Question

the angular velocity of a wheel increases from 1200 to 4500 revolutions per minute in 10 second find its angular acceleration
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Answer 1

Given :

• Initial angular velocity of wheel $\sf\omega_o=1200rpm=\dfrac{1200\times2\pi}{60}=40rad\:s^{-1}$
• Final angular velocity of wheel $\sf\omega=\dfrac{4500\times2\pi}{60}=150\:rad\:s^{-1}$
• Time $\sf\:t=10sec$

To Find :

Angular accelaration of wheel

Theory :

1) Average 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 Angular Accelaration of wheel

By Using Equations of motion

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

Put the given values

$\sf\implies\:150=40+\alpha\times\:10$

$\sf\implies\:150-40=10\alpha$

$\sf\implies\alpha=\dfrac{110}{10}$

$\sf\implies\alpha=11\:rad\:sec^{-2}$

Therefore , Angular Accelaration of wheel is 11 rad /s².

Answer 2

$\Large\bf{\color{indigo}GiVeN,}$

• Initial angular velocity $\bf{(\omega_1)}$ = 1200 rev/min = $\bf{\dfrac{1200}{60}\:=\:20\:rev/sec\:}$

• Final angular velocity $\bf{(\omega_2)}$ = 4500 rev/min = $\bf{\dfrac{4500}{60}\:=\:75\:rev/sec\:}$

• Time period = 10 sec

$\bf\pink{We\:know\:that,}$

$\red\bigstar\:\:\bf{\color{coral}\omega_2\:=\:\omega_1\:+\:\alpha{t}\:}$

$:\implies\:\:\bf{\alpha{t}\:=\:\omega_2\:-\:\omega_1\:}$

$:\implies\:\:\bf{\alpha\:=\:\dfrac{\omega_2\:-\:\omega_1}{t}\:}$

$\bf\red{Where,}$

• $\bf\blue{\alpha\:=\:Angular\:acceleration\:}$

$:\implies\:\:\bf{\alpha\:=\:\dfrac{75\:-\:20}{10}\:}$

$:\implies\:\:\bf{\alpha\:=\:\dfrac{55}{10}\:}$

$:\implies\:\:\bf\green{\alpha\:=\:5.5\:rev/sec^2\:}$

$\Large\bold\therefore$ Angular acceleration of the wheel is $\bf\purple{5.5\:rev/sec^2\:}$.