Question

A particle starts from rest moves along a circle of radius 20 cm with constant angular acceleration of $2\mathrm{rad}/{\mathrm{s}}^2$. Determine the time in which the magnitude of tangential acceleration of particle is equal to half of the radial acceleration of the particle.

Answer 1

Step 1: Given data:
The radius of the circle $\mathrm{r}=20\mathrm{cm}=0.2\mathrm{m}$
Constant angular acceleration $\alpha =2rad/s^2$
Step 2: Tangential acceleration:

The magnitude of the tangential acceleration of the particle is equal to half of the radial acceleration of the particle, which is,
$a_t=\frac{a_r}{2}$ where $a_r$ is the radial acceleration.
The tangential acceleration $a_t=r\alpha$, $r$ is the radius of the circular path.
The radial acceleration $a_r={\omega }^{2}r$, $\omega$ is the angular velocity.
Therefore,
$r\alpha =\frac{{\omega }^{2}r}{2}$

Step 3: Angular velocity:

The angular velocity $\omega$ is given by
$\omega =\alpha t$

Therefore, substituting for $\omega$, we get,
$r\alpha =\frac{{\left(\alpha t\right)}^{2}r}{2}$
$\alpha =\frac{{\alpha }^2{t}^2}{2}$
$t^2=\frac{2}{\alpha }$
$t^2=\frac{2}{2rad/s^2}$
$t=1s$

The time in which the magnitude of the tangential acceleration of the particle is equal to half of the radial acceleration of the particle is.$t=1s$