Question

A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding

Answer 1

0.6

Step-by-step explanation:

Given that :

Radius, R = 7m

Period, T = 6.9s

The Coefficient of static friction, μs can be obtained using the relation :

μs = v² / 2gR

Recall, v = 2πR/T

μs becomes ;

μs = (2πR/T)² / 2gR

μs = (4π²R² / T²) ÷ 2gR

μs = (4π²R² / T²) * 1/ 2gR

μs = 4π²R / T²g

μs = 4π²*7 / 6.9^2 * 9.8

μs = 28π² / 466.578

μs = 276.34892 / 466.578

μs = 0.5922887

μs = 0.6