Question

A car is moving at a speed of $72\, km / hr$. The diameter of its wheels is $0.5\, m$. If the wheels are stopped in $20$ rotations applying brakes. Then angular retardation produced by the brake is:

• A. $ -45.5\,rad/s^{2}$
• B. $ -33.5\,rad/s^{2}$
• C. $ -29.5\,rad/s^{2}$
• D. $ -25.5\,rad/s^{2}$

Answer 1

Answer: (D)

Here : Speed of car = $72\, km / hr$
$=\frac{72 \times 5}{18}=20\, cm / s$
Diameter of each wheel $=\frac{0.5}{2} m$
Number of rotation $=20$
The relation for angular speed is
$\omega=\frac{v}{r}=\frac{20}{0.25}=80\, rad / \sec$
Angular displacement
$Q_{1}=2 \pi \times 20=40\, \pi\, rad$
The relation for angular retardation is given by
$\omega^{2}=\omega_{0}^{2}+2 a Q$
$0=(80)^{2}+2 a(40 \pi)$
$a=\frac{(80)^{2}}{80 \pi}=-25.5\, rad / s ^{2}$