Bead between two rings .

Question

samigupta8 · Answer

@parthkohli

samigupta8 · Answer

Do we consider the normal reaction between two rings in calculation of torque on one ring ?

samigupta8 · Answer

The question says that 
Two identical uniform large rings of mass m are connected through a bead of same mass , which can move freely.When bead is released it starts sliding down.The large rings roll apart over a sufficiently rough horizontal surface.Whole system is released from rest,acceleration of bead at initial moment is?
@irishboy123

vincent-lyon.fr · Answer

Can we have a sketch, please?

samigupta8 · Answer

In the sketch of it, a bead is placed just at the uppermost contact of the two rings .

gollywolly · Answer

Well the acceleration of an object when released is g. Sliding down a slanted plane you have the normal force split. Forces can always be split into their x,y components. For going down a plane you have the equation g*sin(theta)=a. For this you have rotational and translational kinetic energy. So you should have something like $$\Delta U=K_rot+Ktrans$$ I'm tired and im sorry if I mislead you haha

samigupta8 · Answer

Hey, i just need to know did you take into account the normal force between the two rings for purpose of rotation of the rings (besides the normal force provided by bead)

samigupta8 · Answer

Btw just for your info, the surface is sufficiently rough that means no sliding will take place thereby.

gollywolly · Answer

Everything that is rolling down a plane is also translating to so you always have take both into account. Well then just neglect the coefficient.

gollywolly · Answer

Did that answer your question?

samigupta8 · Answer

Pls be clear.I didn't understand where is the inclined plane here in this question?

gollywolly · Answer

ohhh Im sorry. I thought i was rolling down. I guess a picture would have helped haha. Well it appears as if the gravitational potential energy of the bead is turned into Krot and Ktran. So that equation still applys.

gollywolly · Answer

$$Ug=krot+Ktrans$$

irishboy123 · Answer

Sami

i too can't for the life of me visualise this 

but i think that with your specific question, if there is no friction between the bead and the ring, then the force is the normal.  

and thus should pass through the centre of each. 

which means no torque on the ring, just a translational force.  just my 2 cents.

samigupta8 · Answer

Sir, but we can have a torque about the bottommost point (on the horizontal surface) 
That too of normal force.
I just wonder why don't we consider the normal force by one ring on other for calculation of torque about the bottommost point. Can you explain it?

irishboy123 · Answer

Sami

I can't get a picture in my head.  of what you are trying to do.  

all i can say is that you will have a friction torque about the centre of the ring.  that is why it will rotate.

but that's so cryptic.

samigupta8 · Answer

Actually my question is like a bead entrapped in between the two rings.

irishboy123 · Answer

so the rings are concentric to start with?

samigupta8 · Answer

Nope,they are very close to each other in same plane and having contact at two points with a bead at one of the contact point.

samigupta8 · Answer

Did you get the figure in your mind?

irishboy123 · Answer

lol! i typed the first few words of your question into Google and got this http://physics.stackexchange.com/questions/253828/find-the-acceleration-of-the-bead so now we have a drawing. and the answer too, one hopes! not sure about the advice given there though. i'd just use an energy approach. much less fiddly. get a DE from that and you should have your answer.