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# Theorems on moment of inertia

Module by: Sunil Kumar Singh. E-mail the author

Summary: Theorems on moment of inertia help to calculate it about additional relevant axes without any integral evaluation.

The use of integral expression of moment of inertia (MI) for a rigid body can be employed to calculate MI about different axes of rotation. However, such consideration is tedious because of difficulty in integral evaluation. There are two theorems, which avoid integral evaluation for calculating MI about certain other axes, when moment of inertia about an axis is known. The theorems are (i) theorem of parallel axes and (ii) theorems of perpendicular axes.

## Theorem of parallel axes

This theorem enables us to calculate MI about any axis, parallel to the axis passing through center of mass (COM). The mathematical expression of this theorem is given as :

I = I C + M d 2 I = I C + M d 2
(1)

where I C I C is MI of the body about an axis passing through center of mass, "C"; I is MI of the body about an axis parallel to the axis passing through center of mass; "M" is the mass of the rigid body and "d" is the perpendicular distance between two parallel axes.

We shall here drive expression for the MI about parallel axis in a general case of arbitrarily shaped three dimensional rigid body. This derivation requires clear visualization in three dimensional coordinate space. We shall use three figures to bring out relative distances and angles in order to achieve the clarity as stated.

Let "M" be the mass of the rigid body and "dm" be the mass of a small element. Let "C" be the center of mass and "Cz" be an axis of rotation as shown in the figure. For convenience, we consider that "Cz" is along z-coordinate of the reference system.

In general, the elemental mass "dm" may be situated anywhere in relation to the axis "cz". Here, we consider two cross sections of the rigid body in parallel planes, which are perpendicular to z - axis (axis of rotation). The two cross sections are drawn here such that one (on the rear) contains center of mass "C" and other (on the front) contains the elemental mass "dm". We designate the plane containing elemental mass "dm" as xy-plane of the coordinate system.

Now we concentrate on the plane (the plane in the front) containing elemental mass "dm". The element of mass "dm" being in the xy-plane has z = 0 and its position is given by two coordinates "x" and "y".

The parallel axis through center of mass intersects the plane (xy) at point "O". Let Az' be the parallel axis about which, we have to calculate MI of the rigid body. This axis intersects the plane (xy) at point "A" as shown in the figure below.

In order to comprehend three dimensional aspect of perpendicular lines, we should realize that all lines drawn on the plane (xy) are perpendicular to two parallel axes (Oz or Az') as xy plane is perpendicular to them. The line BA in the xy plane, therefore, represents the perpendicular distance of elemental mass "dm" from the axis Az'. Similarly, the line BO represents the perpendicular distance of elemental mass "dm" from the axis Oz. Also, the line OA in xy-plane is perpendicular distance between two parallel axes, which is equal to "d".

Now, MI of the rigid body about axis Az' in the form of integral is given by :

I = r 2 đ m = ( AE 2 + BE 2 ) đ m = { ( x - a ) 2 + ( y - b ) 2 } đ m I = r 2 đ m = ( AE 2 + BE 2 ) đ m = { ( x - a ) 2 + ( y - b ) 2 } đ m

I = ( x 2 + a 2 - 2 x a + y 2 + b 2 - 2 x b ) đ m I = ( x 2 + a 2 - 2 x a + y 2 + b 2 - 2 x b ) đ m

Rearranging, we have :

I = ( x 2 + y 2 ) đ m + ( a 2 + b 2 ) đ m - 2 a x đ m - 2 b y đ m I = ( x 2 + y 2 ) đ m + ( a 2 + b 2 ) đ m - 2 a x đ m - 2 b y đ m

However, the coordinates of the center of mass by definition are given as :

x C = x đ m M y C = y đ m M z C = z đ m M x C = x đ m M y C = y đ m M z C = z đ m M

But, we see here that "x" and "y" coordinates of center of mass are zero as it lies on z - axis. It means that :

x C = x đ m M = 0 x đ m = 0 x C = x đ m M = 0 x đ m = 0

Similarly,

y đ m = 0 y đ m = 0

Thus, the equation for the MI of the rigid body about the axis parallel to an axis passing through center of mass "C" is :

I = ( x 2 + y 2 ) đ m + ( a 2 + b 2 ) đ m I = ( x 2 + y 2 ) đ m + ( a 2 + b 2 ) đ m

From the figure,

x 2 + y 2 = R 2 x 2 + y 2 = R 2

and

a 2 + b 2 = d 2 a 2 + b 2 = d 2

Substituting in the equation of MI, we have :

I = ( x 2 + y 2 ) đ m + ( a 2 + b 2 ) đ m = R 2 đ m + d 2 đ m I = ( x 2 + y 2 ) đ m + ( a 2 + b 2 ) đ m = R 2 đ m + d 2 đ m

We, however, note that "R" is variable, but "d" is constant. Taking the constant out of the integral sign :

I = R 2 đ m + d 2 đ m = R 2 đ m + M d 2 I = R 2 đ m + d 2 đ m = R 2 đ m + M d 2

The integral on right hand side is the expression of MI of the rigid body about the axis passing through center of mass. Hence,

I = I C + M d 2 I = I C + M d 2

## Application of theorem of parallel axes

We shall illustrate application of the theorem of parallel axes to some of the regular rigid bodies, whose MI about an axis passing through center of mass is known.

(i) Rod : about an axis perpendicular to the rod and passing through one of its ends

MI about perpendicular bisector is known and is given by :

I C = M L 2 12 I C = M L 2 12

According to theorem of parallel axes,

I = I C + M d 2 = M L 2 12 + M x ( L 2 ) 2 = M L 2 3 I = I C + M d 2 = M L 2 12 + M x ( L 2 ) 2 = M L 2 3

(ii) Circular ring : about a line tangential to the ring

MI about an axis passing through the center and perpendicular to the ring is known and is given by :

I C = M R 2 I C = M R 2

According to theorem of parallel axes,

I = I C + M d 2 = M R 2 + M R 2 = 2 M R 2 I = I C + M d 2 = M R 2 + M R 2 = 2 M R 2

Since hollow cylinder has similar expression of MI, the expression for the MI about a line tangential to its curved surface is also same.

(i) Circular plate : about a line perpendicular and tangential to the circular plate

MI about an axis passing through the center and perpendicular to the circular plate is known and is given by :

I C = M R 2 2 I C = M R 2 2

According to theorem of parallel axes,

I = I C + M d 2 = M R 2 2 + M R 2 = 3 M R 2 2 I = I C + M d 2 = M R 2 2 + M R 2 = 3 M R 2 2

Since solid cylinder has similar expression of MI, the expression for the MI about a line tangential to its curved surface is also same.

(iii) Hollow sphere : about a line tangential to the hollow sphere

MI about one of its diameters is known and is given by :

I C = 2 M L 2 3 I C = 2 M L 2 3

According to theorem of parallel axes,

I = I C + M d 2 = 2 M R 2 3 + M L 2 = 5 M L 2 3 I = I C + M d 2 = 2 M R 2 3 + M L 2 = 5 M L 2 3

(iv) Solid sphere : about a line tangential to the solid sphere

MI about one of its diameters is known and is given by :

I C = 2 M L 2 5 I C = 2 M L 2 5

According to theorem of parallel axes,

I = I C + M d 2 = 2 M R 2 5 + M R 2 = 7 M L 2 5 I = I C + M d 2 = 2 M R 2 5 + M R 2 = 7 M L 2 5

## Theorem of perpendicular axes

Theorem of perpendicular axes is applicable to planar bodies only. According to this theorem, moment of inertia of a planar rigid body about a perpendicular axis (z) is equal to the sum of moments of inertia about two mutually perpendicular axes (x and y) in the plane of the body, which are themselves perpendicular to the z - axis.

I z = I x + I y I z = I x + I y
(2)

Three axes involved in the theorem are mutually perpendicular to each other and form the rectangular coordinate system. It should be noted that there is no restriction that axes pass through center of mass of the rigid body as in the case of parallel axis theorem. In order to prove the theorem in general, we consider a planar body of irregular shape. We consider an elemental mass "dm" at P (x,y) as shown in the figure above.

The moment of inertia of the body about z - axis is given as :

I = r 2 đ m I = r 2 đ m

We drop two perpendiculars in the plane of body on "x" and "y" axes. Then,

r 2 = x 2 + y 2 r 2 = x 2 + y 2

Substituting in the equation of moment of inertia,

I z = ( x 2 + y 2 ) đ m = x 2 đ m + y 2 đ m I z = ( x 2 + y 2 ) đ m = x 2 đ m + y 2 đ m

The expressions on the right hand side represent moments of inertia of the body about "x" and "y" axes. Hence,

I z = I x + I y I z = I x + I y

## Application of theorem of perpendicular axes

We shall determine MIs of objects about other axes, which are not symmetric to the body. As pointed out earlier, we shall limit the application of the theorem of perpendicular axes to planar objects like rectangular plate, ring and circular plate etc.

(i) Rectangular plate : about a line passing through center of mass and perpendicular to the plane

Moment of inertia about a line passing through center of mass and parallel to one of the side is given as :

I x = M b 2 12 I y = M a 2 12 I x = M b 2 12 I y = M a 2 12

Applying theorem of perpendicular axes, we have :

I z = I x + I y = M b 2 12 + M a 2 12 I z = M ( a 2 + b 2 ) 12 I z = I x + I y = M b 2 12 + M a 2 12 I z = M ( a 2 + b 2 ) 12

(ii) Circular ring : about one of the diameter

Moment of inertia about a perpendicular line passing through center of ring is given as :

I z = M R 2 I z = M R 2

We consider two mutually perpendicular diameters as "x" and "y" axes of the cordinate system. Now, applying theorem of perpendicular axes, we have :

I z = I x + I y I z = I x + I y

However, the circular ring has uniform mass distribution, which is symmetrically distributed about any diameter. Thus, MIs about the axes "x" and "y" are equal,

I x + I y = 2 I x I x + I y = 2 I x

Thus,

2 I x = I z = M R 2 I x = M R 2 2 2 I x = I z = M R 2 I x = M R 2 2

(iii) Circular plate : about one of the diameter

Moment of inertia about a perpendicular line passing through center of circular plate is given as :

I z = M R 2 I z = M R 2

We consider two mutually perpendicular diameters as shown in the figure. These diameters can be treated as two coordinates of the plane of the circular plate. Now, applying theorem of perpendicular axes, we have :

I z = I x + I y I z = I x + I y

However, the circular plate has uniform mass distribution, which is symmetrically distributed about any diameter. Thus, MIs about the planar axes "x" and "y" are equal,

I x + I y = 2 I x I x + I y = 2 I x

Thus,

2 I x = I z = M R 2 2 I x = M R 2 4 2 I x = I z = M R 2 2 I x = M R 2 4

## Summary

1. Theorem of perpendicular axes

The mathematical expression of this theorem is given as :

I = I C + m d 2 I = I C + m d 2

where I C I C is MI of the body about an axis passing through center of mass, "C"; I is MI of the body about an axis parallel to the axis passing through center of mass; "M" is the mass of the rigid body and "d" is the perpendicular distance between two parallel axes.

2. Theorem of parallel axes is valid for all shapes (regular and irregular).

3. Theorem of parallel axes expresses MI about a parallel axis in terms of the MI about a parallel axis passing through COM. Thus, we should ensure that one of the two parallel axes be through the COM of the rigid body.

4. Theorem of perpendicular axes

According to this theorem, moment of inertia of a planar rigid body about a perpendicular axis (z) is equal to the sum of moments of inertia about two mutually perpendicular axes (x and y) in the plane of the body, which are also perpendicular to the axis along z - axis.

I z = I x + I y I z = I x + I y

4. Theorem of perpendicular axes is valid only for all planar shapes (regular and irregular).

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