Question

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. A particle of mass 1 kg moving with a velocity of 5 m/s 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:

• A. 15.0 rad/s
• B. 12.0 rad/s
• C. 10.0 rad/s
• D. 30.0 rad/s

Hint:

To solve angular momentum problems, remember that angular momentum is conserved in the absence of external forces. Use the formula \( L = I \omega \) for both the particle and the rotating body, and set the initial and final angular momenta equal to each other.

Answer 1

The Correct Option is A

The system consists of a solid cylinder and a particle. We will use the principle of conservation of angular momentum to solve this problem. Since there are no external torques, the angular momentum before and after the collision will be conserved. The angular momentum of the particle before the collision is given by: \[ L_{\text{particle}} = m v r \] where: - \( m = 1 \, \text{kg} \) (mass of the particle), - \( v = 5 \, \text{m/s} \) (velocity of the particle), - \( r = 0.2 \, \text{m} \) (radius of the cylinder).
Thus, the angular momentum of the particle before the collision is: \[ L_{\text{particle}} = 1 \times 5 \times 0.2 = 1 \, \text{kg} \cdot \text{m}^2/\text{s} \] The initial angular momentum of the cylinder is: \[ L_{\text{cylinder}} = I \omega \] where: - \( I = \frac{1}{2} M R^2 \) (moment of inertia of the solid cylinder), - \( M = 2 \, \text{kg} \) (mass of the cylinder), - \( R = 0.2 \, \text{m} \) (radius of the cylinder), - \( \omega = 5 \, \text{rad/s} \) (angular velocity of the cylinder). Substituting values, we get: \[ I = \frac{1}{2} \times 2 \times (0.2)^2 = 0.04 \, \text{kg} \cdot \text{m}^2 \] \[ L_{\text{cylinder}} = 0.04 \times 5 = 0.2 \, \text{kg} \cdot \text{m}^2/\text{s} \] Now, the total angular momentum before the collision is: \[ L_{\text{total}} = L_{\text{cylinder}} + L_{\text{particle}} = 0.2 + 1 = 1.2 \, \text{kg} \cdot \text{m}^2/\text{s} \] After the particle sticks to the cylinder, the total angular momentum is: \[ L_{\text{total}} = I_{\text{total}} \omega_{\text{final}} \] where \( I_{\text{total}} = I_{\text{cylinder}} + m r^2 \) (total moment of inertia of the system), and \( \omega_{\text{final}} \) is the final angular velocity.
Thus, \[ I_{\text{total}} = 0.04 + 1 \times (0.2)^2 = 0.04 + 0.04 = 0.08 \, \text{kg} \cdot \text{m}^2 \] Now, using the conservation of angular momentum: \[ 1.2 = 0.08 \omega_{\text{final}} \] \[ \omega_{\text{final}} = \frac{1.2}{0.08} = 15 \, \text{rad/s} \]
Thus, the angular velocity of the system after the particle sticks to it is \( 15.0 \, \text{rad/s} \).