Obtain the equation $\omega = \omega_{0} + \alpha t$ from first principles.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The angular acceleration $\alpha$ is uniform,hence the rate of change of angular velocity $\omega$ with respect to time $t$ is constant:
$\frac{d\omega}{dt} = \alpha$
Rearranging the terms for integration:
$d\omega = \alpha dt$
Integrating both sides:
$\int d\omega = \int \alpha dt$
$\omega = \alpha t + c$,where $c$ is the constant of integration.
Applying the initial condition: at $t = 0$,$\omega = \omega_{0}$.
Substituting these values into the equation:
$\omega_{0} = \alpha(0) + c \implies c = \omega_{0}$
Substituting the value of $c$ back into the integrated equation:
$\omega = \omega_{0} + \alpha t$.

Explore More

Similar Questions

An object moves along a circle with normal acceleration proportional to $t^\alpha$, where $t$ is the time and $\alpha$ is a positive constant. The power developed by all the forces acting on the object will have time dependence proportional to

If the frequency of an object in uniform circular motion is doubled,its acceleration becomes

$A$ particle moves along a circular path of radius $r = \frac{20}{\pi} \ m$ with a constant tangential acceleration. If the velocity of the particle at the end of the second revolution is $80 \ m/s$,what is the tangential acceleration?

Difficult
View Solution

$A$ body is revolving with a uniform speed $v$ in a circle of radius $r$. The tangential acceleration is

If the length of the second's hand in a stop clock is $3 \, cm$,the angular velocity and linear velocity of the tip are:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo