Question

In the diagram below, a 75 kg college student is standing at the perimeter of a merry-go-round initially at rest) which can be considered a uniform solid disk with a mass of 130 kg and a radius of 2.1 m. The frictionless vertical rotation axis is indicated by the dot. A friend throws a 0.42 kg football to the person on the merry-go- round, who catches it when the part of its velocity parallel to the ground is 12 m/s at the angle = 37°, measured as shown. What is the angular speed of the merry-go-round right after the student catches the ball? (Hint: use the general cross-product version of the angular momentum of the ball just before it is caught. In the cross-product equation, the angle o is the angle between the vector f and the vector p.) Ball Person

Answer 1

let
m = 0.42 kg, v = 12 m/s

M1 = 75 kg
M2 = 130 kg
R = 2.1 m

initial angular momentum of the fottball with respect to the center of the merry-go-round,

Li = m*v*R*sin(90- theta)

= 0.42*12*2.1*sin(90-37)

= 8.45 kg.m^2/s

let wf is the angular speed of merry-go-round.
final moment of inertia of the system, If = 0.5*M2*R^2 + (M1 + m)*R^2

= 0.5*130*2.1^2 + (75 + 0.42)*2.1^2

= 619.2 kg.m^2

now apply conservation of angular momentum

Lf = Li

If*wf = Li

wf = Li/If

= 8.45/619.2

= 1.36*10^-2 rad/s <<<<<<<<----------------Answer