# More M2 help!!

Discussion in 'Mathematics' started by Maths_Mike, Sep 1, 2011.

1. ### Maths_MikeNew commenter

Particle, P mass, m moving in a horizotal circle with uniform speed and the smooth inner surface of a hemisphere with base radius a.
A light inextensible string is attached to P and passes through a small hole at the base of the hemisphere. A mass, m hangs at rest at the opposite end to P.
The plane of motion is a/2 below the rim of the hemisphere.
prove that v^2 = 3ga.
I have so far concluded that the angle between normal reaction and the vertical is the same as between the string and the vertical and is 60 degrees.
and that the radius of the horizontal circle is ((root 3/2)a)
I have then tried to resolve horizntally so T-R = ma(v^2/ ((root 3/2)a))
and resolving vertically Tcos 60 + r cos 60 = 2mg.
But I cant get it to work out!

2. ### briceanusNew commenter

Mike, angles are correct, radius is correct.
At other end of string, T = mg (I'm sure you have this)
When you resolve 'horizontally', both R and T are 'inward. and nothing is 'outward'. The mg of the moving particle is perpendicular to the motion.
So you get Tcos30 +Rcos30 = mv (squared)/r where r is the radius of the horizontal circle.
But you now need to resolve vertically at the particle, so mg + Tsin30 = Rsin 30.
(p.s. are you sure you meant 2mg, where is the 2 from? )
Combining these gives the result
Bric

3. ### Maths_MikeNew commenter

oops - for some silly reason I had the tension in the string as 2mg!
Thanks