Exploring Circular Motion: A Bead on a Ring

[RoryP]

Circular Motion!

Homework Statement

A bead of mass m moves on a smooth circular ring of radius a which is fixed in a vertical plane. Its speed at A, the highest point of its path, is v and its speed at B, the lowest part of its path, is 7v.

a) show that v= [square root](ag/12)

b) find the reaction of the ring on the bead, in terms of m and g, when the bead is at A

Homework Equations

F=mv ² /r
E_K= 1/2mv²
E_P= mgh

The Attempt at a Solution

a)
At point A
E_K= 1/2mv²
E_P= 2mga

At point B
GPE= 0
E_K= 1/2m(7v)²

(Conservation of energy)
1/2m(7v)² = 1/2mv² + 2mga
(m's cancel)(multiply through by 2)
49v²= v² + 4ag
48v²= 4ag
v²= 4ag/48 = ag/12
v=[square root](ag/12)

That I am fine with, its part b which confuses me!

b)
mv²/r + mg = R
(r=a)(v²= ag/12)
mag/12a + mg= R
mg/12 + mg= R
R=13mg/12

But that isn't correct! any help would be greatly appreciated =]

 

[Dweirdo]

Is there a sketch to the problem?
I think R= mgcos(P) +...
while we need to find angle P, or I'm imaging the case all wrong.
but I'm not sure, If You could show me a sketch I could be greater help.

 

[Doc Al]

b)
mv²/r + mg = R

Careful with signs. I'll use positive = up and negative = down:

ΣF = ma
-mg + R = -mv²/a

(and so on..)

 

[RoryP]

is that becuase v^2/r = acceleration
ahhh that would give you an answer of 11mg/12 which is the correct answer! Thanks for your help =]