Question

A car is moving with a speed of $72km{h}^{-1}$. The radius of its wheel its $50cm$. If its wheels come to rest after $20$ rotations as a result of application of brakes, then the angular retardation produced in the car will be

• A. $23.5rad{s}^{-2}$
• B. $0.25rad{s}^{-2}$
• C. $6.35rad{s}^{-2}$
• D. $zero$

Answer 1

Answer: (C)

$6.35rad{s}^{-2}$

Given,
$v=72kmph=$$72\ast \frac{5}{18}m{s}^{-1}$
$=20$$m{s}^{-1}$.
Now , angular velocity of wheel intially=$\frac{v}{r}$
where , $r$ is radius of wheel.
$r=50cm$ {given}
$=0.5m$
angular velocity intially=$40rad$$s^{-1}$
and also given, That the wheel is rotated by $20revolutions$.
angle of rotation say ( $\theta$ ) =$20\times 2$$\pi rad{s}^{-1}$
$\theta =40\pi rad{s}^{-1}$
As we know that,
${\omega }_{final}^2-{\omega }_{intial}^2=2\times \alpha \times \theta$ {let $\alpha$ be angular accerlation}
By substituting the known values,
$\Rightarrow 0^2-{40}^2=2\times \alpha \times 40\pi$
$\Rightarrow \alpha =-6.3661rad{s}^{-1}${- sign indicates retardation}
Therefore answer is Option C.