Question

A car is moving at a seed of $72km/hr$. The diameter of its wheel is $0.52m$. If the wheels are stopped in $20$ rotations by applying brakes, the angular retardation produced by the brakes will be-

• A. $25.5rad/s^2$
• B. $2.55rad/s^2$
• C. $0.255rad/s^2$
• D. $255rad/s^2$

Answer 1

The correct option is A $25.5rad/s^2$

Speed of car $=72km/hr=\frac{72\ast 5}{18}=20cm/s$

Diameter of each wheel $=0.52m$

Number of rotation $=20$

The relation for angular speed is $\omega =\upsilon r=200.25=80rad/sec$

Angular displacement $Q=2\pi \ast 20=40\pi rad$

The relation for angular retardation is given by ${\omega }^2={{\omega }_0}^2+2aQ$

$0={\left(80\right)}^2+2a\left(40\pi \right)$ Or $a=\frac{{80}^2}{80\pi }=25.5rad/s^2$