Question

A child places a picnic basket on the outer rim of a merry-go-round that has a radius of 4.6 mand revolves once every 30s. (a) What is the speed of a point on that rim? (b)What is the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride?

Short answer

a) Speed of a point on rim is 0.96m/s

b)${\mu }_s=0.021$

Step-by-step solution

Given

Merry-go-round that has a radius of 4.6 m and revolves once every 30 s

Understanding the concept

The problem is based on the centrifugal force. Centrifugal force is the apparent outward force on a mass when it is rotated. It also deals with the static friction.

Formula:

Centrifugal force is given by

$\mathbf{F}\mathbf{=}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{R}}$

and the maximum value of static friction is given by

${\mathbf{f}}_{\mathbf{s}\mathbf{,}\mathbf{max}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{s}}\mathbf{mg}$.

Calculate the speed of a point on that rim

(a)

The distance traveled in one revolution is

$$\begin{gathered} 2\mathrm{\pi R}=2\mathrm{\pi }\left(4.6\mathrm{m}\right) \\ =29\mathrm{m} \end{gathered}$$.

The speed is

$$\begin{gathered} \mathrm{v}=\left(29\mathrm{m}\right)/\left(30\mathrm{s}\right) \\ =0.96\mathrm{m}/\mathrm{s} \end{gathered}$$

Calculate the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride 

(b)

According to Newton’s second law

$$\begin{gathered} f_s=m\left(v^2/R\right) \\ =m\left(0.20\right) \end{gathered}$$

As N = mg in this situation, the maximum possible static friction is,

$f_{s,max}={\mu }_{s}mg$.

Equating this withlocalid="1660973301753" $$\begin{gathered} f_s=m\left(0.20\right) \\ {\mu }_s=0.20/9.8 \\ =0.021 \end{gathered}$$