Question:

A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x\, = \,0$ in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is $x$ and the velocity is $v$. At that instant, which of the following options is/are correct?

Updated On: Jun 14, 2022
  • The position of the point mass $m$ is: $x=-\sqrt{2}\frac{mR}{M+m}$
  • The velocity of the point mass $m$ is: $V=\sqrt{\frac{2\sqrt{gR}}{1+\frac{m}{M}}}$
  • The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{mR}{M+m}$
  • The velocity of the block $M$ is: $V=-\frac{m}{M}\sqrt{2gR}$
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The Correct Option is C

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Concepts Used:

System of Particles and Rotational Motion

  1. The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
  2. The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
  3. The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.