View Full Version : Uniform circular Motion
physicsman7
Jan31-09, 12:15 AM
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
americanforest
Jan31-09, 01:11 AM
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}
physicsman7
Jan31-09, 01:17 AM
thanks kind of get it
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