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

Explanation:

Given

Time Period $T=5.3 s$

mass of child $m=48.5 kg$

Radius $r=2.7 m$

Velocity $v=\frac{2\pi \cdot r}{T}$

$v=\frac{2\pi \cdot 2.7}{5.3}$

$v=3.20 m/s$

Now Centripetal Force will be Balanced by Frictional Force

Centripetal Force $F_c=m\cdot \frac{v^2}{r}$

$F_c=48.5\cdot \frac{3.2^2}{2.7}$

$F_c=183.94 N$

therefore Friction Force is 183.94 N