# Vertical Circle

• February 23rd 2008, 03:06 AM
free_to_fly
Vertical Circle
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 $\
v^2
\$
in terms of a and g. This part I can do and the answer is $\
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.
$\
2ag\left( {\frac{{\sqrt 3 - 1}}{2}} \right) = a^2 g^2 \left( {2 + \sqrt 3 } \right)^2 - v^2
\$

which simplifies to
$\
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?
• February 23rd 2008, 12:08 PM
topsquark
Quote:

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 $\
v^2
\$
in terms of a and g. This part I can do and the answer is $\
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.
$\
2ag\left( {\frac{{\sqrt 3 - 1}}{2}} \right) = a^2 g^2 \left( {2 + \sqrt 3 } \right)^2 - v^2
\$

which simplifies to
$\
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?

In the absence of friction the net force in the radial direction on the bead is the centripetal force. This will be the sum of the normal force on the bead from the wire plus the component of the weight in the radial direction.
$F_c = \frac{mv^2}{a} = N + w~cos(60^o)$
We can find the magnitude of the normal force from this equation.

The reaction force on the wire is the same magnitude as the centripetal force, but being a Newton's third law pair with the normal force on the bead, it will be in the opposite direction. So the reaction force is radially outward, rather than radially inward.

-Dan
• February 23rd 2008, 11:00 PM
mr fantastic
Quote:

Originally Posted by topsquark
In the absence of friction the net force in the radial direction on the bead is the centripetal force. This will be the sum of the normal force on the bead from the wire plus the component of the weight in the radial direction.
[snip]

A proof that for this problem the normal reaction force is radial can be supplied upon request.