COMEDK2024Evening ShiftPhysicsRotational MotionActual
A solid cylinder of mass 2 ~kg and radius 0.2 ~m is rotating about its own axis without friction with angular velocity 5 ~rad s ⁻¹ . A particle of mass 1 ~kg moving with a velocity of 5 ~ms ⁻¹ strikes the cylinder and sticks to it as shown in figure. The angular velocity of the system after the particle sticks to it will be
Options
- A15.0 ~rad~s ⁻¹
- B10.0 ~rad~s ⁻¹
- C30.0 ~rad~s ⁻¹
- D12.0 ~rad~s ⁻¹
Correct answer
A. 15.0 ~rad~s ⁻¹
Step-by-step solution
The initial angular momentum of the system about the axis of rotation is the sum of the angular momentum of the cylinder and the angular momentum of the particle. The cylinder is rotating about its own axis, so its angular momentum is L_ cyl = I = ( 1 2 M R^2) . Given M = 2 kg , R = 0.2 m , and = 5 rad s ⁻¹ , we have I = 1 2 2 (0.2)^2 = 0.04 kg m ^2 . Thus, L_ cyl = 0.04 5 = 0.2 kg m ^2 s ⁻¹ . The particle of mass m = 1 kg moves with velocity v = 5 m s ⁻¹ at a perpendicular distance R = 0.2 m from the axis. Its ang