# Help with free body diagram for a pendulum

1. Apr 5, 2015

### theone

1. The problem statement, all variables and given/known data
I want to sum the forces perpendicular to the pendulum and sum the moments about the pendulums center of gravity.

2. Relevant equations

3. The attempt at a solution
$P\sin \theta - mg\cos \theta - N\cos \theta = -m\ddot x\cos \theta + ml\ddot \theta$
$-Pl\sin \theta - Nl\cos \theta + K_t\theta= I\ddot\theta$

I want to know if these are correct, especially the plus/minus

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2. Apr 7, 2015

### haruspex

You have a cosine and sine swapped in the first equation.
Your second equation looks doubtful in several respects. What point are you taking moments about? Is it ok to take moments about a point that is accelerating?
Do you even need to take moments, i.e. does the rod have mass?

3. Apr 8, 2015

### theone

The rod does have mass. The point was about the center of gravity of the combined rod/ball thing. l is the the distance from the pivot to the center of gravity. I needed a moment equation because I wanted to eliminate the $$Psin\theta-Ncos\theta$$ in the first equation.

I dont really know if its ok to take moments about that point. I also don't understand the $$ml\ddot\theta$$ term in the first equation...

4. Apr 9, 2015

### theone

for the sum of moments about the pendulum's COG, is the moment due to P negative and the moment due to N positive?

5. Apr 9, 2015

### haruspex

With the usual anticlockwise convention, yes.

6. Apr 9, 2015

### theone

For the first equation, which two terms did you identify as wrong? Is it the P and N terms?

7. Apr 9, 2015

### haruspex

In each of the terms involving a trig function, the other factor is related to acceleration or force in either the vertical or horizontal direction. You should expect that all of one direction take cos, and all of the other direction take sin.

8. Apr 9, 2015

### theone

But for P and mg, which are in opposite directions, I am getting the result that both are sin

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9. Apr 9, 2015

### haruspex

They're both vertical, so should match the same trig function.

10. Apr 9, 2015

### theone

so is the final answer including the signs correct: $$Psin\theta-mgsin\theta-Ncos\theta=m\ddot x cos\theta+ml\ddot\theta$$

11. Apr 9, 2015

### haruspex

Not quite. The left hand side appears to be taking up and.left as positive, the right hand side down and right as positive.

12. Apr 9, 2015

### theone

I meant for the $$mg\ddot x cos\theta$$ to be negative. Is that the only mistake

13. Apr 9, 2015

### haruspex

What about the $ml\ddot \theta$ term? Also, you said the rod has mass, so that should be in the equation, and you need a contribution from the torsion spring.

14. Apr 9, 2015

### theone

for the rod mass, I considered the rod and ball a single object. Is that ok?
I don't know what the contribution from the torsion spring is. and like I said earlier, I don't really understand the $ml\ddot \theta$ term

15. Apr 9, 2015

### haruspex

The $ml\ddot \theta$ term is the acceleration of the mass (in the tangential direction) relative to the acceleration of the block.
The torsion spring applies a torque. What is the relationship between force and torque?
If you are considering the ball and mass as one object, then you need to figure out the mass centre of the combination and see how that affects the $ml\ddot \theta$ term and the torsion spring tem.

16. Apr 9, 2015

### theone

The spring applies ${K_t\theta}$ of torque , so then the force it applies is $\frac{K_t\theta}{l}$, with l being the length of the rod from the pivot to the combined COG?

17. Apr 9, 2015

### haruspex

That's the force applied to the mass, if the rod is massless. But as I wrote, if you consider the mass and rod as a single unit, you need to deal in terms of their common mass centre, and how far that will be from the joint.

18. Apr 9, 2015

### theone

So is it...
$$Psin\theta-mgsin\theta-Ncos\theta+\frac{K_t \theta}{l}=m\ddot x cos\theta-ml\ddot \theta cos\theta$$

with l being the distance from the base of the pivot to the combined object's COG?
The force due to the torsional spring acts perpendicular to the length of the rod, right?

Last edited: Apr 9, 2015
19. Apr 9, 2015

### haruspex

Yes, except that you've made the $m\ddot x$ term positive again (you said it was meant to be negative), and you've introduced a cos factor on the $\ddot \theta$ term.

20. Apr 9, 2015

### theone

thanks
and finally, does the force due to the torsional spring act up and left or down and right? In the equation above, I put it as up and left.
And also, if I were to take a force balance on the cart, would I need to include anything other than P and N to account for the torsional spring?