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Rotational Motion question

2017 · Shift 1 · Q54
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Rotational Motion question

2017 · Shift 1 · Q54

JEE AdvancedPhysicsRotational MotionMultiple correct+4 / −1
A block of mass MMM has a circular cut with a frictionless surface as shown. The block resets on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x=0,x=0,x=0, in a co-ordinate system fixed to the table. A point mass mmm 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 xxx and the velocity is v.v.v. At that instant, which of the following options is/are correct? JEE Advanced 2017 Paper 1 Offline Physics - Rotational Motion Question 48 English
  1. A
    The position of the point mass mmm is : x=−2mRM+mx = - \sqrt 2 {{mR} \over {M + m}}x=−2​M+mmR​
  2. B
    The velocity of the point mass mmm is : v=2gR1+mMv = \sqrt {{{2gR} \over {1 + {m \over M}}}}v=1+Mm​2gR​​
  3. C
    The xxx component of displacement of the center of mass of the block MMM is: −mRM+m- {{mR} \over {M + m}}−M+mmR​
  4. D
    The velocity of the block MMM is: V=−mM2gRV = - {m \over M}\sqrt {2gR}V=−Mm​2gR​
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