Circular Motion Physics

Question

A merry-go-round makes one complete revolution in 8.3 s. A 39.4 kg child sits on the horizontal floor of the merry-go-round 4 m from the center.

Find the child’s acceleration. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s 2

Find the horizontal force of friction that acts on the child. Answer in units of N.

What minimum coefficient of static friction is necessary to keep the child from slipping?

Answer 1

1 revolution is 2π radians so the angular velocity (ω) is 2π/8.3 = 0.75701 rad/sec

The radial acceleration is given by:

a = ω²r = (0.75701)²(4) = 2.3 m/s²

Since F = ma, then F = (39.4)(2.3) = 90.3 N

Since the child is not moving, then friction must be equal to that force, so the force of friction is 90. N.

The Normal Force is equal to the girl's weight so (39.4)(9.81) = 386.514 N. Since F_F = μN then μ = F_F/N = 90/386.514 = 0.23 so the static coefficient is at least 0.23.