Physics - Rotation Question with Solution | TestHub

PhysicsRotationEnergy in rotationMedium2 minQB

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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 velocity of puck at the lowest point of trajectory is :

Options:

Answer:

Solution:

From the momentum relation of Q.13, at the lowest point $v = V + u$ and $2 M V + m v = 0$ ⇒ $V = - \frac{m v}{2 M}$ .

Energy: $m g R = M V^{2} + \frac{1}{2} m v^{2} = \frac{m ^{2} v ^{2}}{4 M} + \frac{m v ^{2}}{2} = \frac{( 2 M + m ) m v ^{2}}{4 M}$ .

$v^{2} = \frac{4 M g R}{2 M + m}$ ⇒ $v = \frac{4 M g R}{2 M + m}$ - (A)

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Energy in rotation

⏱ 2mℹ️ Source: QB

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