A uniform thin rod of length 0.60 m and mass 5.0 kg can rotate in a horizontal p

Question

A uniform thin rod of length 0.60 m and mass 5.0 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 degree 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

using law of conservation of angular momentum

angular momentum before collision = angular momentum after collision

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

Given that m = 3*10^-3 Kg

v = ?

r = 0.6/2 = 0.3 m

theta = 60 deg

I is the moment of inertia of system after collision
I = (1/12)*M*L^2 + m*(L/2)^2 = ((1/12)*5*0.6^2) + (3*10^-3*(0.6/2)^2)

I = 0.15 kg-m^2

w = 11 rad/sec

then

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

(3*10^-3*v*0.3*sin(60)) = 0.15*11

v = 2117 m/sec is the speed of the bullet

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