Question

A car moves in a circular track of radius 100m. The coefficient of friction between the road and the track is 0.2. Find the maximum velocity the car can move with, without skidding.

Answer 1

The car is moving in a circular direction, making the force acting on the car circular and not linear thereby the force acting on the car is centripetal. Multiplying the frictional constant with the circular force, We use the formula as:
$\mu mg=\dfrac{m{{v}_{\max }}^{2}}{R}$
where $\mu$ is the frictional force, $m$ is the mass of the car, $v$ is the velocity of the car moving, $R$ is the radius of the circular track, with gravity $g=10\text{ }m{{s}^{-2}}$.

Complete step by step answer:
The force acting on the car is centripetal force as the car is moving in a circular motion on a circular track. The radius of the circular track is given as: $100\text{ }m$.
The frictional constant acting on the car is given as $\mu$, whose value is $0.2$.
Hence, the force with which the car moves along with frictional force is:
$\Rightarrow \mu mg=\dfrac{m{{v}_{\max }}^{2}}{R}$

Placing the values of $\mu =0.2, R=100\text{ }m$ , in the above formula, we get the value of the force is:
$\Rightarrow 0.2mg=\dfrac{m{{v}_{\max }}^{2}}{100}$

Replacing the value of $F$ as $F=mg$, with $g=10\text{ }m{{s}^{-2}}$ in the above formula:
$\Rightarrow 0.2\times m\times 10=\dfrac{m{{v}_{\max }}^{2}}{100}$
$\Rightarrow 0.2\times m\times 10\times 100=m{{v}_{\max }}^{2}$
$\Rightarrow {{v}_{\max }}^{2}=200$
$\Rightarrow {{v}_{\max }}=10\sqrt{2}$
Therefore, the maximum velocity due to equality of the LHS and RHS is given as ${{v}_{\max }}=10\sqrt{2}$.

Note: The force used in this question uses centripetal force and not centrifugal force as the centripetal force is acting along the circular path that is why the circular force is positive whereas the centrifugal force is negative as the force is acting inside the circular path. Now the velocity has to be maximum as given in the question therefore, the force has to be positive.