Centripetal Acceleration: Solving Physics Circular Motion

  • Thread starter Thread starter jesuslovesu
  • Start date Start date
  • Tags Tags
    Centripetal
Click For Summary
SUMMARY

The discussion focuses on centripetal acceleration and the calculations involved in a vertical circular motion scenario with a 0.10 kg ball on a string, having a radius of 0.80 meters and a tangential velocity of 6.0 m/s at the top. The user questions the accuracy of their calculations for the tangential velocity at the lowest point, which they found to be 2.0 m/s, and the tension at the top of the circular path, estimated at approximately 2.6 N. The calculations utilize the principles of kinetic energy (KE) and potential energy (PE) to derive these values, as well as the formula for tension involving centripetal force (Fc) and gravitational force (Fw).

PREREQUISITES
  • Understanding of centripetal acceleration and forces in circular motion
  • Familiarity with kinetic energy (KE) and potential energy (PE) concepts
  • Knowledge of tension calculations in physics
  • Basic grasp of gravitational force (Fg) and its role in circular motion
NEXT STEPS
  • Review the derivation of centripetal acceleration formulas
  • Learn about energy conservation in circular motion scenarios
  • Study tension calculations at different points in vertical circular motion
  • Explore the effects of varying mass and radius on circular motion dynamics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples of centripetal acceleration and energy conservation principles.

jesuslovesu
Messages
185
Reaction score
0
Here's a picture http://img215.imageshack.us/my.php?image=physics6ry.png

A .10kg ball is on a string in vertical circular motion with a radius .80 meters and the whole system is above the ground .20 meters At the top the tangential velocity is 6.0m/s.

I'm wondering if these are correct:
1)At the lowest point on the circular path the tangential velocity is 2.0 m/s.
2)The tension at the top of the circular path is ~2.6 N

I found the velocity by using KE = PE
I found the tension of the top by: T = Fc + Fw

The reason I'm wondering this is because I figured out the speed, and tensions on the top and bottom and I came up with a higher tension in the top than in the bottom...
 
Physics news on Phys.org
At the top, the tension and weight are acting in the same direction:
T + Fg = mv^2 / r
so then:
T= mv^2/r - Fg
 
In addition, for part A, the ball has both kinetic energy and gpe at the top, while at the bottom it should only have gpe.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
Replies
55
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
922
Replies
5
Views
2K