umep physics thread

[/0]

But the problem is if you take the pivot as the centre, then how do you find the magnitude of the frictional force? The gravitational force comes easily as ... but what is the expression for friction?

[evaporade]

what is the ratio of its final centre-of-mass velocity compared to a hollow cylinder rolling down a hill with friction?

Sqrt2 : 1

[dcc]

But the problem is if you take the pivot as the centre, then how do you find the magnitude of the frictional force? The gravitational force comes easily as ... but what is the expression for friction?

Who says there'll even be a torque at all, without friction?  With friction, its clear to see that the torque being applied is proportional to the normal force (and hence the gravitational force), but without friction, that torque isn't there to turn the object.

The gravitational force on this object (assuming its nice and circular/cylinderish) will act on the centre of mass, and hence will not apply a torque to the object (Think about it, if you have a plank in a gravitational field, the torques on the left of the plank will 'even out' with the torques on the right, so there will be no net torque).

At least, that's what I think it is.  This could be wrong, but that's what I would have written if someone asked me

[/0]

Yeah lol I guess you're right, thanks guys

[evaporade]

what is the expression for friction?

1/2 mgsin@

[kamil9876]

Yes. It's sort of like, an object doesnt spin in free fall because each particle has the same velocity. But if say on the right of the object we place a wall that exerts friction on the right side only then the right side will be slower than the left and so turning happens. However if the wall was frictionless then the affect would be the same as if there was no wall at all. (This is what I heard from some apparent physics expert, not my original idea/analogy)

So maybe this analogy extends to the incline as it is a case of some horizontal component of free fall.
The way I imagine it is to treat the sphere as a polyon. In a rotating polygon, only one point touches the ground for some time, and it must be stationary. In order for that atom to be stationary you need friction.

and /0: as to your question about choosing the pivot point, I now realise the usefulness of choosing the contact as pivot point because it discards the need for knowing the friction. If you remember, this is a common strategy in torque problems i.e: beams being supported by poles etc.
I just chose the contact as pivot for aesthetic reasons because I liked the polygon argument but choosing the pivot as centre still demonstrates how friction is neccesary, just doesn't give the exact number as easily.

[kamil9876]

what is the expression for friction?

1/2 mgsin@

why the ? I thought it should be just mgsin@ since it should give the same answer as when you take the contact point as the pivot. I'm not completely sure myself but I would like some explanation

[evaporade]

grav pot energy changes to transl k.e. and rotat. k.e. 1/2 and 1/2. Rotat.k.e. from the work done by friction.

[/0]

I must experiment with this some day... I love motion experiments...

I wonder how I can make an ice ramp...

[/0]

If a spring is stretched a distance , you get at equilibrium, so



Usually the solution is given as , but I just did a spesh assignment about solving these equations and...

The characteristic equation is so

Since these are distinct roots, does that mean another solution is:

?

[Mao]

No. the characteristic equation to is , which has complex solutions

When you have p as a complex number , the general solution is

Hence in this case, , since , the exponential term equals to 1.


Proof of general solution where p are complex conjugates :



By Euler's formula

[/0]

oh yeah, thanks mao...

I got the sign wrong >_<

Thanks for the proof too mao

[/0]

Is it true that the point joining two ropes of different linear mass density is a node in a standing wave? Is there a way to prove this?


Also, for joined ropes, if you send a pulse down the first rope, then and , where is the velocity in the first rope and in the second.

If you let (i.e. the second rope doesn't exist) then , but that means that and !

How can it be +2?

[/0]

Just learnt about Gauss Law, and it seems to simplify lots of problems...

If you use coulomb's law you can find the field from a non-infinite plate or non-infinite rod but is this possible with Gauss' Law?


Also, just wondering, is there a flux type thing for gravity?

Like, say the gravitational flux was D, then by analogy would this expression for the flux through a sphere hold?

i.e.

[mark_alec]

Yes