Question

A cat is on a merry-go-round that completes 1 full rotation in 6 seconds. The cat sits at a radius of 8.4 metres from the centre.
Find the minimum coefficient of friction to prevent the cat from sliding off.

Answer 1

To solve this problem we will use the concepts related to the uniform circular movement from where we will obtain the speed of the object. From there we will go to the equilibrium equations so that the friction force must be equal to the centripetal force. We will clear the value of the coefficient of friction sought.

The velocity from the uniform circular motion can be described as

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

Here,

r = Radius

T = Period

Replacing,

$v = \frac{2\pi (8.4)}{6}$

$v =8.7964 m/s$

From equilibrium to stay in the circle the friction force must be equivalent to the centripetal force, therefore

$F_f = F_c$

$\mu N = \frac{mv^2}{r}$

Here,

$\mu =$ Coefficient of friction

N = Normal Force

m = mass

v = Velocity

r = Radius

The value of the Normal force is equal to the Weight, then

$\mu(mg) = \frac{mv^2}{r}$

Rearranging to find the coefficient of friction

$\mu = \frac{v^2}{gr}$

Replacing,

$\mu = \frac{(8.7964)^2}{(9.8)(8.4)}$

$\mu =0.9399$

Therefore the minimum coefficient of friction to prevent the cat from sliding off is  0.9399