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The Polar Co-Ordinates Of A Particle Moving In A Plane Are Given By, r=a sin(ω1)t And θ=(ω2)t. Obtain An Expression For The Polar Components Of The Velocity And Acceleration Of The Particle.

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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}

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