Question

The coefficient of friction between a car's wheels and a roadway is 0.5. The least distance in which the car can accelerate from rest to a speed of 72 kmph is (g = 10 ms$^{-2}$)
(A) 10 m
(B) 20 m
(C) 30 m
(D) 40 m

Answer 1

Hint: In order for the wheel to gain some velocity, it has to accelerate against friction. The friction acts in the direction opposite to the bottom of the wheel. The magnitude of frictional deceleration needs to be overcome by the wheel in order to gain the required velocity.
Formula used:
The magnitude of rolling friction is determined as:
$F = \dfrac{\mu mg}{r} $
The torque acting on the wheel is given by:
$F = I \alpha = F r$

Complete step-by-step solution:
If the wheel has a mass m and radius r then, the force of friction that the wheel possesses is given by the relation:
$F = \dfrac{\mu mg}{r} $ .
The wheel overcomes this deceleration and we can say that on the wheel, this force creates a torque. The angular acceleration is written as:
$\alpha = \dfrac{Fr}{I}$ .
The moment of inertia of the wheel is written as:
$I = mr^2$ .
(Assuming that wheel is a ring).
We substitute F and I to get the angular acceleration as:
$\alpha = \dfrac{\mu g}{r^2}$
The angular acceleration of the any rotating body is related to the final angular velocity as given as:
$\omega^2 = 2 \alpha s$
The angular velocity is related to the velocity as:
$v = r \omega $
We know what final velocity we need to attain. It is given as 72 kmph or
$ v = \dfrac{72 \times 1000}{3600} = 20 $ m/s.
Keeping all these values in the equation of rotational motion (just like the equation of motion).
$s = \dfrac{v^2}{2 \alpha r^2} = \dfrac{(20)^2}{2 (0.5)(10)} = 40 $ m.
Therefore, the correct answer is option (D).

Note: We have used the direct result here for rolling. The linear velocity of the wheel is related to the angular velocity. Since the wheel is rolling, we cannot consider the coefficient of kinetic friction, we have to consider the coefficient of rolling friction only.