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Calling all Oxbridge Physicists!

My oh my, hoofbeat, you have been very busy over the holidays haven't you? I'm shaking in my boots now.

Nice to see you going over the Kraus problem sheets. EM seems to be the hardest thing we're doing. Anyway, I'm getting up early these days, so I might see you in the lectures this time. I don't know quite how I would recognise you though...

Reply 301

Hoofbeat

OP

14

LennonMcCartney

My oh my, hoofbeat, you have been very busy over the holidays haven't you? I'm shaking in my boots now.

Nice to see you going over the Kraus problem sheets. EM seems to be the hardest thing we're doing. Anyway, I'm getting up early these days, so I might see you in the lectures this time. I don't know quite how I would recognise you though...

Lol! Well, I know it "appears" I've been busy, but actually I haven't....see I spend most of the time "trying" to do the questions and then giving up, which really isn't a good technique to have. Besides, I'm actually trying to get to the stage that everyone else was at all term...I couldn't do a single question then. Also, haven't actually been going over Kraus' sheets...been using my tutors and exam questions I found...kraus' notes are bad enough, without having to use his questions too! lol. Recognise me...um...I'm not particularly distinctive! Basically, we're normally sitting about 4 rows down from the back in the middle. I'm always with two other guys and sometimes a really tall guy too. I have long brown hair tied back and I wear glasses! I usually cycle to lectures wearing a "Lincoln" splashjacket! lol

Reply 302

LennonMcCartney

Hoofbeat

Recognise me...um...I'm not particularly distinctive! Basically, we're normally sitting about 4 rows down from the back in the middle. I'm always with two other guys and sometimes a really tall guy too. I have long brown hair tied back and I wear glasses! I usually cycle to lectures wearing a "Lincoln" splashjacket! lol

Well I would give you a general indication of where Queen's people sit, but I haven't been for so long I can't guarantee that they are still in the same place. In fact, I don't think I can even remember how many rows down...but usually on the left somewhere. That didn't help much, did it!?

Reply 303

Hoofbeat

OP

14

QUICK, SOMEONE TELL ME EVERYTHING I NEED TO KNOW ABOUT VECTOR CALCULUS/MULTIPLE INTEGRALS/WAVES/NORMAL MODES AND THEN CIRCUIT THEORY (how the **** do I do time dependant problems *cries*) & EM!!!!

I'm going to die *cries*.....can you tell I'm beginning to panic?

Reply 304

LennonMcCartney

Hoofbeat

QUICK, SOMEONE TELL ME EVERYTHING I NEED TO KNOW ABOUT VECTOR CALCULUS/MULTIPLE INTEGRALS/WAVES/NORMAL MODES AND THEN CIRCUIT THEORY (how the **** do I do time dependant problems *cries*) & EM!!!!

I'm going to die *cries*.....can you tell I'm beginning to panic?

Woah woah woah...slow down, chill out. Honestly, there's more to life than this nonsense.

Reply 305

Hoofbeat

OP

14

I know

but when everyone else around me is worrying it's hard (and I don't mean the other physicists who have either revised enough, or in the majority of cases, revised v.little and prepared to accept what is coming to them!). I just feel really bad as I spent the afternoon lying on our quad in the sunshine watching others (damn those lawyers with no exams til 2007 now!) playing croquet when I should have at least been reading through my notes again!

Reply 306

LennonMcCartney

Well good luck hoofbeat, and don't worry too much -- after all they are only collections.

Reply 307

Hoofbeat

OP

14

Grrrrrrr, our maths one this morning was horrible! It probably wouldn't have been that bad if I hadn't forgotten EVERYTHING I know about waves/normal modes! I'm so annoyed, as thats my favourite topic and I can usually do such questions, but my mind went blank!

Anyways, g2g cram for our EM collection at 2pm! :eek:

Hope everyone elses went well!

Reply 308

Hoofbeat

OP

14

Hey guys, I'm having to redo my collection papers this weekend ready for our tutes next week, so I was wondering if could someone have a go at the following question and tell me if they get 64/3 as the answer? Then I know whether I have to redo this question as well or not (I probably have got it wrong, as I'm not sure about the limits). Thanks

=====
Q. Evaluate by direct integration the outward flux of F=(0,2x,z) over the closed surface of a tetrahedron formed by the planes x=0, y=0, z=0 and 2x + y + 2z = 4.
=====

Thanks :smile:

Hope not everyone's collections went as badly as mine!

Reply 309

RichE

15

Hoofbeat

=====
Q. Evaluate by direct integration the outward flux of F=(0,2x,z) over the closed surface of a tetrahedron formed by the planes x=0, y=0, z=0 and 2x + y + 2z = 4.
=====

I got 0 flux out of the x=0 and z=0 planes.

-8/3 out of the y=0 plane

16/3 out of the 2x+y+2z=4 plane.

So I got 8/3 overall.

[or at least second time around I did]

Reply 310

Bezza

2

If you use the divergence theorem then you just get div F = 1 so isn't the flux just the volume which I reckon is 8/3. I haven't tried to do it the long way though.

8/3 it is

Bezza

If you use the divergence theorem then you just get div F = 1 so isn't the flux just the volume which I reckon is 8/3. I haven't tried to do it the long way though.

Yes, you're absolutely right - the volume of the tetrahedron is 8/3.

1/6 of the scalar triple product of (2,0,0),(0,4,0),(0,0,2)

Reply 313

Hoofbeat

OP

14

could someone go through the surface integral in detail? I'm not allowed to do it using the divergence theorem as that's the next part of the question!!!!

Reply 314

RichE

15

Hoofbeat

could someone go through the surface integral in detail? I'm not allowed to do it using the divergence theorem as that's the next part of the question!!!!

We spoil you hoofbeat :p:

see attached

can someone run through the method to solve a question like this:

a cylinder of radius r and mass m rolls down a slope inclined at angle a, until it has dropped vertically by a height h. provided it doesnt slip when it roles and that it starts initially at rest, calculate it's velocity at the bottom and it's angular velocity.
i'm just getting a liittle st uck with dealing with the angular side of things (all that moment stuff etc)

thanks

and sorry if i can t spell properly my fingers wont straighten themselves atm

Reply 316

Morbo

17

Method: energy considerations
Difficulty rating: easy when you know how

GPE lost = (linear KE + rotational KE) gained

mgh = (1/2)mv^2 + (1/2)Iw^2

I = (1/2)mr^2

4mgh = 2mv^2 + mr^2w^2

Rolling condition: r^2w^2 = v^2

4gh = 2mv^2 + mv^2 ----> v^2 = 4gh/3

w^2 = v^2 / r^2 ----> w^2 = 4gh/3r^2

I think.

Reply 317

Hoofbeat

OP

14

RichE

We spoil you hoofbeat :p:

see attached

Thanks Rich. But a few questions:

1) How did you know when you were working in the z=0 (or xy) plane that the limits of z are 0->2? I understand how you got the limits for x.

2) On the second page, I'm not entirely sure what you're doing! Sorry! Are you doing a cross product to get an area? Can't you just define say G = 2x + y + 2z and then find the unit normal from grad G/|grad G|? but then I'm not sure how the limits would work?

Reply 318

Bezza

2

Hoofbeat

Thanks Rich. But a few questions:

1)Do you have to do each side separately? Can't you just define say G = 2x + y + 2z and then find the unit normal from grad G/|grad G|?

That will only give you the flux through that 1 side of the tetrahedron that's bounded by the plane 2x + y + 2z = 4 whereas you need to sum the flux through all the sides.

2) How did you know when you were working in the z=0 (or xy) plane that the limits of z are 0->2? I understand how you got the limits for x.

If you put in x=y=0 then you see the plane crosses the z-axis at 2, then do similar things to get it crosses the x-axis at 2 and the y-axis at 4 then do a sketch.

3) On the second page, I'm not entirely sure what you're doing! Sorry! Are you doing a cross product to get an area?

I'm not too sure what he's done with that cross product either! He must be doing it to find a normal vector to the plane, but personally I would just use the fact that a plane can also be written as r.n=d so the normal vector is just (2,1,2)

One question I have is why don't you have to use a unit normal vector?

Reply 319

Willa

11

1) if you understand that the limits are x,z >= 0 and x+z<=2 then you either chose x:0->2 and hence z=2-x, or you choose z:0->2 and x=2-z. You are free to put it whatever way you want, with area integrals you choose one to be an independant range (the 0->2 is independant of any variables) and the other range must be dependant on this range.

2)He's cross-producted two vectors to obtain a normal vector....although i dont at this moment see where he grabbed those vectors from. It would be very easy to just look at the equation of the plane to see that the normal vector is just (2,1,2) since it's just the coefficients. Either way those two vectors must be vectors lying on the plane...so they might just have come from taking some basic points on the plane and subtracting one from the other to obtain a vector in the plane.
But then you raise an interesting question...i cant see why he's managed to leave the vector as a non-unit vector. Surely dS must be a unit vector?

Edit: I think i understand now: He is using a projection onto the xz plane....so the dxdz is smaller than the true area element ds...it must be multiplied up...and that has been encompassed in the use of (2,1,2) rather than the unit vector in this direction.

Calling all Oxbridge Physicists!

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