In this article, we shall study to solve problems of calculation of moment of inertia, the radius of gyration, and torque acting on a rotating body.

Example – 01:

A thin uniform rod of length 1 m and mass 1 kg is rotating about an axis passing through its centre and perpendicular to its length. Calculate the moment of inertia and radius of gyration of the rod about an axis passing through a point midway between the centre and its edge perpendicular to its length.

Given: Mass of rod = M = 1 kg, length of rod = l = 1m.

To Find: M.I. = I =? and radius of gyration = K =? at h = l/4

Solution:

Moment of inertia of a thin uniform rod about a transverse axis passing through its centre is given by

Now I = MK²

K² = I/M = 0.1458/1 = 0.1458

K = 0.3818 m

Ans: M.I. about the required axis is 0.1458 kg m², and radius of gyration is0.3818 m

Example – 02:

Calculate the M. I. of a thin uniform rod of mass 100g and length 60 cm about an axis perpendicular to its length and passing through (1) its centre and (2) one end.

Given: Mass of rod = M = 100 g = 0.1 kg, length of rod = l = 60 cm = 0.6 m.

Solution:

Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is given by

I = Ml²/12 = 0.1 x (0.6)²/12 = 0.003 kg m²

Moment of inertia of the thin uniform rod about a transverse axis passing through its end is given by

I = Ml²/3 = 0.1 x (0.6)²/3 = 0.012 kg m²

Ans: Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is 0.003 kg m²  and the moment of inertia of the thin uniform rod about a transverse axis passing through its end is 0.012 kg m².

Example – 03:

Find the M. I. of a thin uniform rod 1 m long and weighing 0.024 kg about a transverse axis passing through (1) its centre (2) one end (3) a point 20 cm from one end.

Given: Mass of rod = M = 0.024 kg, length of rod = l = 1 m.

Solution:

Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is given by

I = Ml²/12 = 0.024 x (1)²/12 = 0.002 kg m²

Moment of inertia of the thin uniform rod about a transverse axis passing through its end is given by

I = Ml²/3 = 0.024 x (1)²/3 = 0.008 kg m²

Moment of inertia of the thin uniform rod about a transverse axis passing through a point 20 cm from one end.

Ans: Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is 0.002 kg m², Moment of inertia of the thin uniform rod about a transverse axis passing through its end is 0.008 kg m², Moment of inertia of the thin uniform rod about a transverse axis passing through a point 20 cm from one end is 4.16 x 10-3 kg m²

Example – 04:

A thin ring has mass 0.25 kg and radius 0.5 m. Calculate the moment of inertia about i) an axis passing through its centre and perpendicular to the plane and ii) an axis passing through a point on its circumference, perpendicular to its plane iii) diameter. iii) an axis tangent to ring and its plane

Given : Mass of Ring = M =  0.25 kg, Radius of ring = R = 0.5 m

Solution:

Moment of inertia about a transverse axis passing through its centre is given by

I = MR² = 0.25 x (0.5)² = 0.25 x 0.25 = 0.0625 kgm²

Moment of inertia about a transverse axis passing through its circumference is given by

I = 2MR² = 2 x 0.25 x (0.5)² = 2 x 0.25 x 0.25 = 0.125 kgm²

Moment of inertia about its diameter is given by

I = MR²/2  = 0.25 x (0.5)²/2 =  0.25 x 0.25/2 = 0.03125 kgm²

Moment of inertia about an axis tangent to the ring and its plane is given by

I = 3/2MR² = 3/2 x (0.25 x (0.5)²) = 3/2 x (0.25 x 0.25) = 0.09375 kgm²

Ans: Moment of inertia about a transverse axis passing through its centre is 0.0625 kgm². Moment of inertia about a transverse axis passing through its circumference is 0.125 kgm². Moment of inertia about its diameter is 0.03125 kgm². Moment of inertia about an axis tangent to the ring and its plane is 0.09375 kgm²

Example – 05:

Calculate the moment of inertia of a ring of mass 500 g and radius 0.5 m about an axis of rotation coinciding with its diameter and tangent perpendicular to its plane.

Given: Mass of Ring = M = 500 g = 0.5 kg, Radius of ring = R = 0.5 m

Solution:

Moment of inertia about its diameter is given by

I = MR²/2  = 0.5 x (0.5)²/2 =  0.5 x 0.25/2 = 0.0625 kgm²

Moment of inertia about an axis tangent to ring and perpendicular to its plane is given by

I = 2MR²  = 2 x 0.5 x (0.5)² =  0.25 kgm²

Ans: Moment of inertia about its diameter is 0.0625 kgm². Moment of inertia about an axis tangent to ring and perpendicular to its plane is 0.0625 kgm²

Example – 06:

Calculate the M.I. of a disc of mass 0.5 kg and radius 10 cm about an axis passing through its centre and at right angles to its plane. What would be the M. I. if the axis passes through a point at a distance from the centre equal to half the radius of the disc?

Given : Mass of disc = M =  0.5 kg, Radius of disc = R = 10 cm = 0.1 m

Solution:

Moment of inertia about a transverse axis passing through its centre is given by

I = MR²/2  = 0.5 x (0.1)²/2 = 0.25 x 0.25 = 0.0025 kgm²  = 2.5 x 10^-3 kg m²

By parallel axes theorem

Ans: M.I. about the axis passes through a point at a distance from the centre equal to half the radius of the disc is 3.75 x 10^-3 kg m²

Example – 07:

Calculate the M. I. of a solid brass sphere of radius 12 cm about a tangent to the sphere. Density of brass = 8500 kg/m³

Given: Radius of sphere = R = 12 cm = 0.12 m, density = r = 8500 kg/m³

Solution:

Moment of inertia of a solid sphere about its tangent is given by

Example – 08:

A solid sphere has a radius ‘R’. If its radius of gyration of this sphere about its diameter is √2/5R. show that the radius of gyration about the tangential axis of rotation is √7/5R.

Solution:

By parallel axes theorem

Example – 09:

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

Given: R₂ = 2R₁

To Find: Ratio of M.I. = I₁/I₂ =?

Solution:

Moment of inertia of a solid sphere about its diameter is given by

Ans: The ratio of initial M.I. to the Final M.I. is 1:4

Example – 10:

A solid cylinder of uniform density of radius 2 cm has a mass of 50 g. If its length is 10 cm, calculate the moment of inertia about i) its own axis of rotation passing through its centre ii) an axis passing through its centre and perpendicular to the length.

Given: Mass of cylinder = M = 50 g = 0.05 Kg, Radius of cylinder = R = 2 cm = 0.02 m, length of cylinder = l = 10 cm = 0.1 m

To Find: Moment Of Inertia =?

Solution:

 Moment of inertia of the cylinder about its own axis passing through its centre is given by

I = MR²/2  = 0.05 x (0.02)²/2 =  0.05 x 0.0004/2 = 10^-5 kgm²

Moment of inertia of the cylinder about an axis passing through its centre and perpendicular to the length

Ans: Moment of inertia of the cylinder about its own axis passing through its centre is 10^-5 kgm². Moment of inertia of the cylinder about an axis passing through its centre and perpendicular to the length is 4.67 x 10^-5 kgm²

Example – 11:

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 torque applied.

Given: Diameter of sphere = 25 cm, Radius of sphere = R = 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 seconds.

To Find: Moment of inertia = I =? and torque = τ = ?

Solution:

Moment of inertia of a solid sphere about its diameter by

Ans: Moment of inertia of sphere is 0.1562 kg m² and torque applied  is 0.3124 Nm

Example -12:

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 applied = t = 10 Nm, Initial angular speed w₁ = 0 rad/s, time = t = 6 s.

To Find: final angular speed = ω₂ =?

Solution:

Moment of inertia of disc about an axis passing through its centre and perpendicular to its plane is given by

Ans: Final angular speed is 300 rad/s

Example – 13:

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: Torque acting = τ = 400 Nm, Mass = M = 40 kg, angular acceleration = α = 20 rad/s².

To Find: Moment of Inertia =?, Radius of gyration =?

Solution:

Torque acting on body is given by

τ = I α

I = τ/α = 400/20 = 20 kg m².

Now the moment of inertia is given by

I = MK²

K² = I/M = 20/40 = 0.5

K = √50

K = 0.707 m

Ans: The moment of inertia is 20 kg m² and the radius of gyration is 0.707 m

Example – 14:

The M.I. of a uniform circular disc about an axis passing through its centre and perpendicular to its plane is ½ MR². Find the distance of a parallel axis from the centre of mass about which the M.I. of the disc is MR². Radius of disc = √2 cm.

Solution:

By parallel axes theorem

Ans: The distance of the axis is 1 cm

Example – 15:

A disc has a radius of gyration of 0.02 m when rotating about an axis passing through its centre and at right angles to its plane. What would be its radius of gyration when rotating about an axis coincident with a diameter of its face?

Given: Radius of gyration = K = 0.02 m for axis passing through centre.

To Find: Radius of gyration = K = ?  for axis passing through diametre

Solution:

The M.I. in the first Case

Ans: The radius of gyration is 1.414 x 10^-2 m

Example = 16:

Radius of gyration of a body about an axis at a distance of 0.12 m from its centre of mass is 0.13 m. Find its radius of gyration about a parallel axis through the centre of mass.

Given: Distance between axes = h = 0.12 m, radius of gyration = K = 0.13 m

To Find: Radius of gyration = K = ? for axis passing through centre.

Solution: 

Ans: Radius of gyration is 0.05 m for axis passing through centre.

Previous Topic: Applications of Parallel and Perpendicular Axes

Next Topic: The Concept of Angular Momentum

Science > Physics > Rotational Motion > Numerical Problems on Moment of Inertia and Radius of Gyration

The post Numerical Problems on Moment of Inertia and Radius of Gyration appeared first on The Fact Factor.

The parallel axes theorem states that ” The moment of inertia of a rigid body about any axis is equal to the sum of its moment of inertia about a parallel axis through its centre of mass and the product of the mass of the body and the square of the distance between the two axes.”

I_O = I_G +  Mh²

The perpendicular axes theorem states that ” Moment of inertia of a rigid plane lamina about an axis perpendicular to its plane is equal to the sum of its moment of inertia about any two mutually perpendicular axes in its plane and meeting in the point where the perpendicular axis cuts the lamina.”

l_z    =    l_x  +   l_y

Expression for Moment of Inertia of a Uniform Rod About a Transverse Axis Passing Through its End:

 We know that  moment of inertia of a thin  rod about a transverse axis passing through its centre G is given  by

We have to find the Moment of Inertia about the parallel transverse axis passing through the end of the rod.  Let I_o be the moment of Inertia about this axis. Then by the principle of parallel axes,

This is an expression for moment of inertia of a thin uniform rod about a transverse axis passing through its end.

Expression for Moment of Inertia  of Thin Uniform Disc About its Tangent Perpendicular to its Plane:

The moment of inertia of thin uniform disc about a transverse axis passing through its centre is given by

We have to find the M.I. about a tangent perpendicular to the plane of the disc. These two axes are parallel to each other. By parallel axes theorem,

This is an expression for M.I. of a thin uniform disc about its tangent perpendicular to its plane.

Expression for Moment of Inertia of Thin Uniform Ring About its Tangent Perpendicular to its Plane:

The moment of inertia of thin uniform ring about a transverse axis passing through its centre is given by

I = MR² = I_G

We have to find the M.I. about a tangent perpendicular to the plane of the ring. These two axes are parallel to each other. By parallel axes theorem

I_O =  I_G +  Mh²

∴   I_O =  MR² +  MR²

∴   I_O =  2 MR²

This is an expression for M.I. of a thin uniform ring about its tangent perpendicular to its plane.

Expression for Moment of Inertia of Thin Uniform Disc About its  Diameter:

The moment of inertia of thin uniform disc about a transverse axis passing through its centre is given by

We have to find the M.I. about a diameter of the disc. Let us consider a system of three axes, such that the z-axis is along the geometrical axis of the disc and x-axis and y-axis lie in the plane of the disc such that the centre of mass G lies at the origin of the system of axes.

By perpendicular axes theorem, we have

l_z    =    l_x  +   l_y   …………. (2)

Due to symmetry M.I. of the disc about diameter is

l_d  =  l_x  =   l_y   …………. (3)

Substituting values of equations (1) and (3) in (2)

This is an expression for M.I. of a thin uniform disc about its diameter.

Expression for Moment of Inertia  of Thin Uniform Ring About its  Diameter:

The moment of inertia of thin uniform ring about a transverse axis passing through its centre is given by

I = MR²

We have to find the M.I. about a diameter of the ring. Let us consider a system of three axes, such that the z-axis is along the geometrical axis of the ring and x-axis and y-axis lie in the plane of the ring such that the centre of mass G lies at the origin of the system of axes.

l_z  = MR²  ………(1)

By perpendicular axes theorem, we have

l_z    =    l_x  +   l_y   …………. (2)

Due to symmetry M.I. of the disc about diameter is

l_d  =  l_x  =   l_y   …………. (3)

Substituting values of equations (1) and (3) in (2)

This is an expression for M.I. of a thin uniform ring about its diameter.

Expression for Moment of Inertia of a Thin Uniform Disc About an Axis Tangent to the Disc and in the Plane of the Disc:

The moment of inertia of thin uniform disc about a transverse axis passing through its centre is given by

We have to find the M.I. about a diameter the disc first. Let us consider a system of three axes, such that the z-axis is along the geometrical axis of the disc and x-axis and y-axis lie in the plane of the disc such that the centre of mass G lies at the origin of the system of axes.

By perpendicular axes theorem, we have

l_z    =    l_x  +   l_y   …………. (2)

Due to symmetry M.I. of the disc about diameter is

l_d  =  l_x  =   l_y   …………. (3)

Substituting values of equations (1) and (3) in (2)

Now y axis passes through the centre of mass G of the disc

Now the tangent to the disc in the plane of the disc is parallel to the y-axis.

By parallel axes theorem.

This is an expression for M.I. of a thin uniform disc about its tangent in its plane and in the plane of the disc.

Expression for Moment of Inertia of a Thin Uniform Ring About an Axis Tangent to the Ring and in the Plane of the Ring:

The moment of inertia of thin uniform ring about a transverse axis passing through its centre is given by

l = MR²

We have to find the M.I. about a diameter the ring first. Let us consider a system of three axes, such that the z-axis is along the geometrical axis of the ring and x-axis and y-axis lie in the plane of the ring such that the centre of mass G lies at the origin of the system of axes.

l_z  = MR²  ………(1)

By perpendicular axes theorem, we have

l_z    =    l_x  +   l_y   …………. (2)

Due to symmetry M.I. of the disc about diameter is

l_d  =  l_x  =   l_y   …………. (3)

Substituting values of equations (1) and (3) in (2)

Now the tangent to the ring in the plane of the ring is parallel to the y-axis.

By parallel axes theorem.

This is an expression for M.I. of a thin uniform ring about its tangent in its plane.

Expression for Moment of Inertia  of a Solid Cylinder About a Transverse Axis Passing Through its Centre:

Consider a solid cylinder of mass M, length ‘’ and radius ‘r’ capable of rotating about its geometrical axis. Let ‘m be its mass per unit length.

m = M/l      Hence M = m . l

A solid cylinder can be regarded as a number of thin uniform discs of infinitesimal thickness piled on top of one another. Let us consider one such disc of thickness ‘dx’ at a distance of ‘x’ from the centre C of the cylinder.

Mass of such disc is given by

Mass, dm = m.dx  = (M/l) dx

The M.I. of such disc about a transverse axis (passing through C) is given by

The M.I. of a disc about diameter is given by

Integrating above expression in limits

This is an expression for M. I. of a solid cylinder about a transverse axis passing through its centre.

Expression for Moment of Inertia of a Hollow Cylinder About a Transverse Axis Passing Through its Centre:

Consider a hollow cylinder of mass M, length ‘’ and radius ‘r’ capable of rotating about its geometrical axis. Let ‘m be its mass per unit length.

m = M/l      Hence M = m . l

A hollow cylinder can be regarded as a number of thin uniform rings of infinitesimal thickness piled on top of one another. Let us consider one such ring of thickness ‘dx’ at a distance of ‘x’ from the centre C of the cylinder.

Mass of such ring is given by

Mass, dm = m.dx  = (M/l) dx

The M.I. of such ring about a transverse axis (passing through C) is given by

dI = dm . R²

The M.I. of a ring about diameter is given by

Integrating the above expression in limits

This is an expression for M. I. of a hollow cylinder about a transverse axis passing through its centre.

Expression for Moment of Inertia of a Solid Sphere About its Tangent.

The moment of inertia of a solid sphere about its geometrical  axis (diameter) is given by

We have to find the M.I. about a tangent to the sphere. These two axes are parallel to each other. By parallel axes theorem

This is an expression for M.I. of a solid sphere

Previous Topic: Derivation of Expression for M.I.

Next Topic: Numerical Problems on Moment of Inertia

Science > Physics > Rotational Motion > Applications of Parallel and Perpendicular Axes Theorems

The post Applications of Parallel and Perpendicular Axes Theorems appeared first on The Fact Factor.

In this article, we shall study the method of deriving an expression for moment of inertia of a body.

Expression for Moment of Inertia of Uniform Rod About a Transverse Axis Passing Through its Centre:

Consider a thin uniform rod, of mass M, an area of cross-section A, length l, and density of material ρ. Let us consider an infinitesimal element of length dx at a distance of x from the given axis of rotation.  Then its Moment of Inertia about the given axis is given by

dI = x² . dm,  …… (1)

But,   Mass =  Volume x Density

∴   dm    =  A . dx . ρ  ……   (2)

From Equations (1) and (2)

dI  =    x² A . dx .ρ

∴ dI  = A .ρ . x² . dx

The moment of inertia I of the rod about. the given axis is given by

This is an expression for moment of inertia of a thin uniform rod about a transverse axis passing through its centre.

Expression for Moment of Inertia of a Uniform Rod About a Transverse Axis Passing Through its End:

Method – I:

Consider a thin uniform rod, of mass M, an area of cross-section A, length l, and density of material ρ. Let us consider an infinitesimal element of length dx at a distance of x from the given axis of rotation.  Then its Moment of Inertia about the given axis is given by

dI = x² . dm,  …… (1)

But,   Mass =  Volume x Density

∴   dm    =  A . dx . ρ  ……   (2)

From Equations (1) and (2)

dI  =    x² A . dx .ρ

∴ dI  = A .ρ . x² . dx

The moment of inertia I of the rod about. the given axis is given by

This is an expression for moment of inertia of a thin uniform rod about a transverse axis passing through its end.

Expression for the Moment of Inertia of an Annular Ring:

Consider a uniform thin annular disc of mass M having inner radius R₁, outer radius R₂, thickness t, and density of its material ρ. Let us assume that disc is capable of rotating about a transverse axis passing through its centre. Let us assume that the disc is made up of infinitesimally thin rings.

Consider  one such ring of radius r and width dr.  Moment of Inertia of such element is  given, by,

dI  =   r² . dm   ………….. (1)

But,      Mass   =    volume × density

dm  =  ( 2 π r . dr .  t) ρ   ………  (2)

From Equation (1) and (2)

dI  =   r² . ( 2 π r . dr .  t) ρ

dI  =   2 π t  ρ  r³ . dr .

The moment of inertia I of the annular disc will be given by

Where M is the total mass of the annular ring.

This is an expression for moment of inertia of annular ring about a transverse axis passing through its centre.

Expression for Moment of Inertia of a Thin Uniform Disc About a Transverse Axis Passing Through its Centre and Perpendicular to its Plane:

The moment of inertia of annular ring about a transverse axis passing through its centre is given by

For the solid disc, there is no centre hole, hence R₂ =   R and R₁ = 0

This is an expression for moment of inertia of thin uniform disc about a transverse axis passing through its centre.

Expression for Moment of Inertia of a Thin Uniform Ring About an Axis Passing through its Centre and Perpendicular to its Plane:

The moment of inertia of annular ring about a transverse axis passing through its centre is given by

For ring, the centre hole extends up to its periphery, hence R₂ =   R and R₁ =R

This is an expression for moment of inertia of thin uniform ring about a transverse axis passing through its centre.

Expression for Moment of Inertia  of a Solid Cylinder About its Geometrical Axis:

Consider a solid cylinder of mass M, length ‘’ and radius ‘r’ capable of rotating about its geometrical axis. Let ‘m be its mass per unit length.

m = M/l      Hence M = m . l

A solid cylinder can be regarded as a number of thin uniform discs of infinitesimal thickness piled on top of one another. Let us consider one such disc of thickness ‘dx’ at a distance of ‘x’ from the centre C of the cylinder.

Mass of such disc is given by

Mass, dm = m.dx   =  (M /l). dx

The M.I. of such disc about a transverse axis (passing through C) is given by

Integrating the above expression in limits

This is an expression for M. I. of a solid cylinder about its geometrical axis.

Expression for Moment of Inertia of a Hollow Cylinder About its Geometrical Axis:

Consider a hollow cylinder of mass M, length ‘’ and radius ‘r’ capable of rotating about its geometrical axis. Let ‘m be its mass per unit length.

m = M/l      Hence M = m . l

A hollow cylinder can be regarded as a number of thin uniform rings of infinitesimal thickness piled on top of one another. Let us consider one such ring of thickness ‘dx’ at a distance of ‘x’ from the centre C of the cylinder.

Mass of such ring is given by

Mass, dm = m.dx  = (M/ l) dx

The M.I. of such ring about a transverse axis (passing through C) is given by

This is an expression for M. I. of a solid cylinder about its geometrical axis.

Expression for Moment of Inertia of a Solid Sphere About its Diameter (Geometrical axis):

Let us consider a solid homogeneous sphere of radius ‘R’ and mass ‘M’, capable of rotating about its diameter. Let us consider a circular strip of infinitesimal thickness ‘dx’ at a distance of x from centre ‘O’. The radius of this circular strip is PM, which is given by

This circular strip can be treated as thin disc rotating about a transverse axis passing through its centre.

The M.I. of the disc about a transverse axis passing through its centre is given by

The M.I. of the whole sphere about diameter can be obtained by integrating the above expression.

The mass of the sphere = M. Hence, the M.I. of the solid homogeneous sphere is given by

This is an expression for M.I. of a solid sphere about its diameter (Geometrical axis).

Previous Topic: Principles of Parallel and Perpendicular Axes

Next Topic: Applications of Parallel and Perpendicular Axes

Science > Physics > Rotational Motion Moment of Inertia of Standard Bodies

The post Moment of Inertia of Standard Bodies appeared first on The Fact Factor.

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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Science > Physics > Rotational Motion > Numerical Problems on Kinetic Energy and Angular Momentum

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

Previous Topic: Numerical Problems on Moment of Inertia

Next Topic: Numerical Problems on Angular Momentum

Science > Physics > Rotational Motion > Angular Momentum

The post Angular Momentum appeared first on The Fact Factor.

In this article, we are going to study important physical quantity related to the rotational motion of the body called the moment of inertia and its significance.

Rigid Body:

A rigid body is one whose geometric shape and size remains unchanged under the action of any external force.

Axis of rotation:

When a rigid body performs the rotational motion, the particles of the body moves in circles. The centres of these circles lie on a straight line called the axis of rotation, which is fixed and perpendicular to the plane of circles. The particles on the axis of rotation are stationary.

Moment of Inertia:

The moment of inertia of a rigid body about a given axis is, defined as the sum of the products of the mass of each and every particle of the body and the square of its distance from the given axis.

S.I. unit of moment of inertia is kg m² and C.G.S. system it is g cm². Dimensions of the moment of inertia are [M¹L²T⁰]

Explanation:

If the system of masses consists of a number of point masses m₁, m₂, m₃, m₄, …….., m_n situated at a distance of r1, r2, r3, r4, …….., rn from the axis of rotation, then by definition of the moment of Inertia we have

Instead of assuming the body to be composed of discrete masses, it can be considered to be composed of continuous matter (mass); then the process of summation should be replaced by integration with proper limits.

Physical Significance of Moment of Inertia:

Newton’s first law of motion is also called a law of inertia indicates that a body is unable to change by itself its state of rest or state of uniform motion along a straight line.  This property of inertness is known as inertia. It is an inherent property of matter. It is due to the inertia, a body opposes any change in its state of rest or of uniform motion in a straight line.

In the translational motion, the mass of a body is a measure of its inertia.  Greater the mass, larger is the inertia, greater is the force required to produce a given linear acceleration in it. In rotational motion, the moment of inertia of a body is a measure of its inertia.  Greater the moment of inertia, larger is the torque required to produce a given angular acceleration in it.

Thus the moment of inertia in the rotational motion is analogous to the mass in translational motion because it plays the same role in rotational motion as the mass plays in translational motion. This is clear from the following table

Radius of Gyration:

The radius of gyration of a rigid body about a given axis is defined as the distance from the axis at which the whole mass of the body must be supposed to be concentrated so that this imaginary point mass has the same moment of Inertia as the actual body, about the given axis.

The radius of gyration is denoted by letter ‘K’.  As it is a distance, its S.I. unit is m and c.g.s. unit is cm. Its dimensions are [M⁰L¹T⁰]

The expression for moment of inertia in terms of the radius of gyration is

 I = MK².

Where I = Moment of inertia of a body

M = Mass of the body

K = Radiation of gyration of the body

Physical Significance of Radius of Gyration:

Moment of inertia depends on the mass of the body, the distribution of mass about the axis of rotation and the position of the axis of rotation. These factors can be separated by expressing the moment of inertia as a product of the mass and the square of a particular distance from the axis of rotation.  This particular distance is called a radius of gyration (K) of the body. By definition of the radius of gyration

From the above explanation, we can conclude that the radius of gyration is the measure of the distribution of the mass of the body about the given axis

Numerical Problems on Definition of Moment of Inertia:

Example 01:

Four point masses 1kg, 2 kg, 3 kg and 4 kg are located at the corners A, B, C and D of a square ABCD of side 1m. Find the moment of inertia and radius of gyration in each of the following cases when the axis of rotation is

• passing through A and perpendicular to the plane of ABCD
• passing through O, the centre of the square  and perpendicular to plane of ABCD
• along the side AB
• along diagonal AC

Given: m₁ = 1 kg, m₂ = 2 kg, m₃ = 3 kg, m₄ = 4 kg, AB = BC = CD = AD = 1 m

Solution:

Axis passing through A and perpendicular to the plane of ABCD

I_A = ∑ m_ir_i² = m₁r₁² + m₂r₂² + m₃r₃² + m₄r₄²

∴ I_A =  (1)(0)² + (2)(1)² + (3)(√2)² + (4)(1)²

∴ I_A =  0 + 2 + 6 + 4 = 10 kg m²

M = m₁ + m₂ + m₃ + m₄ = 1 + 2 +3 + 4  = 10 kg

By definition of radius of gyration

I_A = MK_A²

∴  10 = 10 K_A²

∴ K_A² = 1

K_A = 1 m

Ans: Moment of inertia of system about axis passing through A and perpendicular to the plane of ABCD is 10 kg m² and corresponding radius of gyration is 1 m.

Axis passing through O, the centre of the square and perpendicular to the plane of ABCD

I_O = ∑ m_ir_i² = m₁r₁² + m₂r₂² + m₃r₃² + m₄r₄²

∴ I_O =  (1)(√2 /2)² + (2)(√2 /2)² + (3)(√2 /2)² + (4)(√2 /2)²

∴ I_O =  (1)(0.5)² + 2(0.5)² + 3(0.5)² + 4(0.5)² = 10(0.5)² = 10 × 0.25 = 2.5 kg m²

M = m₁ + m₂ + m₃ + m₄ = 1 + 2 +3 + 4  = 10 kg

By definition of radius of gyration

I_A = MK_O²

∴  2.5 = 10 K_O²

∴ K_O² = 0.25

∴  K_O = 0.5 m

Ans: Moment of inertia of system about axis passing through O  and perpendicular to the plane of ABCD is 2.5 kg m² and corresponding radius of gyration is 0.5 m.

Axis passing through Side AB:

I_AB = ∑ m_ir_i² = m₁r₁² + m₂r₂² + m₃r₃² + m₄r₄²

∴ I_AB =  (1)(0)² + (2)(0)² + (3)(1)² + (4)(1)²

∴ I_AB =  0 + 0 + 3 + 4  = 7 kg m²

M = m₁ + m₂ + m₃ + m₄ = 1 + 2 +3 + 4  = 10 kg

By definition of radius of gyration

I_AB = MK_AB²

∴  7 = 10 K_AB²

∴ K_AB² = 0.7

K_AB = 0.837 m

Ans: Moment of inertia of system about side AB is 75 kg m² and corresponding radius of gyration is 0.837m.

Axis passing through Diagonal AC:

I_AC = ∑ m_ir_i² = m₁r₁² + m₂r₂² + m₃r₃² + m₄r₄²

∴ I_AC =  (1)(0)² + (2)(√2 /2)² + (3)(0)² + (4)(√2 /2)²

∴ I_AC =  0 + (2)(0.5)² + 0 + 4(0.5)²  = 7 kg m²

M = m₁ + m₂ + m₃ + m₄ = 1 + 2 +3 + 4  = 10 kg

By definition of radius of gyration

I_AC = MK_AC²

∴  7 = 10 K_AC²

∴ K_AC² = 0.7

K_AC = 0.837 m

Ans: Moment of inertia of system about diagonal AC is 75 kg m² and corresponding radius of gyration is 0.837m.

Example 02:

Three point masses 1 kg, 2 kg and 3 kg are located at the vertices A, B and C of an equilateral triangle ABC of side 1m. Find the moment of inertia and radius of gyration in each of the following cases when axis of rotation is

• passing through A and perpendicular to the plane of triangle ABC.
• passing through centroid G of the triangle and perpendicular to the plane of triangle ABC
• along the side AB

Given: m₁ = 1 kg, m₂ = 2 kg, m₃ = 3 kg, AB = BC = AC = 1 m

Solution:

Axis passing through A and perpendicular to the plane of triangle ABC

I_A = ∑ m_ir_i² = m₁r₁² + m₂r₂² + m₃r₃²

∴ I_A =  (1)(0)² + (2)(1)² + (3)(1)²

∴ I_A =  0 + 2  +3  = 5 kg m²

M = m₁ + m₂ + m₃  = 1 + 2 +3  = 6 kg

By definition of radius of gyration

I_A = MK_A²

∴  5 = 6 K_A²

∴ K_A² = 5/6 = 0.8333

K_A = 0.913 m

Ans: Moment of inertia of system about A is 5 kg m² and corresponding radius of gyration is 0.913 m.

Passing through centroid G of the triangle and perpendicular to the plane of triangle ABC

For equilateral triangle GA = GB = GC = side/√3   = 1 / √3   = 0.577 m

I_G = ∑ m_ir_i² = m₁r₁² + m₂r₂² + m₃r₃²

∴ I_G =  (1)(0.577)² + (2)(0.577)² + (3)(0.577)²

∴ I_G =  1/3 + 2/3  +3/3  = 6/3 = 2 kg m²

M = m₁ + m₂ + m₃  = 1 + 2 +3  = 6 kg

By definition of radius of gyration

I_G = MK_G²

∴  2 = 6 K_G²

∴ K_G² = 2/6 = 0.3333

K_G = 0.577 m

Ans: Moment of inertia of system about G is 2 kg m² and corresponding radius of gyration is 0.577 m.

Along the side AB

I_AB = ∑ m_ir_i² = m₁r₁² + m₂r₂² + m₃r₃²

∴ I_AB =  (1)(0)² + (2)(0)² + (3)(0.866)²

∴ I_AB =  0 + 0  + 1.25  = 1.25 kg m²

M = m₁ + m₂ + m₃  = 1 + 2 +3  = 6 kg

By definition of radius of gyration

I_AB = MK_AB²

∴  1.25 = 6 K_AB²

∴ K_AB² = 1.25/6 = 0.2083

K_AB = 0.456 m

Ans: Moment of inertia of system about G is 1.25 kg m² and corresponding radius of gyration is 0.456 m.

Example 03:

A light thin uniform rod of length 2 m has two bodies stuck to its ends. The mass of each body is 100 g. Find the moment of inertia of the system about a transverse axis passing through i) the centre of mass of the rod ii) one end of the rod.

Case – I: Axis through the centre of mass of the rod

Case – II: Axis through the centre of mass of the rod

Ans: The M.I. of the system about the axis passing through the centre is 0.2 kg m² and about the axis passing through the end is 0.4 kg m².

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Science > Physics > Rotational Motion > Concept of Moment of Inertia

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