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Spot the Stupid Mistake (Dynamics)

Exactly what the title says. I can't seem to get the right answer, so it's either a stupid mistake or complete misunderstanding on my part. I hope it's the former :redface:

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Reply 1

tomkent

Ah i cant wait till uni maths :smile:

sorry cant help, if i understood it then i would gladdy... but alas!

Reply 2

steve10

Just a thought, but ...

You say that R is perp. to r_dot, but isn't R the centripetal force, acting towards the origin, whereas r-dot is perpindicular to the z-axis. So in that case, they won't be perpindicular.

Reply 3

James

steve10

You say that R is perp. to r_dot, but isn't R the centripetal force (tension) in the string, acting towards the origin, whereas r-dot is perpindicular to the z-axis. So in that case, they won't be perpindicular.

That's actually right, although it's my fault that you were confused since I forgot to underline those vectors.

Vector r_dot is the velocity of the particle at any given time, which is in a plane tangent to the hemisphere.

R is the normal reaction force, which is always perpendicular to the surface of the hemisphere.

Reply 4

James

Thank you Worzo, that's a great help. It's very kind of you to answer in such detail. I guess my main mistake was forgetting that dr/dt = constant.

Reply 5

Morbo

Uh oh, I might have made a mistake. I thought it was working in spherical coordinates, but it's defined cylindrincal coordinates for some reason, so dr/dt is definitely not constant then. It's not a very clear question. I'm not even sure which way the ball is projected now. Is it projected around the inside of the hemisphere, like it is "orbiting" inside the hemisphere? Or it is projected towards the base of the hemisphere?

Let me think again.

Reply 6

James

Any luck? I agree that it's seriously odd to be using cylindrical polar coordinates, but I guess that's how they want us to do it.

Reply 7

Morbo

Looking at it again, I think what is that this hemisphere is sitting there like a bowl. It says the particle starts at z=0 - this means it's sitting right next to the rim of the bowl. It is projected in the theta direction - I think this means that it's projected around the rim. Spherical coordinates would seem more natural, but it's probably slightly easier in cylindrical coordinates if that's what they've chosen to use.

Too late to think about it now - just need to find out a general expression for the velocity, which I can't do.

Reply 8

James

I've tried this again and still don't get it. Perhaps, rather than looking through my working, someone could approach the question from scratch. I am trying to imitate a solution I have to an equivalent question set in a cone, but it's not working out.

Reply 9

Popa Dom

I just gave it a quick go, but my dynamics is pretty rusty so its almost certainly wrong. I got it going betwee z=a and z=v^2(a^2-1)/2g. Just used the energy equation, subbed in for r and dtheta/dt in terms of z, and rearranged to get dz/dt=(a^2-z^2)(v^2(1-a^2)+2gz)/a^2

EDIT: wait a minute, just noticed I forgot to square something, so thats wrong, let me get back to you in a mo

Reply 10

Popa Dom

At the end where you have 2agz, are you sure that shouldnt be 2a^2gz, which is what I ended up with, cos that means you can factor by (a^2-z^2) so you get one solution z=a and a quadratic for the other (aswell as z=0). The only reason I say this is cos I got it without ever expanding the (a^2-z^2) (I cant actually physically read through other peoples work and understand it, so I'm not sure)

EDIT: Ok, just thinking about it z cant actually equal a can it, otherwise you'd end up dividing by 0 at some point. I reckon its justthe 2 solutions to the quadratic then, z=1/4g * v^2+-root(16g^2a^2-v^2). I'll go check maple

Reply 11

James

Yes, I lost an 'a' in the very last step. Thanks.

Does it work out, then? Does the second part of the question naturally follow?

Reply 12

Kolya

Christ, what happened to your profile picture Dom? (Or is that what taking Maths at university does to one's mind?)

Reply 13

James

Did you get that yourself, or looking at my work for 'inspiration'? Because I came to exactly the same conclusion, and similarly gave up. And how do you do the second part of question 3?

Reply 14

redfox2

James, I know uve probably handed it in by now, but I think that the mistake you made was when you wrote "Now we have...". In that equation for the r.theta term uve put an r^2 when it is simply r. This means that it is constant since there is no acceleration in the theta direction. I might be wrong, but that looks to be the source of the problem...