Question

Two wheels $A$ and $C$ are connected by a belt $B$ . As denoted in the figure, the radius of $C$ is three times the radius of $A$ . If it is known that both the wheels have the same rotational kinetic energy, compute the ratio of their respective inertia $\left(\frac{I_C}{I_A}\right)$ .

Answer 1

Answer: 9

As the belt does not slip, $v_p=v_Q$ i.e.,
$r_A{\omega }_A=r_C{\omega }_C\dots .(\because v=r\omega )\dots \text{ (i) }$
If $r_A=r$, then $r_C=3r$,
$\therefore {\omega }_A=3{\omega }_C\dots [\mathrm{From}(i)]$
If both the wheels have the same rotational kinetic energy, then
$\begin{array}{l}\frac{1}{2}{I}_A{\omega }_A^2=\frac{1}{2}{I}_C{\omega }_C^2\\ \frac{I_C}{I_A}={\left[\frac{{\omega }_A}{{\omega }_C}\right]}^2={\left[\frac{3}{1}\right]}^2=\frac{9}{1}\end{array}$