A bullet of mass 2 m moving with velocity v strikes tangentially the edge of a s

Question

A bullet of mass 2 m moving with velocity v strikes tangentially the edge of a spoked wheel of radius R, and the bullet sticks to the edge of the wheel. The spoked wheel has a mass 6 m which is concentrated on its rim (neglect the mass of the spokes). The wheel, initially at rest, begins to rotate about its center, which remains fixed on a frictionless axle. What is the angular velocity of the spoked wheel after the collision? (0.142857 )v/R (0.125 )v/R (0.333333 )v/R (0.25 )v/R (0.75 )v/R (0.5 )v/R (1.33333 )v/R (3 )v/R (4 )v/R (0.2 )v/R

Explanation / Answer

This problem is Conservation of Angular Momentum.

Initially, the angular momentum of the bullet is    I = I ( v/R ) = (2m) R2 ( v/R ) = 2 m v R.

After contact with the wheel, the final angular momentum of both the bullet and the wheel is:

                       I wheel + I bullet = (6m)R2 + (2m)R2 = 8 m R2 .

Equating the above two values of initial and final angular momenta, we get :

     2 m v R = 8 m R2 .    The factors of m cancel out. Solve for final angular velocity to obtain:

= 2 v R / [ 8 R2 ]   =   v / (4R)   = 0.25 ( v / R ).   final answer is CHOICE 4.

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