BITSAT2024PhysicsWork, Power and EnergyActual
A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m . The height of the table from the ground is 0.8 m . If the angular speed of the particle is 12 rad s ⁻¹ , the magnitude of its angular momentum about a point on the ground right under the centre of the circle is
Options
- A14.4 ~kg ~m ^2 ~s ⁻¹
- B8.64 ~kg ~m ^2 ~s ⁻¹
- C20.16 ~kg ~m ^2 ~s ⁻¹
- D11.52 ~kg ~m ^2 ~s ⁻¹
Correct answer
A. 14.4 ~kg ~m ^2 ~s ⁻¹
Step-by-step solution
Angular momentum L₀= mvr 90^ =2 0.6 12 1 1 [ . As V=r , . Sin 90^ =1 ] So, L₀=14.4 kgm ^2 / s