A uniform thin rod of length 0.70 m and mass 4.0 kg can rotate in a horizontal p

Question

A uniform thin rod of length 0.70 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 11.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

Explanation / Answer

The axis of rotation is in the middle of the rod, r = 0.35 m from either end

the initial angular momentum of the system (which is just that of the bullet, before impact) is = r*m*v*sinA

m = 0.003 kg and A = 60 deg Relative to the axis, this is counterclockwise and thus (by the common convention) positive

After the collision, the moment of inertia of the system is I = Irod + mr^2

where Irod = M*L^2/12

M = 4.0 kg and L = 0.7 m

Angular momentum conservation leads to

r*m*v*sinA = (ML^2/12 + mr^2)*w

v = [(ML^2/12 + mr^2)*w]/(r*m*sinA) = [(4*0.7^2/12 + 0.003*0.35^2)*11]/(0.003*0.35*1.732/2)

v = 1980.32 m/sec

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