# Circular Motion

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

F=mv 2 /r
EK= 1/2mv2
EP= mgh

## The Attempt at a Solution

a)
At point A
EK= 1/2mv2
EP= 2mga

At point B
GPE= 0
EK= 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 =]

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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.

Mentor

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 =]

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