A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is LA and LB computed about points A and B, respectively, with OB = 2 OA. The value of LALB is: [figure]
- 2
- 14
- 12
- 1
Answer
(4) 1
Solution
The angular momentum of a rotating rigid body about any point P is given by:
LP = Lcm + rcm/P pcm
Since the centre of the disc O is fixed, the centre of mass is at rest, meaning vcm = 0 and pcm = 0.
Therefore, the angular momentum about any point in the plane is simply equal to the angular momentum about the centre of mass:
LA = Lcm = LB
Thus, LALB = 1.
Answer: (4)