11. A uniform thin rod of length 0.65 m and mass 4.0 kg can rotate in a horizont

Question

11. A uniform thin rod of length 0.65 m and mass 4.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet traveling in 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° with the rod (see the figure). If the bullet lodges in the rod and the angular velocity of the rod is 12.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

Explanation / Answer

here,

length , l = 0.65 m

mass , m = 4 kg

mass of bullet , mb = 0.003 kg

angular velocity of rod , w = 12 rad/s

let the bullet velocity vbefore the impact be v

using conservation of angular momentum

mb*v*(l/2)*sin(60) = ( m * L^2 /12) * w

0.003 * v*( 0.65/2)*sin(60) = ( 4 * 0.65^2 /12) * 12

solving for v

v = 2001.48 m/s

the velocity of the bullet before the impact is 2001.48 m/s

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