kg m². Final angular speed = ω₂ = 5 rad/s

To Find: Initial angular speed w₁ =? Final K.E. =?

Solution:

By the principle of conservation of angular momentum

Final K.E. = ½ I w2 = ½ x 60 x (5)² = 750 J

Ans: Initial angular speed = 2 rad/s, Final kinetic energy = 750 J.

Example – 10:

A wheel is rotating with an angular speed of 500 r.p.m. on a shaft. A second identical wheel initially at rest is suddenly coupled to the same shaft. What is the angular speed of the combination? Assume that the M.I. of the shaft is negligible.

Given: Initial speed of rotation of first wheel = N₁ = 500 r.p.m.,

To Find: New frequency of rotation = N =?

Solution:

M.I. of the first wheel = I₁ and M.I. of the second wheel = I₂,

Let N be their final common angular speed.

The two wheels are identical. I₁ = I₂.

By the principle of conservation of angular momentum

Ans: The angular speed of combination is 250 r.p.m.

Example – 11:

Two wheels of moments of inertia 4 kg m² and 2 kg m² rotate at the rate of 120 rev/min and 240 rev/min respectively and in the same direction. If the two are coupled so as to rotate with a common angular velocity, find the speed of revolution.

Given: M.I. of the first wheel = I₁ = 4 kg m², Initial speed of rotation of first wheel = N₁ = 120 r.p.m., M.I. of the second wheel = I₂ = 2 kg m², Initial speed of rotation of second wheel  = N₂ = 240 r.p.m.,

To Find: New frequency of rotation = N =?

Solution:

By the principle of conservation of angular momentum

Ans: The angular speed of combination is 160 r.p.m.

Example – 12:

Two wheels of moments of inertia 4 kg m² each rotate at the rate of 120 rev/min and 240 rev/min respectively and in the opposite direction. If the two are coupled so as to rotate with a common angular velocity, find the speed of revolution.

Given: M.I. of the first wheel = I₁ = 4 kg m², Initial speed of rotation of first wheel = N₁ = 120 r.p.m., M.I. of the second wheel = I₂ = 4 kg m², Initial speed of rotation of second wheel  = N₂ = 240 r.p.m.,

To Find: New frequency of rotation = N =?

Solution:

As the two wheels are rotating in the opposite direction, the total initial angular momentum is

By the principle of conservation of angular momentum

Ans: The angular speed of combination is 60 r.p.m.

Example – 13:

Two wheels A and B can rotate side by side on the same axle. Wheel A of M.I. 0.5 kg m² is set spinning at 600 r.p.m. Wheel B of M.I. 2 kg m² is initially stationary. A clutch now acts to join A and B so that they must spin together. At what speed will they rotate? How does the rotational K.E. before joining compare with the rotational K.E. afterwards?

Given: M.I. of the first wheel = I_A = 0.5 kg m², Initial speed of rotation of first wheel = N_A = 600 r.p.m., M.I. of the second wheel = I_B = 2 kg m², Initial speed of rotation of second wheel  = N_B = 0 r.p.m.,

To Find: New common frequency of rotation = N =?

Solution:

By the principle of conservation of angular momentum

Kinetic energy before coupling

Kinetic energy after coupling

Dividing equation (2) by (1)

Ans: The new angular speed is 120 r.p.m., the ratio of initial K.E. to Final K.E.is 5:1

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