Question

A 8.20 g bullet is fired into a 0.900 kg block attached to the end of a 0.402 m nonuniform rod of mass 0.899 kg. The block-rod-bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertia of the rod alone about A is 0.0740 kg·m2. Treat the block as a particle. (a) What then is the rotational inertia of the block-rod-bullet system about point A? (b) If the angular speed of the system about A just after impact is 3.19 rad/s, what is the bullet's speed just before impact?

Answer 1

Given

Bullet mass , m = 0.0082 kg

Block mass , M = 0.9 kg

Rod mass , M' = 0. 899 kg

Length of rod, L = 0.402 m

Rotational inertia, I_r = 0.074 kgm^2

a)

The  rotational inertia of the block-rod-bullet system about point A

     I = I_r + (I_{bullet +block})

I = 0.074 + (m + M)*L²

I = 0.074 + (0.0082 + 0.9) *(0.402)²

I = 0.220 kg*m²

b) Final angular speed is w = 3.19 rad/s

     By law of conservation of angular momentum

      Li = Lf

    m * v_b* L = I *w

           v_b = (0.220) * (3.19) / (0.0082 * 0.402)

v_b = 213.64 m/s is the speed of the bullet.