Two identical uniform discs of mass \(m\) and radius \(r\) are arranged as shown in the figure. If…
Two identical uniform discs of mass \(m\) and radius \(r\) are arranged as shown in the figure. If \(\alpha\) is the angular acceleration of the lower disc and \(a_{\mathrm{cm}}\) is acceleration of centre of mass of the lower disc, then relation among \(a_{\mathrm{cm}}, \alpha\) and \(r\) is
\(a_{\mathrm{cm}}=\frac{\alpha}{r}\)
\(a_{\mathrm{cm}}=2 \alpha r\)
\(a_{\mathrm{cm}}=\alpha r\)
none of these
Solution
When the lower disc has an angluar acceleration of a the upper disc also accelerates at a, they being identical and the tension on rope is same for both discs
\(\begin{array}{l}
\alpha=\frac{\partial \omega}{\partial \mathrm{t}} \\
\mathrm{a}_{\mathrm{cm}}=\frac{\partial \mathrm{r} \omega}{\partial \mathrm{t}}+\frac{\partial \mathrm{r} \omega}{\partial \mathrm{t}}=2 \mathrm{r} \alpha
\end{array}\)
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Asked in: JEE Mains - Rotational Motion - Chapter Test
