Question

A 50.0 kg child stands at the rim of a merry-go-round of radius 1.95 m, rotating with an angular speed of 2.80 rad/s.

Answer 1

Answer:

a) ac= 15.3 m/s² b) 765 N c) 1.56

Explanation:

a) For the child standing in the rim of the merry-go-round, there exists a force, called centripetal force, that keeps her moving in a circular path, and not going in a straight line at constant speed, as Newton's first law dictates.

The acceleration cosequence of this force is called centripetal acceleration, and it can be showed be equal to the following expression:

ac = ω²*r, where ω is the angular speed, and r is the radius.

Replacing by the givens:

ac= (2.8)² (rad/sec)² * 1.95 m = 15.3  m/s²

b) As we were talking about the centripetal force, we realize that this can't be some type of  an unknown force, so, it must  be one than can be identifiable.

Looking at the external forces on the child, the only one that is applied in horizontal plane, is the friction force.

So, this friction force, is just the centripetal force we were talking about.

The mimimum force between her feet and the floor of the carousel, is just the centripetal force:

Fc = m*ac = 50.0 kg * 15.3 m/s² = 765 N

c) The friction force can take any value, in order to be equal to the centripetal force,  up to a limit value, beyond which , the child will move. This limit value can be expressed as follows:

Fs max = μs*N  = μs*m*g

⇒ m*ac = μs*m*g

Simplifying  common terms, we can solve for the coefficient of static friction ,  μs, as follows:

μs = 15. 3 m/s² / 9.8 m/s² = 1.56