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(A) For a sliding cylinder,the acceleration is $a_s = g \sin 	heta$. The time taken to reach the bottom is $t_s = \sqrt{\frac{2L}{g \sin 	heta}} = \

Vedclass · Accepted Answer

The sliding cylinder will reach the bottom first with greater speed.

Vedclass · Answer

(A) For a sliding cylinder,the acceleration is $a_s = g \sin 	heta$. The time taken to reach the bottom is $t_s = \sqrt{\frac{2L}{g \sin 	heta}} = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g}}$ and the final velocity is $v_s = \sqrt{2gh}$.For a rolling cylinder,the acceleration is $a_R = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}} = \frac{g \sin 	heta}{\beta}$,where $\beta = 1 + \frac{I}{MR^2}$. For a solid cylinder,$I = \frac{1}{2}MR^2$,so $\beta = 1.5$.The time taken is $t_R = \sqrt{\frac{2L}{a_R}} = \frac{1}{\sin 	heta} \sqrt{\beta \frac{2h}{g}}$ and the final velocity is $v_R = \sqrt{\frac{2gh}{\beta}}$.Since $\beta > 1$,it follows that $t_R > t_s$ (the sliding cylinder reaches first) and $v_R < v_s$ (the sliding cylinder has greater speed).

Two identical solid cylinders run a race starting from rest at the top of an inclined plane. If one cylinder slides and the other rolls,which of the following statements is true?

Similar Questions

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