A thin uniform rod of length L=0.582m and mass M=0.603kg that can rotate in a ho

Question

A thin uniform rod of length L=0.582m and mass M=0.603kg that can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a bullet of mass m=3.65g is fired horizontally into one end of the rod. AAs viewed from above, the direction of the bullets velocity makes an angle theta=60 with the rod. if the bullet lodges into the rod and the angular velocity of the rod immediately after the collision is w=7.89 rad/s what must the bullets speed have been in m/s just before impact

Explanation / Answer

The moment of inertia of the rod I = ML^2/12

where M is mass and L is length

impulsive force (int F dt) = m v sin theta

m is bullets mass

v is bullets speed

theta is the angle between bullets trajectory and rod axis

impulsive torque (int T dt ) = mv L/2 sin theta   = int ( I W dt)

then

W = m v L/2 sin theta /I = 6 mv sin theta /ML

v = MLW/ 6 m sin theta

v    = (0.603 * 0.582 * 7.89)/(6 * 0.00365 * sin 60)

v = 0.146 km/s ----------------<<<<<<<<<<<<<Answer

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