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}