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JEE Advanced Physics Rotational Motion 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Physics Question (2022) — Solution

Question

A flat surface of a thin uniform disk A  of radius  R is glued to a horizontal table. Another thin uniform disk  B of mass M  and with the same radius R  rolls without slipping on the circumference of A , as shown in the figure. A flat surface of B also lies on the plane of the table. The center of mass of  B has fixed angular speed ω  about the vertical axis passing through the center of A . The angular momentum of B is n M ω R 2  with respect to the center of A . Which of the following is the value of n ? ​​​​​​​

Options

  1. A. 2
  2. B. 5
  3. C. 7 2
  4. D. 9 2

Answer

B. 5

Step-by-step solution

The centre of mass of  B  has angular velocity  ω  about the centre of  A . Therefore, the velocity of the centre of  B  is V B = ω 2 R . The point of contact of  A  and  B  is at rest therefore, the angular velocity of  B  will be, ω B = v B R = 2 ω Now the angular momentum of B  with respect to centre of  A L = M v B r + I C M ω B ⇒ L = M ω 2 R 2 R + M R 2 2 2 ω L = 5 M R 2 ω Comparing the magnitude with  n M ω R 2 n = 5

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