The movement of a particle in a polar co-ordinate system is described by,

\( r=a\sin{\omega}_1{t} \) and \( \theta={\omega}_2{t} \)

We know that the radial component of velocity is \( v_r=\dot{r} \) and the transverse component of velocity is

\( v_{\theta}=r\dot{\theta} \).

And the radial component of the velocity is

\( a_r=\ddot{r}-r{\dot{\theta}}^2 \)

and the transverse component of the acceleration is

\( a_{\theta}=r\ddot{\theta}+2\dot{r}\dot{\theta} \)

Now,

\( \dot{r}=\frac{dr}{dt}=\frac{d}{dt}( a\sin{\omega}_1{t}) \)

\( or,\ \dot{r}=a{\omega}_1\cos{\omega}_1{t} \)

Again,

\( \ddot{r}=-a{{\omega}_1}^2\sin{\omega}_1{t} \)

Now,

\( \dot{\theta}=\frac{d\theta}{dt}=\frac{d}{dt}({\omega}_2{t}) \)

\( or,\ \dot{\theta}={\omega}_2 \)

Again,

\( \ddot{\theta}=0 \)

So the radial component of the velocity is

\( v_r= a{\omega}_1\cos{\omega}_1{t} \)

and the transverse component of the velocity is

\( v_{\theta}=r\dot{\theta}= (a\sin{\omega}_1{t}){({\omega}_2)} \)

\( or,\ v_{\theta}=a{\omega}_2\sin{\omega}_1{t} \)

Now the radial component of the acceleration is

\( a_r=(-a{{\omega}_1}^2\sin{\omega}_1{t})- a\sin{\omega}_1{t}{{\omega}_2}^2 \)

\( or,\ a_r=-a\sin{\omega}_1{t}[{{\omega}_1}^2+{{\omega}_2}^2] \)

And the transverse component of the acceleration is

\( a_{\theta}=r\ddot{\theta}+2\dot{r}\dot{\theta} \)

\( or,\ a_{\theta}=( a\sin{\omega}_1{t})\times(0)+2(a{\omega}_1\cos{\omega}_1{t})({\omega}_2) \)

\( or,\ a_{\theta}=2a{\omega}_1{\omega}_2\cos{\omega}_1{t} \)