Originally Posted by

**free_to_fly** The question is from Heinemann Modular Mathematics p145

Q6: A bead of mass m is threaded on a smooth circular wire of radius a and centre O which is fixed in a vertical plane. B is released from rest at the point where OB makes an angle of 30 degrees with the upward vertical.

b) Given that the speed of B at the lowest point of its path is v, find $\displaystyle \

v^2

\$ in terms of a and g. This part I can do and the answer is $\displaystyle \

v^2 = ag(2 + \sqrt 3 )

\$

The next part is the bit I'm stuck on:

c) Find in terms of m and g the magnitude and direction of the reaction of the wire on the bead when OB makes an angle of 60 degrees with the upward vertical.

I started off by using the conservation of energy, PE lost=KE gained.

$\displaystyle \

2ag\left( {\frac{{\sqrt 3 - 1}}{2}} \right) = a^2 g^2 \left( {2 + \sqrt 3 } \right)^2 - v^2

\$

which simplifies to

$\displaystyle \

v^2 = a^2 g^2 \left( {2 + \sqrt 3 } \right) - ag\left( {\sqrt 3 - 1} \right)

\$

But I don't know what to do after this. Can someone help please?