Question

A circular road of radius $r$ is banked for a speed $v=40$ km/hr. A car of mass $m$ attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible. Find the correct option(s).

• A. The car cannot make a turn without skidding
• B. If the car turns at a speed less than $40$ km/hr, it will slip down
• C. If the car turns at a correct speed of $40$ km/hr, the force by the road on the car is equal to $\frac{m{v}^2}{r}$
• D. If the car turns at a correct speed of $40$ km/hr, the force by the road on the car is greater than mg as well as greater than $\frac{m{v}^2}{r}$

Answer 1

Answer: (B), (D)

The correct options are
B If the car turns at a speed less than $40$ km/hr, it will slip down
D If the car turns at a correct speed of $40$ km/hr, the force by the road on the car is greater than mg as well as greater than $\frac{m{v}^2}{r}$

Applying Newton's laws in horizontal and vertical directions. we get

$N\sin \theta =\frac{m{v}^2}{r}$ ... $\left(I\right)$ and $N\cos \theta =mg$ ... $\left(II\right)$

from these two equations, we get

$\tan \theta =\frac{v^2}{rg}$

$v=\sqrt{rg\tan \theta }$

This is the speed at which the car doesn't slide down even if there is no friction. Since the car is in horizontal and vertical equilibrium. So the option 'A' is wrong.

If the car's speed is less then the banking speed then It will slip down to reduce the $r$. So the option 'B' is right.

If the car turns at correct speed of $40$ m/s, then the force by the road on the car is given by

$N=\frac{m{v}^2}{r\sin \theta }=\frac{mg}{\cos \theta }$ ....$\left(III\right)$ [from $\left(I\right)$ and $\left(II\right)$]

By looking at equation $\left(III\right)$. we can say that, the force by the road on the car $N$ is equal to $\frac{m{v}^2}{r\sin \theta }$ and not $\frac{m{v}^2}{r}$, So the option 'C' is also wrong.

Since $\theta <\frac{\pi }{2};\Rightarrow \sin \theta <1\&\cos \theta <1\Rightarrow \frac{1}{\sin \theta }>1\&\frac{1}{\cos \theta }>1\Rightarrow \frac{m{v}^2}{r\sin \theta }>\frac{m{v}^2}{r}\&\frac{mg}{\cos \theta }>mg\Rightarrow N>\frac{m{v}^2}{r}\&N>mg$

By looking at above equations, we can deduce that option 'D' is true.