A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of angle alpha. The acceleration of the center of mass is

B. 23 g — For a body rolling without slipping, the acceleration is given by a = g sin alpha / (1 + I/MR^2). For a solid cylinder, the moment of inertia I is 1/2 MR^2.

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$A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of angle α . The acceleration of the center of mass is$

A. $g sin α$
B. $\frac{2}{3} g sin α$
C. $\frac{1}{2} g sin α$
D. $\frac{3}{4} g sin α$

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Correct answer: B — $\frac{2}{3} g sin α$

Solution

The acceleration is a = \frac{g s i n α}{1 + I / M R ^{2}} . Substituting I = \frac{1}{2} M R^{2} gives a = \frac{g sin α}{1 + 1/2} = \frac{2}{3} g sin α .

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A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of angle α. The acceleration of the center of mass is

Solution

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More Rotational Motion practice questions

$A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of angle α . The acceleration of the center of mass is$