A 100 kg solid steel ball with a radius of 5 m is being spun on ice with an angular velocity of 5 rev/s pointed into the ice (clockwise rotation when viewed from above).A student shoots a 10 kg marble at the steel ball. The marble hits the ball along its side as shown with an initial velocity of 5 m/s to the left. If, right after the collision, the final velocity of the marble is 2.5 m/s to the left, what is the angular velocity of the steel ball after the collision?
You need to do conservation of angular momentum about the steel ball's axis. Angular momentum = I.ω The steel ball's initial ω = 5 rev/s = 5*2pi rad/s = 31.4 rad/s The steel ball's moment of inertia I = (2/5)*m*r^2 = 40*25 kg.m^2 = 1000 kg.m^2 The marble's initial ω = v / r where r is the perpendicular distance from the steel ball's axis You need to look at the diagram for this. The marble's I about the steel ball's axis is I = m*r^2 where r is the same as above. (marble is treated as a point mass). Then work out the total initial ang. momentum = final ang. momentum and solve for ω...
