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
AI Thread Summary
The discussion centers on determining the correct tension equation for a pendulum at rest when the string breaks. The user questions whether the tension should be calculated using T = mgCos(theta) + mv^2/r or T = mgSin(theta) + mv^2/r, as the latter is indicated in the provided answers. A vector diagram suggests that the correct equation involves the sine component, leading to T = mgSin(theta) + mv^2/r. The importance of measuring the angle from the horizontal is emphasized to avoid confusion. The consensus is that the tension due to the weight of the bob at the original position (theta = 0) should be clarified in relation to the chosen equation.
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
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...

Similar threads

Replies
3
Views
2K
Replies
21
Views
3K
Replies
8
Views
13K
Replies
9
Views
2K
Replies
21
Views
2K
Replies
1
Views
832
Replies
11
Views
4K
Replies
6
Views
2K
Replies
11
Views
5K
Replies
6
Views
2K
Back
Top