# Tension on a pendulum

• Pochen Liu
In summary, the conversation discusses the tension equation for a pendulum and whether it is correct or not. The tension equation in question is T = mgCos(theta) + mv^2/r or T = mgSin(theta) + mv^2/r. By drawing a vector diagram, it is determined that the correct equation is T = mgSin(theta) + mv^2/r, with the angle (theta) being measured from the horizontal. The conversation concludes with a confirmation and explanation of the correct equation.

## 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.

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.

• Pochen Liu
What's the tension due to the weight of the bob in the original position (##\theta = 0##)?

• Pochen Liu