Question

An automobile moves on a road with a speed of 54 km h $^{-1}$. The radius of its wheel is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg$m^2$. If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to the wheels is:
A. 2.86 kg $m^2 s^{-2}$
B. 6.66 kg $m^2 s^{-2}$
C. 8.58 kg $m^2 s^{-2}$
D. 10.86 kg $m^2 s^{-2}$

Answer 1

All the equations of motions for rotational motion can be written in analogy with the linear equations. The first equation of motion can help us find the rotational acceleration on the body which can be used in finding the torque.

Formula used:
The first equation of motion for rotational motion is:
$\omega_f = \omega_i + \alpha t$ ;
where we have substituted angular velocities in place of linear velocities and angular acceleration in place of linear acceleration.
Torque in rotational motion can be written like force in linear motion or:
$\tau = I \alpha$.

Complete answer:
We first try to find out the angular deceleration due to the torque that results in the automobile coming to rest.
We write the angular velocity as:
$\omega = \dfrac{v}{r}$
For the initial case, we are given that the linear velocity is 54 km/hr before the application of breaks. Therefore,
$\omega_i = \dfrac{54 \times 10^3 m}{3600 s \times 0.45 m}$
$\omega_i = 33.33$rad/s.
The final velocity is obviously zero, so we write the first law of motion (for rotational motion) as:
$0 = 33.33 rad/s - \alpha \times (15s)$.
Negative sign here is for retardation; so we get,
$\alpha = 2.22 rad/s^2$
This is the angular retardation that the tire has to undergo constantly in order to stop.
Now, as the moment of inertia is given to be 3 kg m$^2$, we can write:
$\tau = I \alpha$
$\tau = 3 kg m^2 \times 2.22 rad/s^2 = 6.66 kg m^2 s^{-2}$.

So, the correct answer is “Option B”.

Additional Information:
Moment of inertia plays the same role in rotational motion as mass plays in linear motion. Moment of inertia of any body is found by taking the product of its mass and the square of distance from the centre (or distance from the axis). More the moment of inertia of a body more will be the required torque to produce same angular acceleration.

Note:
Mind the difference between units of linear velocity and angular velocity as linear velocity is m/s and angular velocity is rad/s. The difference between the two units can help anyone at any point if they begin with the wrong formula. The relation between the two velocities can also be cleared up by checking these units. Also negative sign has to be included compulsorily in the laws of motion for the case of retardation.