Question

The maximum speed of a car on a road-turn of radius 30m, if the coefficient of friction between the tyres and the road is 0.4, will be
$\begin{array}{l}
(a)\,10.84\,m{s^{ - 1}}\\
(b)\,9.84\,m{s^{ - 1}}\\
(c)\,8.84\,m{s^{ - 1}}\\
(d)\,6.84\,m{s^{ - 1}}
\end{array}$

Answer 1

Hint: In this question, we have to apply the concept of centripetal force, and the concept of friction and the special case of a car turning from a curved road. And using all these concepts, we have to derive the maximum speed that the car can have.

Complete step by step answer:
We know that when a car is turning from a curved road, the force due to the mass of car, tries to pull it sideways. To cancel this force, a centripetal force needs to act, which will stop the car from going sideways and the car can take the turn smoothly.
This centripetal force has to be produced by the frictional force acting on the tires due to the friction from the road.
For the maximum velocity, the centripetal force must be equal to the friction acting on the car.
The centripetal force on a body of mass m and turning with a radius r and moving with a velocity v is given by:
Centripetal Force - $\dfrac{{m{v^2}}}{r}$
And the frictional force is given by:
Fictional force - $\mu N$, where $\mu $is the coefficient of friction and N is the normal reaction force acting on the car.
The normal reaction force of a body of mass m is given by –
Normal Reaction force - $N = mg$.
After we put the value of N is the equation of frictional force, we get: -
Frictional force- $\mu mg$
And as we have established that the frictional force must be equal to centripetal force, we should equate both the equation,
$\dfrac{{m{v^2}}}{r} = \,\mu mg$,
After we cancel out m from both the sides we get,
$\dfrac{{{v^2}}}{r} = \mu g$,
We have to calculate the velocity of the car therefore, we should make it the subject,
$v = \sqrt {\mu rg} $
Now that we have acquired the relation, we should put the values in the question in the above equation.
$\begin{array}{l}
v = \sqrt {0.4 \times 30 \times 9.8} \\
v = \sqrt {117.6} \\
v = 10.84\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
\end{array}$
Therefore, the maximum speed of the car can be $10.84\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}$

So, the correct answer is “Option A”.

Note:
Students should take care of all the forces when equating two forces which cancel each other, the maximum value of the car should be equal to the centripetal forces. During the calculation, the value of g should be taken as \[9.8\;{{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
 } {\rm{s}}}^2}\] and not \[{{{{\rm{10}}\;{\rm{m}}} {\left/
 {\vphantom {{{\rm{10}}\;{\rm{m}}} {\rm{s}}}} \right.
 } {\rm{s}}}^2}\].