Question

A car is moving with a speed 54 km/hr. The radius of
the wheels is 10 cm. After application of brakes, the
wheels come to rest in 10 rotations. The angular
retardation produced by the brakes is
61 rad/s
260 rad/s
151 rad's
179 rad​

Answer 1

Given :

• Velocity of car , v = 54 km/hr = 15 m/s
• Radius of wheel , r = 10 cm = 0.1m
• After application of brakes the wheels to come to rest in 10 rotations.

To Find :

The angular retardation

Solution :

We have to find the Angular Accelaration of wheel

Let the Angular retardation be -α rad /s²

We have :

$\sf\omega_o=\dfrac{v}{r}=150$

$\sf\omega=0(brakes\:applied)$

$\sf\theta=20\pi$

By Equation of motion

$\rm\:\omega^2=\omega^2_o+2\alpha\theta$

Put the given values

$\sf\implies\:0=(150^2)+2\alpha\times\:20\pi$

$\sf\implies-(150^2)=2\alpha\times\:20\pi$

$\sf\implies\alpha=\dfrac{-150\times150}{2\times20\times\pi}$

$\sf\implies\alpha=\dfrac{-2250}{4\pi}$

$\sf\implies\alpha\approx-179\:rad\:sec^{-2}$

Therefore , Angular retardation produced by the brakes is 179rad /s².

Answer 2

Given :

• Velocity of car , v = 54 km/hr = 15 m/s
• Radius of wheel , r = 10 cm = 0.1 m
• After application of brakes the wheels to come to rest in 10 rotations.

To Find :

The angular retardation produced by the brakes

Theory :

1) Angular Accelaration :

In x- y plane it is the rate at which angular velocity changes with time .

$\sf\:Angular\:Accelaration=\dfrac{\triangle\omega}{\triangle\:t}=\dfrac{\triangle\omega_x}{\triangle\:t}\vec{i}+\dfrac{\triangle\omega_y}{\triangle\:t}\vec{j}$

2) Equations of Motion for constant angular acceleration:

$\bf\:\omega=\omega_o+\alpha\:t$

$\bf\:\theta=\omega_o\:t+\frac{1}{2}\alpha\:t{}^{2}$

$\bf\:\omega^{2}=\omega^{2}_o+2\alpha\theta$

Solution :

We have to find the retardation produced by the brakes.

We know that

$\sf\:v=\omega\:\times\:r$

$\sf\implies\omega=\dfrac{v}{r}$

Put the given values

$\sf\implies\omega=\dfrac{15}{0.1}$

$\sf\implies\omega=150$

And No of rotations = θ/2π

$\sf\implies\theta=10\times2\pi$

$\sf\implies\theta=20\pi$

Let the Angular Acceleration be α rad /s²

Now , We have

$\sf\omega_o=150$

$\sf\omega=0(brakes\:applied)$

$\sf\theta=20\pi$

$\sf\:Angular\:Acceleration=\alpha$

By Equation of motion

$\rm\:\omega^2=\omega^2_o+2\alpha\theta$

Put the given values

$\sf\implies\:0=(150^2)+2\alpha\times\:20\pi$

$\sf\implies-(150^2)=2\alpha\times\:20\pi$

$\sf\implies\alpha=\dfrac{-150\times150}{2\times20\times\pi}$

$\sf\implies\alpha=\dfrac{-2250}{4\pi}$

$\sf\implies\alpha\approx-179\:rad\:sec^{-2}$

Therefore , Angular retardation produced by the brakes is 179rad /s².