Question

A particle is revolving in a circle of radius $1m$ with an angular speed of $12rad/s$. At $t=0$, it was subjected to a constant angular acceleration $\alpha$ and its angular speed increased to $\left(480/\pi \right)$ rotation per minute $\left(rpm\right)$ in $2sec$. Particles then continues to move attained speed. Calculate

1. angular acceleration of the particle
   
2. tangential velocity of the particle as a function of time.
   
3. acceleration of the particle at $t=0.5$ second and at $t=3$ seconds
   
4. angular displacement at $t=3$ seconds
   

Answer 1

Given,

Radius, $r=1m$

Initial angular velocity,${\omega }_o=12rad/s$

Final angular velocity, $\omega =\frac{2\pi }{60}\left(\frac{480}{\pi }\right)=16rad/s$

Apply angular kinematic equation.

$\omega ={\omega }_o+\alpha t$

(a) Angular acceleration, $\alpha =\frac{\omega -{\omega }_o}{t}=\frac{16-12}{2}=2rad/s^2$

(b) Tangential velocity, $V_{\text{tangential}}=\omega r=r({\omega }_o+\alpha t)=12+2t$

(c) Centripetal acceleration $a_c=r{\omega }^2$

Net acceleration, $a_{net}=\sqrt{r{\omega }^2+\alpha r}=\sqrt{r{({\omega }_o+\alpha t)}^2+\alpha r}$

$At,(t=0.5\sec ),a_{net}=\sqrt{1{(12+2\times 0.5)}^2+2\times 1}=13.07m{s}^{-2}$

$At,(t=3\sec ),a_{net}=\sqrt{1{(12+2\times 3)}^2+2\times 1}=18.05m{s}^{-2}$

(d) Angular displacement at $(t=3\sec )$ , $\theta ={\omega }_{o}t+\frac{1}{2}\alpha t^2=12\times 3+\frac{1}{2}\times 2\times 3^2=45rad$