Help with free body diagram for a pendulum

[theone]

Homework Statement

I want to sum the forces perpendicular to the pendulum and sum the moments about the pendulums center of gravity.

Homework Equations

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

 

[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?

 

[theone]

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?

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 don't 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...

 

[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?

 

[haruspex]

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

With the usual anticlockwise convention, yes.

 

[theone]

With the usual anticlockwise convention, yes.

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

 

[haruspex]

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

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.

 

[theone]

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.

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

 

[haruspex]

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

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

 

[theone]

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

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

 

[haruspex]

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

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

 

[theone]

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

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

 

[haruspex]

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

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.

 

[theone]

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.

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

 

[haruspex]

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

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.

 

[theone]

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.

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?

 

[haruspex]

The spring applies ${K_t\theta}$ of torque , so then the force it applies is $\frac{K_t\theta}{L}$, with L being the total length of the rod?

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.

 

[theone]

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.

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?

 

[haruspex]

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?

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.

 

[theone]

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.

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?

 

[haruspex]

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.

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?

Yes, the torque from the spring reacts on the block exactly opposite to its torque on the arm. This is in addition to P and N.