The 1kg ball shown rotates around a circular path of diamete

The 1-kg ball shown rotates around a circular path of diameter d = 1 m at constant rpm. In the state shown, the length L = 1 m. What would the angular velocity be when L is varied such that the diameter of the circle is reduced by a factor of 2?

Solution

Angular momentum J = Iw

MOI = mr^2 = 1*0.5^2 = 0.25 kg-m^2

New Inertia I\' = m*(r/2)^2 = 1*(0.5/2)^2 = 0.0625 kg-m^2

By angular momentum conservation, Iw = I\'*w\'

0.25*w = 0.0625*w\'

w\' = 0.25/0.0625*w

w\' = 4w

Cos theta = r/L = (1/2) / 1 = 0.5

theta = 60 deg

Balancing forces in horizontal direction, T*cos theta = mrw^2

Balancing forces in vertical direction T*sin theta = mg

Dividing both, tan theta = g/(rw^2)

w = sqrt(g / (r*tan theta))

= sqrt (9.81 / (0.5*tan60))

= 3.366 rad/s

So, w\' = 4*3.366 = 13.46 rad/s