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Q.

A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of $1 m s^{- 1}$ goes up an inclined plane which make an angle of $30^{0}$ with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A?

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a

0.57 s

b

0.52s

c

0.60 s

d

0.80 s

answer is C.

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Detailed Solution

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For a solid sphere, $\frac{k^{2}}{R^{2}} = \frac{2}{5}$
$a = \frac{9.8 \sin 30}{1 + \frac{2}{5}}$
$a = 3.5 m / \sec^{2}$
Time of ascent is given by
$\begin{array}{l} v = u + a t \\ 0 = 1 - 3.5 t \\ t = \frac{1}{3.5} s \end{array}$
Time of decent
$t = \frac{1}{3.5} sec.$
Due to symmetry of motion.
Total time, $T = \frac{2}{3.5} = 0.57 s$

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