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

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