# How do I set up this pendulum problem with a pivot point not on edge

1. Oct 24, 2014

### grandpa2390

1. The problem statement, all variables and given/known data
Damped driven oscillator: ruler example. Suppose the ruler used in the classroom demonstration has a length of 12 and 13/16ths inches, a width of 1 ½ inches, is 1/16th inch thick with a density of 1.2 g/cm3. It swings from a pivot point ¼ of an inch from the top end.
a) Find the undamped frequency ω0 for this physical pendulum (ignore the hole except for locating the pivot point).

2. Relevant equations
I don't know

3. The attempt at a solution

I am sure I could find the frequency, I just don't know how to set up the problem. I drew a diagram and marked the forces acting on the pendulum.I am pretty sure since the ruler has uniform mass that the center of mass would be "bob" of the pendulum. but with the pivot point being located away from the edge, I am not sure how to affect the center of gravity.

What I want to say is that I should take the center of gravity to be the middle of the ruler below the pivot. Then I take the force of part of the ruler above the pivot and treat it like a lever. subtract the two forces and then treat the pendulum as if its pivot was on the edge, but it has the altered force at the center of the part of the ruler lieing below the pivot.

Is this the correct approach?

I went ahead with this approach. I found the torques on both sides of the pivot due to gravity. and then I subtracted the torque.

I plugged it into the equation for W{not} and got .776
is this correct?

Last edited: Oct 24, 2014
2. Oct 24, 2014

### haruspex

Why would the location of the pivot point affect the location of the mass centre?
The interesting part is determining the moment of inertia. What theorem or formula can you quote for that?

3. Oct 24, 2014

### grandpa2390

I=mr^2 ?
where r is the distance from the pivot point?
I have never seen that formula before. Do I just

I don't know. I just figure the part of the pendulum above the pivot would exert a downward force of its own opposing the motion of the other side of pivot. I am not sure how to account for that. In my mind I thought the part of the ruler below the pivot was a pendulum to itself with a resistive once being applied by the part of the ruler above the pivot. so when looking at it I would just calculate the force or torque due to the part above the pivot, then I could treat the ruler as if there is no top part. I guess not.

do I calculate the moment of inertia for both side using that formula and then add or subtract them somehow? I don't know. I really need to understand this.

or maybe when I plug it into the formula, the fact that the pivot is offset will automatically be factored in?

edit: I just watched Kahn. and he called it a moment what I did. I calculated the force of one side of the pivot times the distance. then I moved it to the other side, and I moved it out to the same distance as the middle of the other side of the pivot. but because it had to go out further, the force became smaller. and then I subtracted it from the moment on the other side.

Last edited: Oct 25, 2014
4. Oct 25, 2014

### haruspex

I'm sure it can be done by considering the portions above and below the pivot separately, but there's no need.
The centre of mass of the ruler as a whole stays where it is, in the centre of the ruler. Then you just have to find the moment of inertia of the ruler (as a rectangular block) about its mass centre, and apply the parallel axis theorem to find the moment about the pivot point. Armed with the distance from pivot to mass centre and the moment of inertia you can write down the differential equation.

5. Oct 25, 2014

### grandpa2390

This is over my head, but I think I may have found something.
so in another thread on this forum they say it is

I = I_cm + M(d^2)
I is the moment around the pivot point I think
I_cm is the moment about the center of mass
M is the mass of the bar
and d is the distance between the center of mass and the pivot point.

so if I followed this correctly.
I_cm would be I_cm = mr^2?
so what we are doing is adding two moments of inertia together?

so I_cm would be
m=94.48 grams
r (I assume it makes sense that r is the distance to the center of mass, not to the end of the ruler) = 16.27 m
so I_cm = 25016.5 grams*cm^2

and then for M(d^2)
m= 94.48
d (distance of center of mass to pivot) = l/2 - d{which is 1/4 of an inch converted to cm} = 15.6369 cm
so this is 23102.1 g*cm^2

48118.6 g*cm^2 is the inertia around the pivot.

but what does this mean? how do I used this with the sqrt(g/L) formula?

6. Oct 25, 2014

### haruspex

Yes. Note that this only works when one of them is the M of I about an axis through the mass centre. The theorem tells you how to adjust it for moment about any axis parallel to it.
No. You are probably supposed to know the I_cm for a rectangular plate, or be able to calculate it. See "Thin rectangular plate of height h and of width w and mass m" at http://en.wikipedia.org/wiki/List_of_moments_of_inertia.
Strictly speaking, what you have here is a rectangular block, but the dimension along the axis (i.e. the thickness of the ruler) has no effect.
How do you get that? It's too much.
You mean cm, right?
You can't. That formula only works for a simple pendulum, in which all the mass is taken to be at the mass centre. That is not the case here. You need the more general formula for a compound pendulum. In this, the length of the pendulum and its radius of inertia feature separately.

7. Oct 25, 2014

### grandpa2390

the density is in g/cm^3 so I multiplied lengthxwidthxdepth and then multiplied that times 1.2.
yes cm. We never discussed all of this in class ever... I have no clue. My physics textbooks don't even talk about this stuff. I am getting a bit annoyed (not at you). Are you sure there isn't another way to do this?

or maybe you know of a similar problem that was solved and I can follow it through the steps?

8. Oct 25, 2014

### haruspex

As I wrote, I don't get the same mass. Please post your working.
The problem describes a classic compound pendulum, so you must use that formula (or derive it yourself from scratch). There is no shortcut.
I note that the problem intro says damped oscillator, but part (a) is undamped, so I'm guessing part (b) gets into the damped case. That is more advanced than having to deal with an undamped compound pendulum. If your textbook and course notes do not cover the parallel axis theorem and {either the compound pendulum formula, or, how to derive the formula from a differential equation} then you have not been equipped to deal with this problem.

9. Oct 25, 2014

### grandpa2390

Thankyou. I am going to have to go visit my professor and ask him what he thinks. It is possible that he doesn't realize how difficult the problem is. He is teaching three different universities at once, and he is scouring the internet for problems. He hasn't even solved them himself yet. So sometimes he gives us problems without realizing they are way over our heads. This might be one of them with the most brilliant student in the class being able but none of the rest of us.

The homework is past due, I am just trying to figure this out so I can do it if it is on the exam this friday. I have started to just do problems in the chapter review of the textbook and I am finding that those are pretty easy. So I am not sure if they are just too easy, and if I will not be fine on the test, or if the test will be like the homework and the chapter review won't help.

I still want to figure this out anyway. Believe it or not, I am majoring in physics and I actually enjoy these mechanics problems... when I know how to do them. But I feel like I should be able to do this if I am in intermediate mechanics. I'm checking my math for the mass now.

10. Oct 25, 2014

### grandpa2390

Ok I should have gotten 23.621 gram for mass
so using the formula
I_cm = m(h^2 + w^2) / 12 = 2113.28
So I would equal 5775.53+2113.28 = 7888.81 g cm^2

and now in case I haven't made another idiotic mistake, I will be looking for the compound pendulum formula. so far we are in ch3 of Thornton and Marion Classical Mechanics. That book is a little difficult for me to understand sometimes, so I supplement with the book by Taylor. I read on this forum it was highly recommended and so far I like it better.

11. Oct 25, 2014

### haruspex

12. Oct 25, 2014

### grandpa2390

that's a lot different than what this guy says:
http://farside.ph.utexas.edu/teaching/301/lectures/node141.html
I think I am going to go with the link you posted.

so 1/(2π([k2 + h2]/gh)1/2 ) multiplied by 2pi

edit: I think since I already have I calculated, I should probably use the other formula
1 / ((I/mgh)^(1/2))

I got .06825Hz .

Last edited: Oct 25, 2014
13. Oct 25, 2014

### haruspex

14. Oct 25, 2014

### grandpa2390

because that formula calculates the period. I'm looking for angular frequency. w=v0*2pi and v0 = 1/T
so period^-1 *2pi = angular frequency right?

15. Oct 25, 2014

### grandpa2390

I can solve part b I think, I just need to make sure my understanding is correct.

1. The problem statement, all variables and given/known data
Damped driven oscillator: ruler example. Suppose the ruler used in the classroom demonstration has a length of 12 and 13/16ths inches, a width of 1 ½ inches, is 1/16th inch thick with a density of 1.2 g/cm3. It swings from a pivot point ¼ of an inch from the top end.

b) Damping: held between a finger and thumb, it swings about 5 times before coming to a rest. If this is taken to be 5 e-folding times (characteristic time for the exponential envelope), find the damping constant β, the “oscillation” frequency ω1, and the decrement of the motion. Include appropriate units in all answers.

2. Relevant equations

Xen = A0e-βt1

ω1 = √(ω02 - β2)

3. My attempt

Well according to my research, an e-fold is the time it takes for an increase by a factor of e. since there are 5 e-folds, I set e^-Bt = 1/(e^5)
however, I have two unknowns. I don't know how long it took for it to complete. I am thinking that once I find the period using part a, that I just multiply the time of each period by 2.5 (1 period = 2 swings?), and then I could plug that in for t and solve for B?

but I also feel like the damping would affect the time of each swing.

but then again, the period is not dependent on the amplitude and that is what is changing...
So I am thinking that could work... or maybe not?

16. Oct 25, 2014

### haruspex

Oh, sorry - I see you have two lots of 2pi, which will cancel.

17. Oct 25, 2014

### grandpa2390

so is that correct?

18. Oct 25, 2014

### haruspex

It's not clear whether one swing is supposed to be one oscillation or half an oscillation.
Yes. In $e^{-Bt} = e^{-5}$, t is the time for 5 swings at the altered period, i.e. at rate $\omega_1$. Combine that with your second equation.

19. Oct 25, 2014

Yes.