# Tension During Swing of Pendulum

1. Nov 26, 2008

### x^2

1. The problem statement, all variables and given/known data

The diagram shows a simple pendulum consisting of a mass M suspended by a thin, massless string. The magnitude of the tension is T. The mass swings back and forth between +/- theta_0. Which of the following are true statements?

A. T is the largest at the bottom (theta = 0 deg).
B. T = Mg at some angle between zero and theta_0.
C. The vertical component of tension is constant.
D. T equals Mg when theta = theta_0
E. T is greater than Mg when theta = theta_0
F. T is smallest when theta = +/- theta_0

2. Relevant equations

T = (mv^2)/r + mg

3. The attempt at a solution

Using the above equation, the tension should be greatest at the bottom of the swing (v reaches it's maximum) and a minimum at +/- theta_0 when v = 0. Therefore:

A. T is the largest at the bottom (theta = 0 deg).
True: This is where v reaches a maximum so the tension will be maximum.

B. T = Mg at some angle between zero and theta_0.
False: T = Mg only when (mv^2)/r = 0, which occurs at theta_0.

C. The vertical component of tension is constant.
True: The vertical component is Mg which remains constant.

D. T equals Mg when theta = theta_0
True: v = 0 so T = m0^2/r + Mg ==> T = Mg

E. T is greater than Mg when theta = theta_0
False: Can't be true if T = Mg from above statement

F. T is smallest when theta = +/- theta_0
True: There is no velocity at +/- theta_0, so this is where T must be a minimum.

I've tried the above answer set, also with C as false (I wasn't sure if maybe the vertical component related to more than just Mg), but both attempts were wrong. If my pendulum tension equation accurately describes the above scenario, I'm not sure where I went wrong in my reasoning.

Thanks,
x^2

2. Nov 26, 2008

### x^2

Solved it using T = (mv^2)/r + mgcos(theta)

Just consider the velocity and angle at each point:
A: True. v will be the greatest and cos(theta) will be the greatest (1)
B: True. At theta_0 when the bob is released, v = 0 and cos(theta) < 1 so the tension will be less than Mg. At theta = 0, v will be at its max and cos(theta) = cos(0) = 1 so the tension will be greater than Mg. Somewhere between the tension will equal Mg.
C: False. The vertical component is greatest when theta = 0. It decreases as the bob swings to the left and right.
D: False. T = Mg somewhere between theta_0 and theta = 0.
E: False. T is less than Mg at theta_0.
F: True. At theta_0 T = mgcos(theta_0). As v = 0 at theta_0, mv^2/r ==> 0 at theta_0.

Hope this helps someone :D

3. May 3, 2010

### |<ings

What...?

I always thought

T = -mgcos(theta) (i.e. the radial component of the Weight force)

in a pendulum... not T = mgcos(theta) + mac.

T and the radial component of the Weight force cancel out. The only force acting on the system is the tangential (real word?) component of the Weight force:

Fnet = -mgsin(theta)

Besides I don't think the pendulum experiences uniform circular motion (tangential velocity is not constant), in which case ac = v2/r does not apply. Unless ac = v2/r applies to any circular motion and not just uniform circular motion. If it does apply, then:

Fnet = mac
= mv2/r​
mv2/r = -mgsin(theta)

....

Last edited: May 3, 2010
4. May 3, 2010

Hello?

5. May 3, 2010

Help?