Physics - Rotation Question with Solution | TestHub

PhysicsRotationEnergy in rotationHard2 minQB

PhysicsHardsingle choice

Passage / Comprehension

A small puck of mass $m$ is carefully placed onto the inner surface of the thin hollow cylinder of mass $M$ and of radius $R$ . Initially, the cylinder rests on the horizontal plane and the puck is located at the height $R$ above the plane as shown in the figure. Assume that the friction between the puck and the inner surface of the cylinder is absent, and the cylinder moves on the plane without slipping. The free fall acceleration is $g$ .

The interaction force F between the puck and the cylinder at the moment when the puck passes the lowest point of its trajectory is :-

Options:

Answer:

Solution:

Let $V$ be the cylinder's speed and $u$ the puck's speed relative to the cylinder's centre at the lowest point. No external horizontal impulse on (puck + cylinder) in the rolling constraint direction gives $2 M V + m (V + u) = 0$ ... (i), and energy $m g R = M V^{2} + \frac{1}{2} m v^{2}$ with $v = V + u$ ... (ii).

From (i) and (ii): $u^{2} = \frac{( 2 M + m ) g R}{M}$ .

At the lowest point the puck's upward acceleration is $\frac{u ^{2}}{R}$ : $N = m g + \frac{m u ^{2}}{R} = m g (1 + \frac{2 M + m}{M}) = 3 m g (1 + \frac{m}{3 M})$ - (C)

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Energy in rotation

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...