Question

A cat dozes on a stationary merry-go-round in an amusement park, at a radius of${}^{\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{m}}$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${}^{\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}}$.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 (or the cat clinging with its claws)?

Short answer

The least coefficient of static friction between the cat and the merry-go-round is${}^{0.60}$.

Step-by-step solution

Given

$$\begin{gathered} R=5.4\mathrm{m} \\ \mathrm{T}=6.0\mathrm{s} \end{gathered}$$

Determining the concept

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity.

Formula:

The velocity in uniform circular motion is given by,

$v=\frac{2\pi R}{T}$

where, v is the velocity, R is the radius andTis the time period

Determining the least coefficient of static friction between the cat and the merry-go-round

As the cat moving in the circular motion, it has centripetal accelerati ${}^{a_c=\frac{v^2}{R}}$, and frictional force is also directed to the center of the circle.

By using the Newton’s 2nd law along the horizontal direction,

$-f_s=-m\left(\frac{v^2}{R}\right)$

${\mu }_{s}mg=m\left(\frac{v^2}{R}\right)$

$$\begin{gathered} {\mu }_s=\frac{v^2}{gR} \\ =\frac{{\left({\displaystyle \frac{2\pi R}{T}}\right)}^2}{gR} \\ =\frac{4{\pi }^{2}R}{g{T}^2} \\ =\frac{4\times {\pi }^2\times 5.4}{9.81\times 6^2} \\ =0.60 \end{gathered}$$

Hence, the least coefficient of static friction between the cat and the merry-go-round is 0.60.