Mechanics

Original post by Student403

Ouch

If you think that's bad, then it doesn't get any better: after things got messy, I read up on the derivation of curvature in two separate books, to check that I hadn't forgotten something important ..

Reply 62

TeeEm

OP

Original post by atsruser

The most annoying thing is that the problem looks fairly straightforward when done properly.

Here's a question though, that your problem caused me to ponder: what type of problems in mechanics are intractable or very difficult if you don't try to use intrinsic coordinates? I think that this would be tricky without them (I had a go but it looked messy) but in general, I'm not sure what characterises such problems.

Intrinsics:
constrained motion in a path (or constrained until it flies off)
path is not circular or straight (otherwise equations greatly simplify)
no central force (otherwise it reduces to polars)

typical example motion of a bead sliding on a vertical wire bent to some known shape

PS
When I was making this question I wanted to use a parabola like y = -x²but the intrinsic equation for such a simple curve was horrendous.

(edited 8 years ago)

Reply 63

A Slice of Pi

13

Original post by TeeEm

I hope that @A Slice of Pi also puts up his solution as he got the same answer as I

Sorry I'm late with this - just got back from college. Attached is my solution, not as neat as yours, obviously :h:

.

OP

Original post by A Slice of Pi

Sorry I'm late with this - just got back from college. Attached is my solution, not as neat as yours, obviously :h:

.

excellent
it is the same ...
I used the tangential equation to get an expression for v² via calculus, but energies work just as well

Reply 65

Student403

Original post by atsruser

If you think that's bad, then it doesn't get any better: after things got messy, I read up on the derivation of curvature in two separate books, to check that I hadn't forgotten something important ..

I'm sorry

Reply 66

Euclidean

Original post by A Slice of Pi

Sorry I'm late with this - just got back from college. Attached is my solution, not as neat as yours, obviously :h:

.

I have no idea what most of it means but it's impressive, well done! :biggrin:

Reply 67

A Slice of Pi

13

Original post by Euclidean

I have no idea what most of it means but it's impressive, well done! :biggrin:

Thanks. It's actually not that bad, but it's not taught at A-Level anymore it seems...

Reply 68

Euclidean

Original post by A Slice of Pi

Thanks. It's actually not that bad, but it's not taught at A-Level anymore it seems...

Yeah, it's a shame. Looks like an interesting question. Hopefully I'll get round to studying the content and give it a go

Posted from TSR Mobile

Reply 69

Original post by atsruser

Oh my god, no wonder I went astray. The first line of my working after writing down the equations of motion is

\psi = \frac{dy}{dx} = -\sinh x

. It's time to shut up the maths books for good.

Did you not have to do this in your A-levels? (I guess not if you didn't do further mechanics.)

This would've been a standard M6 question (not sure why TeeEm thinks it's harder...)

Reply 70

TeeEm

OP

Original post by shamika

Did you not have to do this in your A-levels? (I guess not if you didn't do further mechanics.)

This would've been a standard M6 question (not sure why TeeEm thinks it's harder...)

I have every question on this topic which has appeared in an EDEXCEL exam and from those questions the hardest is too easy compared to this particular question. (this topic appeared from 1993 to 2006)

What makes it hard is the lack of structure.

Also practically every question in exams the curve is given in intrinsic form

Reply 71

Original post by TeeEm

I have every question on this topic which has appeared in an EDEXCEL exam and from those questions the hardest is too easy compared to this particular question. (this topic appeared from 1993 to 2006)

What makes it hard is the lack of structure.

Also practically every question in exams the curve is given in intrinsic form

Ok. I thought M6 was a lot harder than this. Maybe I'm misremembering it! I'm going to see if I can find my M6 papers now lol!

Reply 72

TeeEm

OP

Original post by shamika

Ok. I thought M6 was a lot harder than this. Maybe I'm misremembering it! I'm going to see if I can find my M6 papers now lol!

Sure ....
I might make a thread on M6 as I have all the papers and upload them as some students might have an interest in them

Reply 73

Original post by TeeEm

Sure ....
I might make a thread on M6 as I have all the papers and upload them as some students might have an interest in them

Lol my paper was quite easy! I didn't remember you were given curves in intrinsic form already. I forgot I did it in 2003, the 2002 paper had a 10 mark question which was an easy M1 question :s-smilie:

I'm a bit disappointed now lol! There was 16 marks for a simple projectile question on my paper!

Reply 74

Euclidean

Original post by TeeEm

Sure ....
I might make a thread on M6 as I have all the papers and upload them as some students might have an interest in them

I'd be interested in a thread like this! :u:

Reply 75

Original post by shamika

Did you not have to do this in your A-levels? (I guess not if you didn't do further mechanics.)

I'm not sure what "this" is. If it's "intrinsic coordinates" in general, then the answer is no, I didn't cover these at A level - they weren't on the A level or FM syllabus of any board at the time, AFAIK, though I can't be certain about that. Degree level only in my day.

Reply 76

Original post by Euclidean

Yeah, it's a shame. Looks like an interesting question. Hopefully I'll get round to studying the content and give it a go

Posted from TSR Mobile

The question relies on using the so-called "intrinsic coordinates" of a curve. The idea is that we know how to handle the motion of a particle constrained to circular motion, and we extend this to more general curvilinear motion.

The physics is as follows: a particle constrained to a smooth curve (say a bead on a wire) knows only about its local environment - it can feel only the forces exerted on it by the wire and external forces at its current location, P.

At any instant in time, the particle has a velocity tangent to the wire, and feels an component of external force accelerating it in the direction of tangent. Also, it feels a force exerted by the wire normal to the tangent at that point. Locally the wire is curved, and over an infinitesimal distance before and after P, that curvature is constant. So at P, we can model the wire as a circle of radius

\rho

, chosen to give the same curvature as that of the wire at P. So when the particle is at P, we can write down normal and tangential equations of motion, if we know the curvature, by treating the particle as being on a circle, instantaneously.

Of course, an instant later the bead is at another point and the tangent direction has changed, and so has the radius of curvature

\rho

, so we need to be able to write down the equations of motion in a general form, taking into account the fact that the tangent and

\rho

are functions of the position, and that from the POV of the bead, it's now on a circle with a different curvature.

To do so, we measure the position of the particle with coords

P(s,\psi)

where

s

is the arc length along the curve measured from some origin, and

\psi

is the angle that the tangent at P makes with the x-axis. These coords are used because it turns out that there are nice formulae relating them to the radius of curvature of the curve at any point - in fact, you can show that

\rho=\frac{ds}{d\psi}

, and since, if we have the equation of the constraining curve in the form

y=f(x)

(as we usually do), then also

\frac{dy}{dx}=\tan \psi

, so with a bit of work we can use the cartesian eqn of the curve to calculate

\rho

at any point

(x,y)

. [Fun exercise: work out the details yourself].

Having done all that, we can write down the radial and tangential equations of motion by noting that

\dot{s}=v

, the tangential velocity, and

\ddot{s}

is the tangential acceleration so we have, by Newton II:

F_t = m\ddot{s}

F_n = m\frac{\dot{s}^2}{\rho}

by using the results from circular motion, generalised to our situation where we don't have a constant radius of curvature

r

, but a variable one, namely

\rho

.

Reply 77

Original post by atsruser

I'm not sure what "this" is. If it's "intrinsic coordinates" in general, then the answer is no, I didn't cover these at A level - they weren't on the A level or FM syllabus of any board at the time, AFAIK, though I can't be certain about that. Degree level only in my day.

Sorry, I did mean intrinsic co-ordinates. This thread has made me realise that I have a romanticised view of my A-level days, where I thought the material was more advanced than it actually was!

Reply 78

Original post by shamika

Sorry, I did mean intrinsic co-ordinates. This thread has made me realise that I have a romanticised view of my A-level days, where I thought the material was more advanced than it actually was!

Not sure I'm getting the point, TBH. You now think that the material is *less* advanced than you previously thought?

Reply 79

Sciatic