Two wheels have the same mass. First wheel is in the form of a solid disc of radius $\mathrm{R}$ while the second is a disc with inner radius $r$ and outer radius $R$. Both are rotating with same angular velocity $\omega_0$ about transverse axes through their centres. If the first wheel comes to rest in time $t_1$ while the second comes to rest in time $t_2$, are $t_1$ and $t_2$ different? Why?
Q 104
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The moments of inertia of the two wheels about transverse axes through their centres are
$
I_1=\frac{1}{2} M R^2, \quad I_2=\frac{1}{2} M\left(R^2+r^2\right)
$
( $\because$ they have the same mass)
Assuming the same (frictional) torque, $\tau$, acts on both the wheels,
$
\tau=I_1 \alpha_1=I_2 \alpha_2
$
Since $I_2>I_1, \quad \alpha_1>\alpha_2$.
$
\omega=\omega_0+\alpha t
$
Since the final angular velocity $\omega=0$,
$
\begin{aligned}
& \alpha_1=-\frac{\omega_0}{t_1} \text { and } \alpha_2=-\frac{\omega_0}{t_2} \\
\therefore & \frac{\omega_0}{t_1}>\frac{\omega_0}{t_2} \\
\therefore & t_1<t_2
\end{aligned}
$
$
I_1=\frac{1}{2} M R^2, \quad I_2=\frac{1}{2} M\left(R^2+r^2\right)
$
( $\because$ they have the same mass)
Assuming the same (frictional) torque, $\tau$, acts on both the wheels,
$
\tau=I_1 \alpha_1=I_2 \alpha_2
$
Since $I_2>I_1, \quad \alpha_1>\alpha_2$.
$
\omega=\omega_0+\alpha t
$
Since the final angular velocity $\omega=0$,
$
\begin{aligned}
& \alpha_1=-\frac{\omega_0}{t_1} \text { and } \alpha_2=-\frac{\omega_0}{t_2} \\
\therefore & \frac{\omega_0}{t_1}>\frac{\omega_0}{t_2} \\
\therefore & t_1<t_2
\end{aligned}
$
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