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Question

(A) For slipping down the plane without friction,the acceleration is $a = g \sin 	heta$. The time taken to cover distance $l$ is $t = \sqrt{\frac{2l}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) For slipping down the plane without friction,the acceleration is $a = g \sin 	heta$. The time taken to cover distance $l$ is $t = \sqrt{\frac{2l}{g \sin 	heta}}$.For rolling down the plane,the acceleration is $a' = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$. For a circular disc,the radius of gyration $k$ is $\frac{R}{\sqrt{2}}$,so $\frac{k^2}{R^2} = \frac{1}{2}$.Thus,$a' = \frac{g \sin 	heta}{1 + 1/2} = \frac{2}{3} g \sin 	heta$.The time taken for rolling is $t' = \sqrt{\frac{2l}{a'}} = \sqrt{\frac{2l}{\frac{2}{3} g \sin 	heta}} = \sqrt{\frac{3}{2}} \sqrt{\frac{2l}{g \sin 	heta}} = \sqrt{\frac{3}{2}} t$.Comparing this with $t' = \left(\frac{\alpha}{2}ight)^{1/2} t$,we get $\sqrt{\frac{\alpha}{2}} = \sqrt{\frac{3}{2}}$,which implies $\alpha = 3$.

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

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