A small 0160 kg sphere is suspended by a light string from a

A small 0.160 kg sphere is suspended by a light string from a vertical post mounted at the edge of a merry-go-round of radius R = 0.440 m. The equilibrium position of the sphere is shown in the figure, with a = 0.430 m and b = 0.730 m. The axis of rotation is indicated by the dashed line. Calculate the angular velocity of the merry-go-round.

Solution

tan theta = a / b

r = R + a , F_y = m g

At equilibrium, F_x = F_y tan theta = m g a / b = m w v =m r w²

w² = g * a / b / (R + a)

= 9.80 * (0.43 / 0.73) / (0.440 + 0.430)

w = 2.58 rad/s