A uniform rod of length L rests vertically on a frictionless horizontal surface.

Question

A uniform rod of length L rests vertically on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. You're answers may depend on L and v What is the final angular speed of the rod after the collision? What is the final angular speed of the rod once it has fallen to the ground?

Explanation / Answer

We will conserve angular momentum

initial angular momentum=final angular momentum

(m/4)*v*(L/2)=I*w

I=(1/3)*m*l^2+(m/4)*(l/2)^2=ml^2/3+ml^2/16=19ml^2/48

so we got

w=6v/19l=6v/19L

(b) conserving energy

0.5*(m/4)*v^2=0.5*I*w^2

I=19ml^2/48

we got

w=0.8v/L

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