Question

A car is moving at a speed of 72 km/hr. The diameter of its wheels is 50 centimeters. If its wheels come to rest after 20 rotations as a result of the application of the brakes, then the angular retardation produced in the car will be
$
  A. - 25.5{\text{ rad/}}{{\text{s}}^2} \\
  B.0.25{\text{ rad/}}{{\text{s}}^2} \\
  C.2.55{\text{ rad/}}{{\text{s}}^2} \\
  D.0 \\
 $

Answer 1

Hint:In this question, we need to determine the angular retardation produced in the car. For this, we will use newton’s equation of motion along with the angular properties.

Complete step by step answer:

First of all, let us convert all the given data in the SI unit as:

$
  {v_0} = 72{\text{ km/hr}} \\
   = \dfrac{{72 \times 1000{\text{ m}}}}{{60 \times 60{\text{ sec}}}} \\
   = \dfrac{{72000}}{{3600}}{\text{ m/sec}} \\
   = 20{\text{ m/sec}} \\
 $

Also,

$
  d = 50{\text{ cm}} \\
   = \dfrac{{50}}{{100}}{\text{ m}} \\
   = 0.5{\text{ m}} \\
 $

The product of the angular velocity of the car and the radius of the wheel of the car is the velocity of the car with which it is moving. Mathematically, $v = \omega r$

Here, the radius of the car is $r = \dfrac{d}{2} = \dfrac{{0.50}}{2} = 0.25{\text{ m}}$ , and the velocity of the car is ${v_0} = 20{\text{ m/sec}}$.

So, substituting the value of $r = 0.25{\text{ m}}$ and ${v_0} = 20{\text{ m/sec}}$ in the formula $v = \omega r$ to determine the angular velocity of the wheel of the car.

$
  v = \omega r \\
  20 = \omega \times 0.25 \\
  \omega = \dfrac{{20}}{{0.25}} \\
   = 80{\text{ rad/sec}} - - - - (i) \\
 $
According to the question, the car stops after 20 rotations when the brakes are applied so, the final angular velocity of the wheel of the car is ${\omega _f} = 0{\text{ rad/sec - }} - - - (ii)$

The product of the number of rotations and the central angle of the wheel is the angular displacement of the car during its journey. Here, 20 rotations are involved.

So, the angular displacement of the car is given as:

$
  \theta = 20 \times 2\pi \\
   = 40\pi - - - - (iii) \\
 $

Now, following the newton’s third equation of motion, ${\omega _f}^2 = {\omega _0}^2 + 2\alpha \theta $ where ${\omega _f}$ is the final angular velocity ${\omega _0}$ is, the initial angular velocity $\theta $ is the angular displacement and $\alpha $is the angular acceleration.

Substitute ${\omega _f} = 0,{\omega _0} = 80{\text{ rad/sec and }}\theta = 40\pi $ in the formula ${\omega _f}^2 = {\omega _0}^2 + 2\alpha \theta $ to determine the angular acceleration of the car.

$
  {\omega _f}^2 = {\omega _0}^2 + 2\alpha \theta \\
  {0^2} = {\left( {80} \right)^2} + 2\alpha \left( {40\pi } \right) \\
  80\pi \alpha = - {(80)^2} \\
  \alpha = \dfrac{{ - 6400}}{{80 \times 3.14}} \\
   = - 25.47{\text{ rad/se}}{{\text{c}}^2} \\
   \approx - 25.5{\text{ rad/se}}{{\text{c}}^2} \\
 $

Hence, the angular retardation produced in the car will be -25.5 radians per square seconds.

Option A is correct.

Note:Here, we can see that the angular acceleration of the car is negative as the brakes are applied to the car so as to stop it. Moreover, the negative acceleration is known as the retardation produced by the car.