First slide

General Planar motion

Question

In the figure, a cylinder A is initially rolling with velocity v on the horizontal surface of the wedge B (of same mass as A).All surfaces are smooth and B has no initial velocity. Then maximum height reached by cylinder on the wedge will be

Difficult

A

$v^{2} / 4 g$
B

$v^{2} / g$
C

$v^{2} / 2 g$
D

$v^{2} / 8 g$

Solution

At the maximum height, vertical velocity of cylinder is zero. But horizontal velocity of the wedge and cylinder will be same.

In the absence of friction between the cylinder and the wedge surface, angular velocity of cylinder remains constant. From energy conservation:
$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v^{' 2} + \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{' 2} + m g h . . . . . . . . . (i)$
By the conservation of linear momentum
$m v = 2 m v^{'} \Rightarrow v^{'} = v / 2 . . . . . . (i i)$
$From (i) and (ii): h = v^{2} / 4 g$

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