T = mgcosQHow is tension related to circular motion in a vertical circle?

Click For Summary

Homework Help Overview

The discussion revolves around the relationship between tension and circular motion in a vertical circle, specifically analyzing the forces acting on a rock being whirled at the end of a rope. Participants are exploring the components of forces and their implications on tension and acceleration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the expression for tension and the total acceleration, questioning how to derive these from the forces acting on the rock. There is a focus on breaking down the weight into components and understanding their directions relative to tension.

Discussion Status

The discussion is active, with participants attempting to clarify the components of weight and their relationship to tension. Some guidance has been offered regarding the direction of forces, but there is no explicit consensus on the overall approach or solution.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is uncertainty regarding the integration of velocity and how it relates to the forces involved.

joemama69
Messages
390
Reaction score
0

Homework Statement



A rock is whirled at the end of a rope in a vertical circle

Find a general expression fo the Tension

What is the magnitude of the total acceleration

Homework Equations





The Attempt at a Solution



[tex]\sum[/tex]Fr = ma = T + mgcos[tex]\theta[/tex]

T = ma - mgcos[tex]\theta[/tex] not sure

a = (T + mgcos[tex]\theta[/tex])/m not sure
 
Physics news on Phys.org
At an angle θ to the vertical, what is the component of the weight along the line of tension?
 
not sure what u mean
 
joemama69 said:
not sure what u mean

Draw the mass at angle θ to the vertical.


Can you split the weight mg into two components?
 
into two components, would that be the tangental compenent in the direction of the velocity and the second pointing towards the center
 
joemama69 said:
into two components, would that be the tangental compenent in the direction of the velocity and the second pointing towards the center

actually it points opposite to the velocity and away from the center. What are these two components in terms of the angle?
 
Component along line of tension ma = mgcosQ

how do i do tat for the velocity, do i integrate that
 
joemama69 said:
Component along line of tension ma = mgcosQ

how do i do tat for the velocity, do i integrate that

the component is mgcosθ.

So the T points towards the center and the mgcosθ points away from the center. What is the resultant force equal to ?
 
Fr = T - mgcosQ = 0
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
15K
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
940