Question

The radius of wheels of a car moving with a speed of $72km{h}^{-1}$ is $0.25m$. If by applying brakes the wheels are stopped in $20$ rotations, then the angular retardation produced by the brakes is

• A. $-25.5rad{s}^{-2}$
• B. $-20.5rad{s}^{-2}$
• C. $-29.5rad{s}^{-2}$
• D. $-33.0rad{s}^{-2}$

Answer 1

The correct option is A $-25.5rad{s}^{-2}$

By using the equation
${\omega }^2={\omega }_0^2+2\alpha \theta$
angular retardation,
$\alpha =\frac{{\omega }^2-{\omega }_0^2}{2\theta }....\left(i\right)$
We know that, ${\omega }_0=\frac{v}{r}=\frac{72\times 5/18}{0.25}$
$=80rad{s}^{-1}....\left(ii\right)$
When, $\omega =0,\theta =2\pi n$
i.e., $\theta =2\pi \times 20=40\pi rad....\left(iii\right)$
Substituting, values of Eqs. (ii) and (iii) in Eq. (i), we get
$\alpha =\frac{0^2-(80{)}^2}{2\times 40\pi }=\frac{-80\times 80}{80\times 3.14}=-25.5rad{s}^{-2}$.