A puck (mass m1 = 2.20 kg) slides on a frictionless table as shown in the figure below. The puck is tied to a string that runs through a hole in the table and is attached to a mass m2 = 6.6 kg. The mass m2 is initially at a height of h = 6.6 m above the floor with the puck traveling in a circle of radius r = 1.76 m with a speed of 6.6 m/s. The force of gravity then causes mass m2 to move downward a distance 0.66 m.

Question

Damon · Answer

angular momentum is conserved since there is no torque in this problem

I omega = constant
m R^2 omega = constant
m R v = constant
m is constant
so
R v = constant

original R = 1.76
final R = 1.76 - .66 = 1.1

so what ever your question might be (you did not ask)

1.76 (6.6) = 1.1 (v)
where v is the final tangential velocity