- #1

- 373

- 3

- #1

- 373

- 3

- #2

Doc Al

Mentor

- 45,051

- 1,362

- #3

- 373

- 3

So lets say the mass is at rest at the highest position, and is just about to fall. So its potential energy would be say mg(51) instead of mg(50)? And at the zero reference point (ground), its potential energy is still mg(1)?

- #4

Doc Al

Mentor

- 45,051

- 1,362

If you measure PE from ground level, when the object is resting on the ground its PE = mgd (say). Then when it's at the highest point, resting on top of the track at height h, its PE would be mg(h+d). For some things it doesn't matter, since theSo lets say the mass is at rest at the highest position, and is just about to fall. So its potential energy would be say mg(51) instead of mg(50)? And at the zero reference point (ground), its potential energy is still mg(1)?

It depends on exactly what you are trying to calculate.

- #5

- 373

- 3

I see, I just find it weird that its potential energy would be non-zero at ground level. When would one want to use the center of mass to calculate PE?If you measure PE from ground level, when the object is resting on the ground its PE = mgd (say). Then when it's at the highest point, resting on top of the track at height h, its PE would be mg(h+d). For some things it doesn't matter, since thechangein PE will still be mgh.

It depends on exactly what you are trying to calculate.

- #6

Doc Al

Mentor

- 45,051

- 1,362

Realize that the zero level is arbitrary. OnlyI see, I just find it weird that its potential energy would be non-zero at ground level.

Depends on how accurate you want to be and what you are asked to calculate.When would one want to use the center of mass to calculate PE?

- #7

- 373

- 3

In the classic problem, where it asks to calculate the energy needed to get pass the loop, is this required? I'm having trouble thinking up with situations where it would want to be done. What is the size difference to make this matter?Realize that the zero level is arbitrary. Onlychangesin gravitational PE have meaning.

Depends on how accurate you want to be and what you are asked to calculate.

- #8

Doc Al

Mentor

- 45,051

- 1,362

Think about it. What's the PE initially? What's the PE of the object when it passes the top of the loop? (Note that the object is nowIn the classic problem, where it asks to calculate the energy needed to get pass the loop, is this required?

It depends how accurate you want to be. Many versions of the problem don't even give the dimensions of the object, so you are expected to ignore it.I'm having trouble thinking up with situations where it would want to be done. What is the size difference to make this matter?

- #9

- 373

- 3

For without accounting for the center of mass:Think about it. What's the PE initially? What's the PE of the object when it passes the top of the loop? (Note that the object is nowbelowthe track.) Since centripetal acceleration is involved, what radius will you use?

y = 2r + 0.5r

For accounting for the center of mass,

y = 2r + 0.5r - 2x

(y is the minimum height needed to pass the loop, r is radius of the loop, and x is the distance to the center of mass)

So it would matter right?

- #10

Doc Al

Mentor

- 45,051

- 1,362

OK.For without accounting for the center of mass:

y = 2r + 0.5r

How did you arrive at this?For accounting for the center of mass,

y = 2r + 0.5r - 2x

Sure. As to how much it matters, that depends on how x compares to r.So it would matter right?

- #11

- 373

- 3

mg(y+x) = mg(2r-x) + 0.5mrgHow did you arrive at this?

y + x = 2r - x + 0.5r

- #12

Doc Al

Mentor

- 45,051

- 1,362

Check that 0.5mrg term. Recall my comment about what radius to use.mg(y+x) = mg(2r-x) +0.5mrg

y + x = 2r - x + 0.5r

- #13

- 373

- 3

0.5mrg should be 0.5m(r-x)g?Check that 0.5mrg term. Recall my comment about what radius to use.

- #14

Doc Al

Mentor

- 45,051

- 1,362

Right.0.5mrg should be 0.5m(r-x)g?

- #15

CWatters

Science Advisor

Homework Helper

Gold Member

- 10,532

- 2,298

- #16

Doc Al

Mentor

- 45,051

- 1,362

That will certainly make things more interesting!

I, of course, was speaking of the simpler case of a rigid object (a cart, for example) with a well-defined center of mass.

- #17

CWatters

Science Advisor

Homework Helper

Gold Member

- 10,532

- 2,298

With a cloud the centre of mass might be the only think you can estimate the position of.

- Last Post

- Replies
- 6

- Views
- 10K

- Last Post

- Replies
- 3

- Views
- 4K

- Last Post

- Replies
- 1

- Views
- 591

- Last Post

- Replies
- 2

- Views
- 1K

- Replies
- 2

- Views
- 296

- Last Post

- Replies
- 8

- Views
- 10K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 8

- Views
- 2K