Question

A merry-go-round makes one complete revolution in 5.3 s. A 48.5 kg child sits on the horizontal floor of the merry-go-round 2.7 m from the center. Find the horizontal force of friction that acts on the child. Answer in units of N.

Answer 1

Answer:183.94 N

Step-by-step explanation:

Given

Time Period
`T=5.3 s`

mass of child
`m=48.5 kg`

Radius
`r=2.7 m`

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

`v=(2\pi \cdot 2.7)/(5.3)`

`v=3.20 m/s`

Now Centripetal Force will be Balanced by Frictional Force

Centripetal Force
`F_c=m\cdot (v^2)/(r)`

`F_c=48.5\cdot (3.2^2)/(2.7)`

`F_c=183.94 N`

therefore Friction Force is 183.94 N