Question

A toy car revolves in a circular path of radius 1m
on a horizontal rough plane. It's speed varies as
V = 2t2. It starts sliding at t= 1s. Find the value of
coefficient of friction between ground and the wheels
of the car. Use g = 10 m/s2.​

Answer 1

Answer:

The coefficient of friction between the ground and the wheels=0.4

Explanation:

Given:

The radius of the circular path=1 m

Variation of speed with time, $v=2t^2$

The centripetal force is being provided by the frictional force acting on the toy car at any time.

The friction force acting on the toy car$=\mu mg$

where

• $\mu$ is the coefficient of friction between the ground and the tires.

The velocity at t=1 , v=2$\times 1^2$=2

The centripetal force acting =$\dfrac{mv^2}{R}$

$\mu mg=\dfrac{mv^2}{R}\\\mu =\dfrac{4}{10}=0.4$