Question

A bullet with a mass m = 4.83 g and initial speed of 330 m/s passes through a wheel which is initially at rest as in the figure. The wheel is a solid disk with mass M = 2.29 kg and radius R = 18.6 cm The wheel rotates freely about an axis through its center and out of the plane shown in the figure. The bullet passes through the wheel at a perpendicular distance 14.8 cm from the center. After passing through the wheel it has a speed of 201 m/s.

1) What is the angular speed (rad/s) of the wheel just after the bullet leaves it?

2) How much kinetic energy (J) was lost in the collision?

Answer 1

Initial angular momentum of bullet around axis of wheel = I w = m r^2 w = m r v = .00483 * .148 * 330 = .2359

Final angular momentum of bullet around axis of wheel = .00483 * .148 * 201 = .1437

change in angular momentum of bullet = .2359 - .1437 = .0922

that is the angular momentum of the wheel
I w = .0922
I = (1/2) M R^2 =.5*2.29*.186^2 = .03961
so
w = .0922/.03961 = 2.328 rad/s

Ke original = (1/2) m v^2 of bullet

Ke final =(1/2)mvfinal^2 of bullet + (1/2)I w^2 of wheel

Answer 2

Awesome!

Thanks a bunch