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 gure. 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 gure. 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?
Physics Angular Momentum - Damon, Monday, July 11, 2011 at 7:13pm
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
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
Physics Angular Momentum - jhay, Monday, July 11, 2011 at 10:32pm
Thanks a bunch