Practice perpendicular axis theorem example. Learn how to calculate moment of inertia of a ring of mass and radius about the given tangent.

- For a planar body, consider that X and Y-axes are in the plane of the body and Z-axis is perpendicular to the plane of the body. The three axes chosen should be mutually perpendicular.
- Then, the theorem states that -

Moment of inertia along Z-axis = Sum of Moment of inertia along X and Y-axes

i.e., \(I_Z = I_X + I_Y\)

**Note : ** This theorem is applicable only to the planar bodies

A A cuboid

B A sphere

C A disk

D A disk

- For a planar body, consider that X and Y-axes are in the plane of the body and Z-axis is perpendicular to the plane of the body. The three axes chosen should be mutually perpendicular.
- Then, the theorem states that -

Moment of inertia along Z-axis = Sum of Moment of inertia along X and Y-axes

i.e., \(I_Z = I_X + I_Y\)

**Note : ** This theorem is applicable only to the planar bodies

A \(I_1-I_2\)

B \(I_1+I_2\)

C \(2I_1+I_2\)

D \(I_1-2I_2\)

- Consider a ring of radius R and mass M.

- Moment of inertia of the ring about OZ, \(I_Z = MR^2\)....(1)
- The mass distribution about the axes OX and OY is same. Thus, moment of inertia about OX and OY will also be same.

Hence, \(I_X=I_Y=I\) ....(2)

- Applying perpendicular axis theorem-

\(I_Z=I_X+I_Y\)

\(MR^2=2I\) [ From equation (1) and (2) ]

\(\therefore\) \(I=\dfrac {MR^2}{2}\)

- Consider a disk of mass M and radius R.

- Moment of inertia of the disk about OZ, \(I_Z=\dfrac {MR^2}{2}\)...(1)
- The mass distribution about the axes OX and OY is same. Thus, moment of inertia about OX and OY will also be same.

Hence, \(I_X=I_Y=I\) ....(2)

- Applying perpendicular axis theorem-

\(I_Z=I_X+I_Y\)

\(\dfrac {MR^2}{2}=2I\)

\(I=\dfrac {MR^2}{4}\)

A 14 kg m2

B 16 kg m2

C 12 kg m2

D 18 kg m2

- Consider a square plate of length L and mass M.

- The square plate is a rod extended parallel to its axis.
- If \(I_X\) be the moment of inertia about X-axis and \(I_Y\) be the moment of inertia about Y-axis then

Applying perpendicular axis theorem,

\(I_Z=I_X+I_Y\)

\(I_Z=\) \(\dfrac {ML^2}{12}+\dfrac {ML^2}{12}\)

\(I_Z=\)\(\dfrac {ML^2}{6}\)

where, M = Mass of square plate

L = length of square plate

- Consider a square plate of length L. The X and Y-axes are cutting the plane, as shown in figure.

- The plate does not seem to be symmetric about X and Y-axes, but the four parts into which X and Y-axes cut the plane are exactly same.
- Since mass distribution is same along X and Y-axes.

Therefore, \(I_X=I_Y=I\)

Applying perpendicular axis theorem

\(I_Z=I_X+I_Y\)

\(\Rightarrow\dfrac{ML^2}{6}=2I\)

\(\Rightarrow \;I=\dfrac {ML^2}{12}\)

where, M = Mass of square plate

L = length of square plate

A 3 kg m2

B 2 kg m2

C 8 kg m2

D 6 kg m2

- Consider a ring of mass M and radius R.

- Moment of inertia of the ring about an axis passing through its center of mass and perpendicular to the plane is

? \(I_Z = MR^2\) ...(1)

- Moment of inertia of the ring about X and Y-axes are equal.

Thus, \(I_X=I_Y=I\)

- Applying perpendicular axis theorem

\(I_Z=I_X+I_Y\)

\({MR^2}=2I\) [ From equation (1) ]

\(\dfrac {MR^2}{2}=I=I_X=I_Y\)

- For the calculation of \(I_A\)
_{ };

use parallel axis theorem,

\(I=I_0+Md^2\)

\(I_A=I_Y+MR^2\)

\(I_A=\dfrac {MR^2}{2}+MR^2\)

\(I_A=\dfrac {3MR^2}{2}\)

- Consider a disk of mass M and radius R.

- Moment of inertia of the disk about an axis passing through its center of mass and perpendicular to the plane is

\(I_Z = \)\(\dfrac {MR^2}{2}\)...(1)

- Moment of inertia of the disk about X and Y-axes are equal.

Thus, \( I_X = I_Y = I\)

- Applying perpendicular axis-theorem

\( I_Z = I_X + I_Y \)

\(\dfrac {MR^2}{2}=2I\) [ From equation (1) ]

\(\dfrac {MR^2}{4}=I=I_X=I_Y\)

- For the calculation of \(I_A\);

use parallel-axis theorem-

\( I = I_0 + Md^2 \)

\( I_A = I_Y +MR^2 \)

\(I_A=\dfrac {MR^2}{4}+MR^2\)

\(I_A=\dfrac {5}{4}MR^2\)

A 750 kg m2

B 250 kg m2

C 500 kg m2

D 225 kg m2