Two identical rings are to be rotated about different axes of rotation as shown by applying torques so as to produce the same angular acceleration in both. How is it possible ?
Q 103
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The $\mathrm{MI}$ of ring 1 about a transverse tangent is $\mathrm{I}_1=2 \mathrm{MR}^2$
The $\mathrm{MI}$ of ring 2 about its diameter is
$
\mathrm{I}_2=\frac{1}{2} \mathrm{MR}^2
$
Since, torque $\mathrm{T}=\tau=I \alpha$, to produce the same angular acceleration in both,
$
\begin{aligned}
\frac{\tau_1}{I_1} & =\frac{\tau_2}{I_2} \Rightarrow \frac{\tau_1}{\tau_2}=\frac{I_1}{I_2}=4 \\
\text { i.e., } \tau_1 & =4 \tau_2
\end{aligned}
$
$\therefore$ It will be possible to produce the same angular acceleration in both the rings only if
$
\tau_1=4 \tau_2
$
The $\mathrm{MI}$ of ring 2 about its diameter is
$
\mathrm{I}_2=\frac{1}{2} \mathrm{MR}^2
$
Since, torque $\mathrm{T}=\tau=I \alpha$, to produce the same angular acceleration in both,
$
\begin{aligned}
\frac{\tau_1}{I_1} & =\frac{\tau_2}{I_2} \Rightarrow \frac{\tau_1}{\tau_2}=\frac{I_1}{I_2}=4 \\
\text { i.e., } \tau_1 & =4 \tau_2
\end{aligned}
$
$\therefore$ It will be possible to produce the same angular acceleration in both the rings only if
$
\tau_1=4 \tau_2
$
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