Question

A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 5.9 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

Answer:0.53

Step-by-step explanation:

Given

Radius of merry-go-round
`r=4.6 m`

Time Period
`T=5.9 s`

Velocity of merry-go-round
`=(distance\ moved)/(time)`

`v=(2\pi \cdot r)/(T)`

`v=(2\pi \cdot 4.6)/(5.9)`

`v=4.89 m/s`

Now Friction Force will nullify the centripetal Force

i.e.
`(mv^2)/(r)=f_r`

`(mv^2)/(r)=\mu mg` , where
`\mu`coefficient of static friction

`\mu g=(v^2)/(r)`

`\mu =(v^2)/(rg)`

`\mu =(4.89^2)/(4.6* 9.8)`

`\mu =0.53`