Question

A solid circular cylinder has a diameter ‘d'. A hollow circular cylinder has an internal diameter ‘d' and has the same cross sectional area as the solid cylinder. Then the ratio of the moment of inertia of the hollow cylinder to that of solid cylinder, is equal to

Answer 1

The ratio of the moment of inertia of the hollow cylinder to that of solid cylinder, is equal to 3.

Explanation:

1. Mas solid cylinder= Mass of hollow cylinder =M

2. Solid cylinder:-

 Diameter of solid cylinder = d

 Cross sectional area of solid cylinder = $A_{s}=\frac{\pi d^{2}}{4}$  ...1)

 Moment of inertia of solid cylinder = $I_{s}=\frac{1}{8}Md^{2}$  ...2)

3. Hollow cylinder:-

  Internal diameter of Hollow cylinder= d

  External diameter of Hollow cylinder= D

  Cross sectional area of Hollow cylinder =$A_{h}=\frac{\pi(D^{2}-d^{2})}{4}$      ...3)

 Moment of inertia of solid cylinder = $I_{h}=\frac{1}{8}M(D^{2}+d^{2})$    ...4)

4. Given that cross sectional area of hollow and solid is same. So from equation 1) and eqaution 3)

      $A_{s}=A_{h}$

  $\frac{\pi d^{2}}{4}=\frac{\pi(D^{2}-d^{2})}{4}$  

So

        $D^{2}$=2×$d^{2}}$     ...5)

5. From equation 2) and equation 4) ,we can find out the ratio of moment of inertia of hollow cylinder to solid cylinder

  $\frac{I_{h}}{I_{s}}=\frac{\frac{1}{8}M(D^{2}+d^{2})}{\frac{1}{8}Md^{2}}$      

$\frac{I_{h}}{I_{s}}=\frac{(D^{2}+d^{2})}{d^{2}}$         ...6)

6. Now from equation 5)

    $\frac{I_{h}}{I_{s}}=\frac{(2d^{2}+d^{2})}{d^{2}}$

So  

  $\frac{I_{h}}{I_{s}}=3$