# Uniform circular Motion

1. Jan 31, 2009

### physicsman7

1. The problem statement, all variables and given/known data

how does velocity and acceleration change in circular moton

2. Relevant equations

3. 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

2. Jan 31, 2009

### americanforest

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 $$\frac{v^{2}}{R}$$

Last edited: Jan 31, 2009
3. Jan 31, 2009

### physicsman7

thanks kind of get it