Ballistic Pendulum A uniform thin rod of length 1.7 m and mass 2.2 kg can rotate

Question

Ballistic Pendulum A uniform thin rod of length 1.7 m and mass 2.2 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 2.2 - g bullet travelling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet's velocity makes an angle of theta = 35.0 degree with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 15.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before the impact?

Explanation / Answer

Let

L = 1.7 m

M = 2.2 kg

m = 2.2 g = 0.0022 kg

theta = 35 degrees

w = 15 rad/s

Let v is the initial speed of bullet.

Apply conservation of angular momentum

Initial angular momentum of bullet = final angular momentum of rod and bullet

m*v*r*sin(theta) = I*w

m*v*(L/2)sin(35) = (M*L^2/12 + m*(L/2)^2)*w

0.0022*v*(1.7/2)*sin(35) = (2.2*1.7^2/12 + 0.0022*(1.7/2)^2)*15

v*0.00107 = 7.97

v = 7.97/0.00107

= 7449 m/s <<<<<<--------Answer

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