Understanding Velocity and Acceleration in Uniform Circular Motion

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
physicsman7
Messages
5
Reaction score
0

Homework Statement



how does velocity and acceleration change in circular moton

Homework Equations





The Attempt at a Solution


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

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

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.

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

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

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

[tex]\vec{\ddot{r}}=-\frac{v^{2}}{R^{2}}\vec{r}=-\frac{v^{2}}{R^{2}}R\hat{r}=-\frac{v^{2}}{R}\hat{r}[/tex]

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