Question

A motorcycle is travelling on a curved track of radius $500m$. If the coefficient of friction between road and tyres is $0.5$, the speed avoiding skidding will be

• A. $50m/s$
• B. $75m/s$
• C. $25m/s$
• D. $35m/s$

Answer 1

The correct option is A

$50m/s$

Step 1: Given data.

The radius of the curved track is $500m$.

The coefficient of friction between the tyres and track is $0.5$.

Step 2: Formula used.

Assume that, the static friction force at the corner is $F_s$.

Since the static friction force balances the centripetal force which is acting in the opposite direction of the static friction force.

The formula for static friction force is $F_s={\mu }_{s}mg$ and the formula of centrifugal force is $\frac{m{v}^2}{r}$.

Where, $v$ is the velocity of the object, $m$ is the mass of the object, $r$ is the radius of the track, ${\mu }_s$ is the coefficient of static friction and $g$ is the gravitational acceleration.

So, the maximum speed at which the motorcycle can take turns is given by $\frac{m{v}^2}{r}=F_s$.

$$\begin{gathered} \frac{m{v}^2}{r}={\mu }_{s}mg \\ \Rightarrow \frac{v^2}{r}={\mu }_{s}g \\ \Rightarrow v^2={\mu }_{s}gr \\ \Rightarrow v=\sqrt{{\mu }_s\times g\times r} \end{gathered}$$

Therefore, The maximum velocity on a curved track can be given by: $v=\sqrt{{\mu }_s\times g\times r}$.

Step 3: Find the maximum velocity.

It is given that, $r=500m$ and $\mu =0.5$.

So, the maximum velocity $v$ is given by:

$$\begin{gathered} v=\sqrt{\left(0.5\right)\times 500\times 10} \\ \Rightarrow v=\sqrt{2500} \\ \Rightarrow v=50 \end{gathered}$$

Therefore, the speed avoiding skidding will be $50m/s$.

Hence, option (A) is the correct answer.