Question

Two wheels $A$ and $C$ are connected by a belt $B$ as shown in figure. The radius of $C$ is three times the radius of $A$. What would be the ratio of the rotational inertias ($I_C/I_A)$ if both the wheels have the same rotational kinetic energy?

Answer 1

Since there is no slipping , So velocity at the contact surface belt and wheel will be same for both.

$v_A=v_C\Rightarrow {\omega }_A\times r_a={\omega }_C\times r_C$

as we know $r_c=3r_A$ , we get ${\omega }_A=3{\omega }_C$

Now their rotational energy are same so ,$\frac{I_a{\omega }_A^2}{2}=\frac{I_c{\omega }_C^2}{2}$

$\Rightarrow \frac{I_A}{I_C}=\frac{1}{9}$