Question

The plane of the figure represents a horizontal plane. A thin rigid rod at rest is pivoted without friction about a fixed vertical axis passing through O. Its mass moment of inertia is equal to 0.1 kg·cm² about O. A point mass of 0.001 kg hits it normally at 200 cm/s at the location shown, and sticks to it. Immediately after the impact, the angular velocity of the rod is ________________ rad/s (in integer).

Hint:

Use conservation of angular momentum to solve problems involving collisions of point masses and rigid bodies.

Answer 1

Correct Answer: 10

The problem involves a collision between a point mass and a rod. To find the angular velocity of the rod immediately after the collision, we can use the principle of conservation of angular momentum. Since there is no external torque, the total angular momentum before the impact is equal to the total angular momentum after the impact. The initial angular momentum of the system is given by the point mass's angular momentum about the pivot point. The angular momentum of a point mass is calculated as: \[ L = m v r \] where:
- \( m = 0.001 \, \text{kg} \) is the mass of the point mass,
- \( v = 200 \, \text{cm/s} = 2 \, \text{m/s} \) is the velocity of the point mass,
- \( r = 0.1 \, \text{m} \) is the distance from the pivot point to where the point mass strikes the rod.
Thus, the initial angular momentum is: \[ L = 0.001 \times 2 \times 0.1 = 0.0002 \, \text{kg·m/s}. \] After the impact, the total angular momentum is the sum of the angular momentum of the rod and the point mass. The point mass sticks to the rod, so the moment of inertia of the system after the impact is: \[ I_{\text{total}} = I_{\text{rod}} + I_{\text{point mass}} = 0.1 \, \text{kg·cm}^2 + (m r^2). \] Converting to kg·m²: \[ I_{\text{total}} = 0.1 \times 10^{-4} + 0.001 \times 0.1^2 = 1 \times 10^{-5} + 0.000001 = 1.1 \times 10^{-5} \, \text{kg·m}^2. \] Using conservation of angular momentum: \[ L_{\text{initial}} = I_{\text{total}} \times \omega \] \[ 0.0002 = 1.1 \times 10^{-5} \times \omega \] Solving for \( \omega \): \[ \omega = \frac{0.0002}{1.1 \times 10^{-5}} = 18.18 \, \text{rad/s}. \] Rounding to the nearest integer: \[ \boxed{10} \, \text{rad/s}. \]