JEE Main 2022 — Circular Motion Question with Solution

For a car to move safely around a curved road without skidding, the centripetal force needed is provided by the frictional force between the tires and the road. The relationship governing this scenario can be expressed as:

$$f_{\text{friction}}=f_{\text{centripetal}}$$

The frictional force is given by:

$$f_{\text{friction}}=\mu \cdot m\cdot g$$

And the centripetal force is given by:

$$f_{\text{centripetal}}=\frac{m\cdot v^2}{r}$$

Where:

$\mu$ is the coefficient of friction,

$m$ is the mass of the car,

$g$ is the acceleration due to gravity (approximately $9.8{\text{m/s}}^2$),

$v$ is the speed of the car,

$r$ is the radius of the curve.

Since the frictional force equals the centripetal force, we have:

$$\mu \cdot m\cdot g=\frac{m\cdot v^2}{r}$$

Cancelling $m$ from both sides (assuming $m\ne 0$) gives:

$$\mu \cdot g=\frac{v^2}{r}$$

Rewriting for $v$ gives the speed:

$$v=\sqrt{\mu \cdot g\cdot r}$$

For the first scenario with $r=75\text{m}$ and $v=30\text{m/s}$:

$$v_1^2=\mu \cdot g\cdot r_1$$

Thus,

$$\mu \cdot g=\frac{v_1^2}{r_1}=\frac{30^2}{75}=\frac{900}{75}=12$$

For the second scenario with $r_2=48\text{m}$, the new speed $v_2$ is:

$$v_2=\sqrt{\mu \cdot g\cdot r_2}=\sqrt{12\times 48}$$

Calculating inside the square root:

$$12\times 48=576$$

So,

$$v_2=\sqrt{576}=24$$

Hence, the maximum allowed speed for a curve with a radius of $48\text{m}$ is 24 m/s.