Question

A particle initially at rest is moving along a circle of radius 3m with constant angular acceleration of $2\mathrm{rad}/{\mathrm{s}}^2$. Determine its linear velocity and angular velocity at $t=5s$. Also determine its radial, tangential and total acceleration at $t=5s$.

Answer 1

Step 1: Given data:
The radius of the circle $\mathrm{r}=3\mathrm{m}$
Constant angular acceleration $\alpha =2rad/s^2$
Time $t=5s$

Step 2: To find the angular velocity of the particle:

The angular velocity of the particle $\omega$,
$\omega =\alpha t$
$\mathrm{\omega }=2\mathrm{rad}/{\mathrm{s}}^2\times 5\mathrm{s}$
$\mathrm{\omega }=10\mathrm{rad}/{\mathrm{s}}^2$
The linear velocity of the particle is,
$v=r\omega$
$v=3m\times 10rad/s$
$\mathrm{v}=30\mathrm{m}/\mathrm{s}$
Step 3: To find the radial acceleration of the particle :

The radial acceleration of the particle is,
${\mathrm{a}}_{\mathrm{r}}={\mathrm{\omega }}^2\mathrm{r}=(10\mathrm{rad}/\mathrm{s}{)}^2\times 3\mathrm{m}$
${\mathrm{a}}_{\mathrm{r}}=300\mathrm{m}/{\mathrm{s}}^2$

Step 4: To find the tangential acceleration:

The tangential acceleration is,
${\mathrm{a}}_{\mathrm{t}}=\mathrm{r\alpha }=3\mathrm{m}\times 2\mathrm{rad}/{\mathrm{s}}^2$
${\mathrm{a}}_{\mathrm{t}}=6\mathrm{m}/{\mathrm{s}}^2$
Step 5: To find the total acceleration of the particle:

The total acceleration of the particle is,
$\mathrm{a}=\sqrt{{\mathrm{a}}_{\mathrm{r}}^2+{\mathrm{a}}_{\mathrm{t}}^2}=\sqrt{{\left(300\right)}^2+6^2}$
$\mathrm{a}=300.06\mathrm{m}/{\mathrm{s}}^2$

The linear velocity and angular velocity at $t=5s$ is $30\mathrm{m}/\mathrm{s}$ and $10\mathrm{rad}/{\mathrm{s}}^2$respectively. The radial, tangential and total acceleration at $t=5s$ is $300\mathrm{m}/{\mathrm{s}}^2$, $6\mathrm{m}/{\mathrm{s}}^2$ and $300.06\mathrm{m}/{\mathrm{s}}^2$.