Question

A car goes on a horizontal circular road of radius R. The speed increases at a constant rate $a=\dfrac{dv}{dt}$. The friction coefficient between the tire and road is $\mu$. What is the speed at which the car will skid?

& \text{A}\text{. }\left( \mu g-a \right){{R}^{2}} \\\ & \text{B}\text{. }\\!\\![\\!\\!\text{ }\left( {{\mu }^{2}}{{g}^{2}}-{{a}^{2}} \right){{R}^{2}}{{\text{ }\\!\\!]\\!\\!\text{ }}^{\dfrac{1}{4}}} \\\ & \text{C}\text{. }\\!\\![\\!\\!\text{ }\left( {{\mu }^{2}}{{g}^{2}}-{{a}^{2}} \right)R{{\text{ }\\!\\!]\\!\\!\text{ }}^{\dfrac{1}{4}}} \\\ & \text{D}\text{.}\left( \mu g-a \right){{R}^{4}}\text{ } \\\ \end{aligned}$$

Answer 1

First we need to draw the free body diagram according to the question. Write the formula for resultant acceleration in a uniform circular motion. We know how acceleration is shown in calculus. Then take the free body diagram and form an equation of force from it and apply the formula for resultant acceleration in a uniform circular motion. Then substitute the values required to obtain the required result.

Complete step by step answer:
The diagram on the left is of a moving car with a force. And the diagram on the right represents the free body diagram.
We know that for uniform circular motion, the resultant acceleration is
${{a}_{R}}=\sqrt{{{a}_{n}}^{2}+{{a}_{t}}^{2}}$ ,
Now, we know that,
${{a}_{n}}=\dfrac{{{V}^{2}}}{R},{{a}_{t}}=\dfrac{dv}{dt}=a$ , here a is the acceleration.

Now, from free body diagram of car on a rough road we can say that,
As,
$f=m{{a}_{r}}$ ,
$\mu N=m\sqrt{\left( \dfrac{{{V}^{2}}}{R} \right)+{{\left( \dfrac{dv}{dt} \right)}^{2}}}$ ,
${{\mu }^{2}}{{N}^{2}}={{m}^{2}}\left( \dfrac{{{V}^{2}}}{{{R}^{2}}}+{{a}^{2}} \right)$,
(on squaring both sides, and we know that ${{a}_{n}}=\dfrac{{{V}^{2}}}{R},{{a}_{t}}=\dfrac{dv}{dt}=a$)
${{V}^{2}}=\dfrac{{{\mu }^{2}}{{N}^{2}}{{R}^{2}}}{{{m}^{2}}}-{{a}^{2}}{{R}^{2}}$
${{V}^{2}}=\dfrac{{{\mu }^{2}}{{N}^{2}}{{g}^{2}}{{R}^{2}}}{{{m}^{2}}}-{{a}^{2}}{{R}^{2}}$
$V={{[\left( {{\mu }^{2}}{{g}^{2}}-{{a}^{2}} \right){{R}^{2}}]}^{\dfrac{1}{4}}}$

Hence, the correct answer is option B.

Additional Information
When two bodies slide over each other then there is a force acting between them that force opposes the body to move this force is known as friction.
There are four types of friction:
Static friction, rolling friction, fluid friction, sliding friction.
The movement of a body along the circumference of a circle or rotation along a circular path this movement is known as circular motion. We can say that this circular motion can be uniform, with constant speed and constant angular rate of rotation.

Note:
While doing substitution students must be careful to avoid silly mistakes. In uniform acceleration, the body’s displacement is zero while distance is increased. The force acting on the car is a frictional force where $\mu$ is the friction coefficient between the car and the road.