- #1

RoryP

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**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}= 1/2mv

^{2}

E

_{P}= mgh

## The Attempt at a Solution

a)

__At point A__

E

_{K}= 1/2mv

^{2}

E

_{P}= 2mga

__At point B__

GPE= 0

E

_{K}= 1/2m(7v)

^{2}

(Conservation of energy)

1/2m(7v)

^{2}= 1/2mv

^{2}+ 2mga

(m's cancel)(multiply through by 2)

49v

^{2}= v

^{2}+ 4ag

48v

^{2}= 4ag

v

^{2}= 4ag/48 = ag/12

v=[square root](ag/12)

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

b)

mv

^{2}/r + mg = R

(r=a)(v

^{2}= 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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