A circular disc reaches from top to bottom of an inclined plane of length $l$. When it slips down the plane, if takes $t \mathrm{~s}$. When it rolls down the plane then it takes $\left(\frac{\alpha}{2}\right)^{1 / 2} t \mathrm{~s}$, where $\alpha$ is _________

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Aicharya · Accepted Answer

For slipping $\begin{aligned}& \mathrm{a}=\mathrm{g} \sin 	heta \ & \ell=\frac{1}{2} \mathrm{at}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \ell}{\mathrm{g} \sin 	heta}}\end{aligned}$ For rolling $\begin{aligned}& \mathrm{a}^{\prime}=\frac{\mathrm{g} \sin 	heta}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}\left[\mathrm{k}=\frac{\mathrm{R}}{\sqrt{2}}ight] \& \Rightarrow \mathrm{a}^{\prime}=\frac{2 \mathrm{~g} \sin 	heta}{3} \& \ell=\frac{1}{2} \mathrm{a}^{\prime}\left(\mathrm{t}^{\prime}ight)^2 \& \Rightarrow \mathrm{t}^{\prime}=\sqrt{\frac{6 \ell}{2 \mathrm{~g} \sin 	heta}}=\sqrt{\frac{\alpha}{2}} \sqrt{\frac{2 \ell}{\mathrm{g} \sin 	heta}} \ & \Rightarrow \alpha=3\end{aligned}$

A circular disc reaches from top to bottom of an inclined plane of length $l$. When it slips down the plane,…

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