Is the angular acceleration and the resultant acceleration of tangential and centripetal acceleration the same thing in non-uniform rotational motion? If not, what's the difference?
2 Answers
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If we have a centre of rotation $O$ then we can divide the velocity $\vec v$ of a particle into a tangential component $v_{\theta}$ and a radial component $v_{r}$ so that
$\vec v = v_r \hat r + v_{\theta} \hat \theta$
where $\hat r$ and $\hat \theta$ are unit radial and tangent vectors (with respect to $O$). If the path of the particle is a circle about $O$ then $v_r=0$ and we have
$\vec v = v_{\theta} \hat \theta$
where $v_{\theta}$ may vary with time. The acceleration of the particle is then
$\displaystyle \frac {d \vec v} {dt} = \frac {d v_{\theta}}{dt} \hat \theta + v_{\theta} \frac {d \hat \theta} {dt} = \frac {d v_{\theta}}{dt} \hat \theta - \frac {v_{_\theta}^2}{r} \hat r $
The first term $\frac {d v_{\theta}}{dt} \hat \theta$ is the tangential acceleration and the second term $- \frac {v_{_\theta}^2}{r} \hat r$ is the radial or centripetal acceleration. The angular acceleration is the tangential acceleration divided by $r$, so it is $\frac 1 r \frac {d v_{\theta}}{dt} \hat \theta$.
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The most basic acceleration in a circular motion would be the centripetal acceleration because without it the motion could not be circular at all. It is always directed towards the center. The other kinds of accelerations may or may not be present. If your angular velocity is changing steadily, you have angular acceleration $\alpha=d\omega /dt$. Since angular velocity is linked with tangential velocity as $v=r\omega$, there will necessarily be a tangentially acceleration of magnitude $r\alpha$.
All in all, centripetal acceleration causes the circular motion and the angular and tangential accelerations arise when you want to change the tangential and angular velocities. The resultant acceleration is the vector sum of centripetal and tangential acceleration which is clearly not towards the center. For some systematic mathematics, see @gandalf61 's answer.
