A bullet of mass mb = 0.01 kg is moving with a speed of 100 m/s when it collides

Question

A bullet of mass mb = 0.01 kg is moving with a speed of 100 m/s when it collides with a rod of mass mr = 5 kg and length L = 1 m. The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass which is located exactly halfway along the rod. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating together. What is the magnitude of the angular velocity of the rotation (in rad/s)?

Explanation / Answer

For a point mass, angular momentum (L) = m * v * r. For anything more complex, you need the moment of inertia and L = I * rotational velocity. Angular momentum must be conserved before and after the collision. Before, all the momentum is in the bullet. L = m * v * r L = .01kg * 100m/s * .25m (bullet hits L/4 or .25 m from axis) L = .25 kg m^2/s After the collision same must be true, but of the whole system. L = (I(rod) I(bullet)) * w I for a rod about the center is 1/12 m L^2 I for point mass is m r^2 L = (1/12 * 5kg * (1m)^2 .01kg * (.25m)^2) * w .599 rad/s = w

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