A uniform thin rod of length 0.75 m and mass 3.5 kg can rotate n a horizontal pl

Question

A uniform thin rod of length 0.75 m and mass 3.5 kg can rotate n a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet traveling n the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet velocity makes an angle of 60 degree with the rod (see the figure). If the bullet lodges in the rod and the angular velocity of the rod is 9.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

Explanation / Answer

I = Irod + m r2

Irod =  M L2 + 12

Angular momentum conservation leads to

r m v sin phi = [1/12 M L2 + m r2 ] w

v = [1/12 * 3.5 * 0.752 + 0.003 * 0.3752 ] (9) / [0.375 * 0.003 * sin 60]

v = 1.52* 103 m/s

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