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** 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 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 transverse axis passing through its end.

**Method – II**

- 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 principle of parallel axes,

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

**Expression for the Moment of Inertia of an Annular Ring:**

- Consider uniform, thin annular disc of mass M having inner radius R
_{1}, outer radius R_{2}, 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 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 solid disc, there is no centre hole, hence R_{2} = R and R_{1} = 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 R_{2} = R and R_{1} =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 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 rimg. 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 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 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 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 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 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 a 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 a diameter is given by

Integrating 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 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 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).

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

** **The moment of inertia of 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 solid sphere

Science > Physics > Rotational Motion > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |