Physics Help Please?

Question

1) a string under a tension of 74 N is used to whirl a rock in a horizontal circle of radius 3.4 m at a speed of 26.79 m/s. The string is pulled in and the speed of the rock increases. When the string is 0.627 m long and the speed of the rock is 55.4 m/s, the string breaks.
What is the breaking strength of the string?
Answer in units of N

2) A merry-go-round makes one complete revolution in 16.4 s. A 35.7 kg child sits on the horizontal floor of the merry-go-round 5.52 m from the center
The acceleration of gravity is 9.8 m/s/s
What minimum coefficient of static friction is necessary to keep the child from slipping?

Answer 1

#1) First off, you need to find the mass of the rock.
You know the tension in the string is equal to the centripetal force, so you have

Fc = (mv^2) / r
74 = (m(26.79)^2) / 3.4
m = 0.35 kg

Use this mass to calculate the centripetal force of the second information given.

Fc = (mv^2) / r
Fc = (0.35(55.4)^2) / 0.627
Fc = 1713 N

Therefore the breaking force of the string is 1713 N

#2) You are given the period, or T, 16.4 s, and the mass, 35.7 kg, and the radius, 5.52 m.

You want the net force to equal 0, so that the child does not move.

Fnet = Fc - Ff
0 = 4pi^2mr / T^2 - muFn
mu = 28.925 / Fn
Fn is equal to -Fg in this case, so Fn = mg
mu = 28.925 / (35.7 * 9.8)
mu = 0.083

Answer 2

In both of these, the force F required to keep a mass m moving in a circle of radius r at velocity v or angular velocity ω is

F = m v² / r = m r ω²

First problem: you know F ( 74 N ), r ( 3.4 m ), and v ( 26.79 m/s). Solve for the mass. Then change r to .627 m and v to 55.4 m/s and solve for the new force, which is the breaking strength of the string.

Second problem: You know the angular velocity ( 2π / 16.4 radians / sec ), the mass ( 35.7 kg ), and the radius ( 5.52 m ). Solve for the force F. That force must be provided by friction, which is equal to μmg. Solve for the coefficient of friction μ.