What is the correct tension equation for a pendulum at rest?

  • Thread starter Thread starter Pochen Liu
  • Start date Start date
  • Tags Tags
    Pendulum Tension
Click For Summary
SUMMARY

The correct tension equation for a pendulum at rest, when the angle is measured from the horizontal, is T = mgSin(theta) + mv^2/r. This conclusion is supported by vector analysis, which indicates that the tension in the string is influenced by the sine component of the gravitational force acting on the bob. The discussion clarifies that the tension will break when it reaches twice the weight of the bob, confirming the importance of understanding the angle's reference point in calculations.

PREREQUISITES
  • Understanding of basic pendulum mechanics
  • Knowledge of vector diagrams in physics
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Basic principles of forces and tension in physics
NEXT STEPS
  • Study the derivation of tension equations in pendulum motion
  • Learn about the effects of angle on tension in circular motion
  • Explore advanced vector analysis techniques in physics
  • Investigate the relationship between tension and gravitational forces in different scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and pendulum dynamics, as well as educators seeking to clarify concepts related to tension and forces in pendulum systems.

Pochen Liu
Messages
52
Reaction score
2

Homework Statement


The pendulum cord is released from rest when the angle = 0 (from a horizontal)
If the string breaks when the tension is twice the weight of the bob at what angle does it break?

Im from NZ and our NCEA system is often riddled with mistakes and I want to clear this up.
Is the tension equation this:

T = mgCos(theta) + mv^2/r
or
T = mgSin(theta) + mv^2/r (What the answers say)

By drawing a vector diagram I believe the test is wrong as the component in Fg providing tension is mgCos(theta)

Can anyone confirm for me? and provide reasons why so that I don't confuse myself.
 
Physics news on Phys.org
Pochen Liu said:
T = mgCos(theta) + mv^2/r
or
T = mgSin(theta) + mv^2/r (What the answers say)

By drawing a vector diagram I believe the test is wrong as the component in Fg providing tension is mgCos(theta)

Can anyone confirm for me? and provide reasons why so that I don't confuse myself.

Don't get confused...Just make sure that the angle(theta) is measured from the horizontal.
Then by the vector diagram, you will get T=mgSin(theta) + mv^2/r.
 
  • Like
Likes Pochen Liu
What's the tension due to the weight of the bob in the original position (##\theta = 0##)?
 
  • Like
Likes Pochen Liu

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 8 ·
Replies
8
Views
14K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 1 ·
Replies
1
Views
946
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K