# Homework Help: Two beads and a ring.

1. Sep 20, 2006

### MathematicalPhysicist

a ring of mass M hangs from a thread, and two beads of mass m slide on it without friction.
the beads are released simultaneously from the top of the ring and slide down opposite sides.
show that the ring will start to rise if m>3M/2, and find the angle at which this occurs.
what i did:
i got the displacement x=R(1-cos(a))
by energies we have 2mgR=(2R-x)mg+mv^2/2
so v^2=2xg
now, i think because it rises, the increase in the normal force equals the change of forces, because both beads have a centripatl force acting on them, then the change in normal force is (2mv^2/R)cos(a) (by the third law of newton, action reaction force), so i think to myself this increase should be greater than Mg, but from here im stuck (if im even right on this claim), can someone point where im wrong if im wrong, and how to correct this, becuase i dont know from here to conclude that m>3M/2 and the angle when this occurs.

2. Sep 20, 2006

### NotMrX

In my oppinion I think you have the right idea. The energy of a bead can be written with the center of the circle as 0 for the starting point with theta being the angle between the vertical and the bead. The Y component of the normal forces must equal the weight of the ring it seems. We can use the energy eqn to subsitute into the centripetal acceleration. This is a long problem, I am not sure where to go from here. I don't formally study physics, but this was my interpretation.

Last edited: Sep 20, 2006
3. Sep 20, 2006

### NotMrX

$$E_i=mgR$$

$$E_f=mgR*cos\Theta +\frac{mv^2}{2}$$

$$E=mgR=mgR*cos\Theta +\frac{mv^2}{2}$$

$$m\frac{v^2}{R}=2mg(1-cos\Theta)$$

$$\sum F=ma_c=m\frac{v^2}{r}=N-mg*(-cos\Theta )=N+mg*cos\Theta$$

$$N=m\frac{v^2}{r}-mg*cos\Theta$$

$$N_y=-N*cos\Theta$$

$$Mg=2*N_y=-2N*cos\Theta =-2(m\frac{v^2}{r}-mg*cos\Theta )*cos\Theta$$

$$Mg=-2(2mg(1-cos\Theta)-mg*cos\Theta )*cos\Theta$$

Last edited: Sep 20, 2006
4. Sep 21, 2006

### MathematicalPhysicist

first off, i spotted an error:
it should be 2mgR cause the displacement is 2R, and i dont understand your decompsition of N, i dont think you did it right, cause N in here is directed upwards in the direction opposite to the weight of the individual bead.

5. Sep 21, 2006

### NotMrX

I took the center to be 0 so at the top it is R and at the bottum is negative R.

N is perpendicular to the plane it is resting on. So only at the top and bottum is N only along the y component.

I would like to draw a free body diagram but my boss could walk in at any minute.

I think there is a mistake with the negative sign at the end.

Last edited: Sep 21, 2006
6. Sep 21, 2006

### MathematicalPhysicist

then how from this i conclude that m must satisfy m>3M/2 in order to rise?
btw, ithink it should be:
N=mv^2/R+mgcos(a)
and N_y=mgcos(a).

7. Sep 22, 2006

### NotMrX

I still think it should be:

N=mv^2/R-mgcos(a)

I might have made mistakes previously. This is how I think of the problem. A ball rolling off a spherical surface eventually leaves the surface. The ring pulls it back down. So the normal force is pointing to the center of the circle. A component of gravity is pulling it to the center also and solving that we get the eqn for N.

N_y=N*cos(a) = [mv^2/R-mg*cos(a)]*cos(a)

According to energy considerations:

mv^2/R=mg(1-cos(a))

Then

N_y=[mg(1-cos(a))-mg*cos(a)]*cos(a) = mg*cos(a)*[1-2*cos(a)]

Note that for values bigger than pi/6 N_y is negative. This makes sense because it is being pulled down. The equal but opposite force on the ring is upward.

So for the whole system we have:

-2*mg*cos(a)*[1-2*cos(a)] - Mg=0

Take the derivative of N_y with respect the angle a. Find the angle for the maximun N_y. Plug that angle into the eqn that relates N_y and Mg

Last edited: Sep 22, 2006
8. Sep 23, 2006

### MathematicalPhysicist

why should i take here a derivative?
the N force is perpendicular to the tangent of circle, and the radial force points to the centre of the ring, the radial force is directed in the same direction as the radial force, i dont understand why you think it's directed inwards to the centre of the circle.
if i take the derivative i get -2mg(-sin(a)+4cos(a)sin(a))=0
and i get that cos(a)=1/4 and when i plug it in the equation, i get:
-2m*1/4(1-1/2)-M=0
m=4M which is different than what i need to prove.

9. Sep 23, 2006

### NotMrX

Try taking the derivative again, only this time be a little more carefull. The normal force is perpendicular to the plane. In this case the ring is sometimes making it not fall straight down do to gravity and at other times keeping the bead from flying off. In any case it is either pointing away from or towards the center.

10. Sep 24, 2006

### MathematicalPhysicist

i think i took the derivative from this equation:-2*mg*cos(a)*[1-2*cos(a)] - Mg=0 as i should have taken, if we know that (cos(a))'=-sin(a).
unless you would kindly correct me, or show me where im wrong here?

11. Sep 26, 2006

### MathematicalPhysicist

so what is wrong with my derivatives?

12. Apr 20, 2008

### pisi

I'm trying to solve this problem now, is the solution above right? I found something with square root 3:(

13. Apr 20, 2008

### kamerling

It should be (1/2)mv^2/R = mg(1-cos(a)).

The rest seems OK.

14. Apr 20, 2008

### pisi

Thanks, I've solved it, I realized that I forgot a - for the rings' mass:) it is ok now:)