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Oxbridge Physics Prelims Revision

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Reply 740
Avatar for RichE
RichE
15
Willa
no i am after a linear expression, but related to angular velocity!!

So far I have this:

Iw = Qr
Q = mwL/2
given that I = 1/3 mL^2, r=2L/3
which looks right. Can you agree on my second equation used there!?


All seems fine to me. (Haven't done this stuff in years though.)
Reply 741
Avatar for Willa
Willa
11
RichE
All seems fine to me. (Haven't done this stuff in years though.)


so when considering the instanteous linear motion of a rotating body you DO condense mass down onto the centre of mass?
Yes subject to a condition I can't remember (but one which usually is always satsified for questions).

Sorry for this sh*t answer, the details should be in any good mechanics textbook.
Reply 743
Avatar for RichE
RichE
15
Willa
so when considering the instanteous linear motion of a rotating body you DO condense mass down onto the centre of mass?


how does this relate to the previous example - that was a purely rotating system?

If you wish to consider a moving and rotating frame you get evil equations of motion like

a = a' + w' x r' + 2w x r' + 2w x v' + w x (w x r')

But I still don't see how what you're asking relates to the previous problem
Reply 744
Avatar for Willa
Willa
11
RichE
how does this relate to the previous example - that was a purely rotating system?

If you wish to consider a moving and rotating frame you get evil equations of motion like

a = a' + w' x r' + 2w x r' + 2w x v' + w x (w x r')

But I still don't see how what you're asking relates to the previous problem


because the previous problem was exactly the same situation! a rod pivotted about one end falling to the vertical! we concluded you must use 0.5Iw^2 cos it is rotating, but arguably couldnt you say it has 0.5mLw momentum, which means it has p^2/2m = energy = mgL/2. p^2 = m^2gL = (0.5mLw)^2 = m^2L^2w^2/4. So w^2 = 4g/L

that is my problem...using instantenous linear momentum you dont get the right answer!
Reply 745
Avatar for RichE
RichE
15
Willa

that is my problem...using instantenous linear momentum you dont get the right answer!


well if you could apply a linear momentum argument like that the whole system potentially could start moving and the pendulum is fixed at a pivot

if everything was moving freely then I think you could apply your argument
Reply 746
Avatar for Hoofbeat
Hoofbeat
OP
14
Bezza
Thanks, it's much better this morning, though a funny colour! Hopefully I won't be limping too much!

Thanks guys for the help with the complex no's. What did you do to your ankle Nick?
Reply 747
Avatar for Bezza
Bezza
2
Hoofbeat
Thanks guys for the help with the complex no's. What did you do to your ankle Nick?

Fell off my bike while messing around in the science aera :redface: Theres a thread in H&R if you want to see a photo of it last night!
Bezza
Fell off my bike while messing around in the science aera :redface: Theres a thread in H&R if you want to see a photo of it last night!


This country's academic elite... :rolleyes: :p: :eek:
Reply 749
Avatar for Willa
Willa
11
i often worry when i think that the people around me may well go on to run the country or something to that effect!
Reply 750
Avatar for RichE
RichE
15
~Raphael~
This country's academic elite... :rolleyes: :p: :eek:


I'm sure Bezza would make a better PM than Blair ...

... and Oxford is the place to go if you wish to run the country - so Willa you have nothing to worry about :rolleyes: :p:
Reply 751
Avatar for Willa
Willa
11
RichE
I'm sure Bezza would make a better PM than Blair ...

... and Oxford is the place to go if you wish to run the country - so Willa you have nothing to worry about :rolleyes: :p:


that makes me worry even more!!!!!!!
Reply 752
Avatar for RichE
RichE
15
Willa
that makes me worry even more!!!!!!!


success breeds envy I guess :rolleyes:

Well you know what to do with your angular momentum next time, Willa :biggrin: :eek:
Reply 753
Avatar for Willa
Willa
11
RichE
success breeds envy I guess :rolleyes:

Well you know what to do with your angular momentum next time, Willa :biggrin: :eek:


no i still dont...cos i still dont understand why it's unacceptable to consider instantaneous linear momentum!
Reply 754
Avatar for RichE
RichE
15
Willa
no i still dont...cos i still dont understand why it's unacceptable to consider instantaneous linear momentum!


well certainly you can't to a pivoted system 'cause it can't move freely!
Is this the equivalent of writer's block? :p:
Reply 756
Avatar for Willa
Willa
11
but the thing is we applied linear momentum to the impulse stopping the rod rotating...and it works there! why can it work there and not elsewhere! i just dont get it! The linear momentum of the body is definitely mwl/2 if we are to be able to use that in the second question. But then surely you can say that is the instanenous momentum of the whole body...and hence work out it's instanteous energy from that!?
Reply 757
Avatar for RichE
RichE
15
Willa
but the thing is we applied linear momentum to the impulse stopping the rod rotating...and it works there! why can it work there and not elsewhere! i just dont get it! The linear momentum of the body is definitely mwl/2 if we are to be able to use that in the second question. But then surely you can say that is the instanenous momentum of the whole body...and hence work out it's instanteous energy from that!?


well if we applied a greater impulse then consider the difference in what would have happened between a free system and a constrained system

If the system were free at that instant onwards the rod would start moving backwards in the vertical. If constrained such a motion is impossible and I guess the pivot gets jarred and takes some of the momentum.

But this is an educated guess, as I'm arguing on physical grounds and as a (mainly pure) mathematician usually prefer to deal with the numbers.

Does the SR issue make more sense to you now?
Reply 758
Avatar for Willa
Willa
11
RichE


Does the SR issue make more sense to you now?


unfortunately i have still not resolved that issue to my satisfaction. I devised an unpleasent way of making it work, which involves a way of laying out a length in space as time goes by. But I dont like it as an explanation because it is time dependant. I want to look at the situation from a purely time independant case.

E.g: Stationary observer, S, says "That object is distance x away from me" (this is an object stationary in S frame). S' (moving observer) says "what you on about, it's much closer than that...it's only x' away from me". Hence x = gx'....which i fully understand. And that is what we were taught to be lorentz contraction.

I then imagine another situation, this time something stationary in S' frame. This time the roles are reversed, x' = gx. I fully understand that situation as well.

If you can try and explain how these situations relate to the lorentz transformation then that might help. Because I fail to understand how time comes into these situations

edit: As a guide to the way i think, LM tried to do some maths a few pages back but ended up with x' = gx and tried to claim that x' = x/g from that...which didnt work. If you can fix his working then that would solve my problem, becuase my working (and thinking) is identical to the working he has presented on that page!

LM
we find l′=x1′−x0′l'=x_1'-x_0'. This gives us:
γ(l−βcT)−γ(0−βcT)=\gamma(l-\beta cT)-\gamma(0-\beta cT)=
γ(l−βcT−0+βcT)=\gamma(l-\beta cT-0+\beta cT)=
γl\gamma l.

The point here is that you (in frame S) measure the length of a rod at rest, and it measures l. Now it is moving past you at speed v, it will measure l/γl/\gamma


That!
lol Willa you owe so much rep to RichE!

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