Energy loss of damped oscillator

[vel]

Homework Statement

A 2.66-kg rock is attached at the end of a thin, very light rope 1.45 m long and is started swinging by releasing it when the rope makes an 11.9 ∘ angle with the vertical. You record the observation that it rises only to an angle of 4.10 ∘ with the vertical after 10.5 swings.

Relevant Equations

To be honest, I have no idea. The section of our chapter had nothing on equations and we didn't talk about this in class. My best guess would be E = 1/2mv^2 + 1/2kx^2.

yeah, I don't even know how to start

 

[haruspex]

My best guess would be E = 1/2mv^2 + 1/2kx^2.

Where k and x are what, here?

 

[vel]

I'm thinking x would be replaced with the angles, but I have no idea for k, especially as this is a rope and not a spring. The only other thing I'm given is that it's E-inital - E-final.

 

[haruspex]

I'm thinking x would be replaced with the angles, but I have no idea for k, especially as this is a rope and not a spring. The only other thing I'm given is that it's E-inital - E-final.

Treating the rope as elastic, x would be its change in length. Since you are not given any elasticity data for it, assume the length does not change.

You also quoted a formula for KE. You could use that, but since you are not given a velocity that's a roundabout route.
What other form of energy changes in this problem?

 

[vel]

Where k and x are what, here?

I'm thinking x would be replaced with the angles, but I have no idea for k, especially as this is a rope and not a spring. The only other thing I'm given is that it's E-inital - E-final.

Treating the rope as elastic, x would be its change in length. Since you are not given any elasticity data for it, assume the length does not change.

You also quoted a formula for KE. You could use that, but since you are not given a velocity that's a roundabout route.
What other form of energy changes in this problem?

None, as far as I'm aware

 

[haruspex]

I'm thinking x would be replaced with the angles

Reread post 4. Or did you accidentally repeat post #3?

None, as far as I'm aware

Then why does the rock move when released?

 

[vel]

Reread post 4.

Then why does the rock move when released?

(sorry that posted again, I have no idea why it did)

Oh! that! yeah, I didn't even think of that lol. The rock would start at rest and gravity, so -9.8 m/s^2, would be its acceleration.
... I could try to use that to find velocity? Probably not what I need to be doing but it's all my brain is going to for the moment.

 

[haruspex]

I could try to use that to find velocity?

You could, but there's a more direct method. As a body falls, what form of energy is it losing?

 

[vel]

You could, but there's a more direct method. As a body falls, what form of energy is it losing?

Potential, right? I remember that PEgrav = mgh

 

[haruspex]

Potential, right? I remember that PEgrav = mgh

Right. So how much energy was lost over the 10.5 swings?

 

[vel]

Right. So how much energy was lost over the 10.5 swings?

I did 2.66 x -9.8 x (11.9 - 4.1) = -203.3
Using the rope length, it would be -37.8. (the first makes more sense to me tho)

 

[vel]

I did 2.66 x -9.8 x (11.9 - 4.1) = -203.3
Using the rope length, it would be -37.8. (the first makes more sense to me tho)

never mind i put both answers in and they're both wrong

 

[haruspex]

I did 2.66 x -9.8 x (11.9 - 4.1) = -203.3

Now there's a wild guess.
Gravitational Potential Energy is related to vertical displacement. How much lower is the rock after 10.5 swings than when it started?

 

[vel]

Now there's a wild guess.
Gravitational Potential Energy is related to vertical displacement. How much lower is the rock after 10.5 swings than when it started?

I thought that was the vertical displacement--The starting angle is 11.9, the ending is 4.1. I have no idea what else I would do to get vertical displacement within this problem.

 

[haruspex]

I thought that was the vertical displacement--The starting angle is 11.9, the ending is 4.1. I have no idea what else I would do to get vertical displacement within this problem.

Drawing a diagram should be your first step.
Draw a circle of radius rope length. Centre, O, is where the rope is attached to a support.
Draw a vertical radius OA down from the centre to represent the rope at the bottom of its swing, and two more radii, OP, OQ from the centre making those two angles to the vertical.
Finally, horizontal lines from P and Q, meeting OA at P', Q'.
You need the vertical displacement from PP' to QQ'.

 

[vel]

I'm sorry, I drew it out but it's not making sense.

 

[haruspex]

I'm sorry, I drew it out but it's not making sense.

Please post your diagram.

 

[vel]

Please post your diagram.

 

[haruspex]

View attachment 288883

I can't see where P' and Q' are in your thumbnail sketch.
Remember, PP' and QQ' are horizontal lines, so PP'O should be a right angle.
Try again, making the diagram larger.

You can put P and Q on opposite sides of OA since the rock makes 10.5 complete oscillations, i.e. it finishes on the opposite side from where it started.

 

[vel]

I can't see where P' and Q' are in your thumbnail sketch.
Remember, PP' and QQ' are horizontal lines, so PP'O should be a right angle.
Try again, making the diagram larger.

You can put P and Q on opposite sides of OA since the rock makes 10.5 complete oscillations, i.e. it finishes on the opposite side from where it started.

 

[haruspex]

The distances OA, OP, OQ should all be the same. They lie on the circle (which you are now leaving out of the diagram). So A, P' and Q' will be at different heights

 

[vel]

The distances OA, OP, OQ should all be the same. They lie on the circle (which you are now leaving out of tge diagram). So A, P' and Q' will be at different heights

 

[vel]

More like these, then?

 

[haruspex]

More like these, then?View attachment 288885

Unfortunately the image you posted only shows part of the best-looking diagram.

 

[vel]

The very bottom one? that one doesn't have a circle, it's one I sent earlier (#20)

 

[haruspex]

The very bottom one? that one doesn't have a circle, it's one I sent earlier (#20)

Yes, I see now it is just a truncated version of the same sketch. Oddly enough, the portion in post#23 is ok, it's the cut off part that's wrong.
But the other sketches in post #23 are still quite wrong. Why is it so hard for you to draw a HORIZONTAL line from P? You seem to insist on P' and Q' being the same point.

 

[vel]

Yes, I see now it is just a truncated version of the same sketch. Oddly enough, the portion in post#23 is ok, it's the cut off part that's wrong.
But the other sketches in post #23 are still quite wrong. Why is it so hard for you to draw a HORIZONTAL line from P? You seem to insist on P' and Q' being the same point.

I didn't even realize I had it on an angle, lol.

 

[vel]

I didn't even realize I had i

I didn't even realize I had it on an angle, lol.
View attachment 288891

this would be a sine problem, so I went ahead and did that. See any issues there (or with the diagram?)?

 

[haruspex]

I didn't even realize I had it on an angle, lol.
View attachment 288891

Ok, now use trig to find distances OP' and OQ' in terms of the rope length and the angles.
(You really should make bigger diagrams - or learn to draw them much more accurately.)

 

[vel]

Ok, now use trig to find distances OP' and OQ' in terms of the rope length and the angles.
(You really should make bigger diagrams - or learn to draw them much more accurately.)

I think we posted at the same time, lmao. I have the sine equations in #28. If I did it that all correctly and my line of thought is right, it should be PEgrav = 2.66(-9.8)(.401) = -10.45

 

[haruspex]

I think we posted at the same time, lmao. I have the sine equations in #28. If I did it that all correctly and my line of thought is right, it should be PEgrav = 2.66(-9.8)(.401) = -10.45

I cannot decipher those scribblings. Please take the trouble to type in your working (as specified by the forum rules). Use the labels I made for the points to represent line lengths, e.g. OP'.

 

[vel]

I cannot decipher those scribblings. Please take the trouble to type in your working (as specified by the forum rules). Use the labels I made for the points to represent line lengths, e.g. OP'.

PP': sin(11.9) = opposite (PP')/1.45 -> 1.45sin(11.9) = .298
QQ': 1.45sin(4.1) = .103
.298 + .103 = 4.01

 

[haruspex]

PP': sin(11.9) = opposite (PP')/1.45 -> 1.45sin(11.9) = .298
QQ': 1.45sin(4.1) = .103
.298 + .103 = 4.01

PP' and QQ' are horizontal distances. You are trying to find the vertical distance between P and Q.

 

[vel]

PP' and QQ' are horizontal distances. You are trying to find the vertical distance between P and Q.

OP': 1.45cos(11.9) = 1.42
OQ': 1.45cos(4.1) = 1.45
OP' - OQ': 1.42 - 1.45 = -.003

 

[haruspex]

OP': 1.45cos(11.9) = 1.42
OQ': 1.45cos(4.1) = 1.45
OP' - OQ': 1.42 - 1.45 = -.003

Better, but that's a bit inaccurate because you are taking the difference of two numbers that are rather close together. Keep more digits through the calculation.