Understanding Velocity and Acceleration in Uniform Circular Motion

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In uniform circular motion, the velocity of an object is always tangential to its path, while the acceleration is centripetal, directed towards the center of the circle. The position vector can be expressed as \(\vec{r}=Rcos(\omega t)\hat{x}+Rsin(\omega t)\hat{y}\), where \(\omega\) is the angular frequency and \(R\) is the radius. Velocity and acceleration are derived by taking the first and second time derivatives of the position vector. The tangential velocity \(v\) is related to the radius and angular frequency by \(R\omega = v\). The acceleration's magnitude is \(\frac{v^{2}}{R}\) and is directed towards the center of the circular path.
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Homework Statement



how does velocity and acceleration change in circular moton

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The Attempt at a Solution


I know when a object is circular motion the velocity is tangential to the motion also, acceleration centripital, sum of the forces which points to a center seeking force
 
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In Uniform Circular Motion the position vector can be expressed as

\vec{r}=Rcos(\omega t)\hat{x}+Rsin(\omega t)\hat{y}

where omega is the frequency of oscillation, t is time , and R is the radius of the circle.

We calculate velocity and acceleration by taking first and second derivatives with respect to time.

\vec{\dot{r}}=-\omega Rsin(\omega t)\hat{x}+\omega Rcos(\omega t)\hat{y}

\vec{\ddot{r}}=-\omega ^{2} Rcos(\omega t)\hat{x}-\omega ^{2}Rsin(\omega t)\hat{y}=-\omega ^{2}\vec{r}

Also, R\omega = v where v is the tangential velocity (To show this use Rd\theta =dS where dS is an infinitesimal tangential distance and divide both sides by dt) so

\vec{\ddot{r}}=-\frac{v^{2}}{R^{2}}\vec{r}=-\frac{v^{2}}{R^{2}}R\hat{r}=-\frac{v^{2}}{R}\hat{r}

So the acceleration is anti parallel to the radius vector (ie. towards the center of the circle) and has a magnitude of \frac{v^{2}}{R}
 
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thanks kind of get it
 
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