Understanding Velocity and Acceleration in Uniform Circular Motion

Click For Summary
SUMMARY

In uniform circular motion, the velocity of an object is always tangential to the path, while the acceleration is centripetal, directed towards the center of the circular path. The position vector is defined as \vec{r}=Rcos(\omega t)\hat{x}+Rsin(\omega t)\hat{y}, where \omega is the angular frequency, t is time, and R is the radius. The velocity is calculated as \vec{\dot{r}}=-\omega Rsin(\omega t)\hat{x}+\omega Rcos(\omega t)\hat{y}, and the acceleration as \vec{\ddot{r}}=-\omega^{2}\vec{r}, indicating that acceleration has a magnitude of \frac{v^{2}}{R} and is directed towards the center of the circle.

PREREQUISITES
  • Understanding of vector calculus
  • Familiarity with angular frequency and its relation to circular motion
  • Knowledge of derivatives and their application in physics
  • Basic concepts of centripetal force and motion
NEXT STEPS
  • Study the relationship between angular velocity and linear velocity in circular motion
  • Learn about the derivation of centripetal acceleration and its applications
  • Explore the effects of varying radius on velocity and acceleration in circular motion
  • Investigate real-world applications of uniform circular motion in engineering and physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in the principles of motion in circular paths.

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

\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:
thanks kind of get it
 

Similar threads

Replies
55
Views
3K
Replies
1
Views
916
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
11
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 9 ·
Replies
9
Views
8K