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Circular Motion: Mechanics Help!!! Urgent

So this is the question:

Screen Shot 2013-11-17 at 2.18.50 PM.png

And this is the solution:
Screen Shot 2013-11-17 at 2.20.30 PM.png

Isn't the the GPE for the bead when it's on the top

mga(1+cosx)

?

Why is the KE=GPE?
All in all can someone please explain the solution.

(edited 10 years ago)

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Reply 1

TenOfThem

Original post by MAyman12

Isn't the the GPE for the bead when it's on the top

mga(1+cosx)

yes

Reply 2

MAyman12

Original post by TenOfThem

yes

Ok, then why is it the opposite in the markshceme?

Reply 3

TenOfThem

Original post by MAyman12

Ok, then why is it the opposite in the markshceme?

where

Reply 4

MAyman12

Original post by TenOfThem

where

Reply 5

TenOfThem

Original post by MAyman12

Oh, I looked at the second line :smile:

I have not done this recently so will hope for someone more current to answer :frown:

Reply 6

ghostwalker

Original post by MAyman12

So this is the question:

Screen Shot 2013-11-17 at 2.18.50 PM.png

And this is the solution:
Screen Shot 2013-11-17 at 2.20.30 PM.png

Isn't the the GPE for the bead when it's on the top

mga(1+cosx)

?

Why is the KE=GPE?
All in all can someone please explain the solution.

GPE at A is whatever, it doesn't matter.

They're using "gain in KE" = "loss in GPE", not KE = GPE.

At the top angle theta (I'll call it x) we have loss in GPE = mg ( a -acosx)
Hence mga(1-cosx)

Reply 7

brianeverit

Original post by MAyman12

So this is the question:

Screen Shot 2013-11-17 at 2.18.50 PM.png

And this is the solution:
Screen Shot 2013-11-17 at 2.20.30 PM.png

Isn't the the GPE for the bead when it's on the top

mga(1+cosx)

?

Why is the KE=GPE?
All in all can someone please explain the solution.

Gain in KE =Loss of GPE

Reply 8

MAyman12

Original post by ghostwalker

GPE at A is whatever, it doesn't matter.

They're using "gain in KE" = "loss in GPE", not KE = GPE.

At the top angle theta (I'll call it x) we have loss in GPE = mg ( a -acosx)
Hence mga(1-cosx)

Ok. What about the GPE when it's at the bottom?

Isn't the GPE at the bottom mga(1-cosx) when x<pi/2?

(edited 10 years ago)

Reply 9

MAyman12

Original post by brianeverit

Gain in KE =Loss of GPE

But what about the GPE of the bead when it's at the bottom?

Reply 10

ghostwalker

Original post by MAyman12

Ok. What about the GPE when it's at the bottom?

Isn't the GPE at the bottom mga(1-cosx) when x<pi/2?

No, the loss in GPE is mg ( a +a cosx) at the bottom position.

Edit: Position B2 in the diagram should be on the left, but I'm not redoing the whole thing.

(edited 10 years ago)

Untitled.jpg17.4KB

Reply 11

MAyman12

Original post by ghostwalker

No, the loss in GPE is mg ( a +a cosx) at the bottom position.

You said that the loss in GPE energy was mga(1-cosx) in the previous post.

Reply 12

ghostwalker

Original post by MAyman12

You said that the loss in GPE energy was mga(1-cosx) in the previous post.

That's at the top position, B1 in my diagram.

Reply 13

MAyman12

Original post by ghostwalker

That's at the top position, B1 in my diagram.

So with respect to the position A the GPE lost at B1= mga(1-cosx) and at B2 the GPE lost= mga(1+cosx)?

So gain in KE1=mga(1-cosx)

gain in KE2=mga(1+cosx)

Reply 14

ghostwalker

Original post by MAyman12

So with respect to the position A the GPE lost at B1= mga(1-cosx) and at B2 the GPE lost= mga(1+cosx)?

So gain in KE1=mga(1-cosx)

gain in KE2=mga(1+cosx)

Yes, and those are the first two lines of the given solution.

Note: Just realised that my B2 position should be on the left rather than the right, but it's a mirror image, and all the values for KE, GPE are the same, and I'm not doing the diagram again.

Reply 15

MAyman12

Original post by ghostwalker

So for the rest of the question the Kinetic energy at B1 is equal to the GPE at B2 since the conversation of energy right?

Reply 16

ghostwalker

Original post by MAyman12

So for the rest of the question the Kinetic energy at B1 is equal to the GPE at B2 since the conversation of energy right?

Huh! Where did that come from?

Reply 17

MAyman12

Original post by ghostwalker

Huh! Where did that come from?

The loss of GPE at B1+Loss in GPE at B2= Total energy, But since Loss of GPE is equal to gain in KE, then GPE at B1 must be equal to KE at B2. The difference between B1 and B2 is pi.

Reply 18

ghostwalker

Original post by MAyman12

The loss of GPE at B1+Loss in GPE at B2= Total energy.

This bit doesn't make any sense to me.

In going from A to B1 we have a loss in GPE, which was worked out previously.

In going from A to B2 we have a different loss in GPE, again previously worked out.