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

$a_{\theta}=r\ddot{\theta}+2\dot{r}\dot{\theta}$

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

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