Question

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia I_A and I_B is __________ .

Options

• I_A > I_B

• I_A = I_B

• I_A < I_B

• depends on the actual values of t and r

Answer 1

I_A < I_B

 

Moment of inertia of circular disc of radius r:
I = \[\frac{1}{2}m r^2\]

Mass = Volume × Density

Volume of disc = \[\pi r^2 t\]

Here, t is the thickness of the disc.

As density is same for both the rods, we have

Moment of inertia,

\[I \propto\text{ thickness }\times  \left(\text{radius} \right)^4\]

\[\frac{I_A}{I_B} = \frac{t . \left( r \right)^4}{\frac{t}{4} \left( 4r \right)^4} < 1\] 

\[ \Rightarrow \frac{I_A}{I_B} < 1\]

\[\Rightarrow  I_A  <  I_B\]