A uniform thin rod of length 0.35 m and mass 6.5 kg can rotate in a horizontal p

Question

A uniform thin rod of length 0.35 m and mass 6.5 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 9.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

Explanation / Answer

You must find the initial (angular) momentum of the system; that is, of the bullet about the axis. You can either find the component of the velocity that is perpendicular to the rod, or find the distance from the axis of the velocity vector of the bullet. I'll try the latter.
The bullet embeds 0.35 m from the axis and makes an angle of 60º with the rod; then the "radius" of the bullet's trajectory w/r/t the axis is r = 0.35m * sin60º = 0.30 m
Then Li = I = (mr²)(v/r) = mvr = 0.003kg * v * 0.3m = 9.1e-4kg·m * v
Lf = ((ML²/12) + mr²) = (6.5kg*(0.7m)²/12 + 0.003kg*(0.35m)²) * 9rad/s = 2.25 kg·m²/s

Then v = 2.25kg·m²/s / 9.1e-4kg·m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.