Question

If a particle is performing non-uniform circular motion with constant angular acceleration, then the linear acceleration of a particle as a function of time is _____. The particle starts from rest and moves along a circle of radius R.

Answer 1

Step 1: Acceleration in a non-uniform circular motion:

The magnitude of acceleration in non-uniform circular motion is,
$\alpha =\sqrt{{\left({\alpha }_r\right)}^2+{\left({\alpha }_t\right)}^2}\left(1\right)$

Here,
${\alpha }_r$ - radial acceleration.
${\alpha }_t$ - tangential acceleration.

Step 2: To find the linear acceleration:

The radial and tangential acceleration are as follows:
$$\begin{gathered} {\alpha }_r={\omega }^{2}R \\ \left[As\omega =\alpha t\right] \\ {\alpha }_r={\alpha }^2{t}^{2}R \\ \mathrm{and} \\ {\alpha }_t=\alpha R \end{gathered}$$

Here, R is the radius, $\omega$ is the angular velocity, t is the time.

Substitute both in equation (1),

$\sqrt{{\left({\alpha }^2{t}^{2}R\right)}^2+{\left(\alpha R\right)}^2}$
Thus, the linear acceleration of a particle as a function of time is $\sqrt{{\left({\alpha }^2{t}^{2}R\right)}^2+{\left(\alpha R\right)}^2}$.