# Circular motion vector question

1. Feb 1, 2008

### joker_900

1. The problem statement, all variables and given/known data
OK this involves vectors but I can't use the normal notation on the internet so i'm using u^ to mean the unit vector along the vector u - i hope you can help me!

A particle of mass m is constrained to slide on the inside of a vertical smooth semi-circular ring of radius r. The position of the particle is described by a polar coordinate system whose origin is at the centre of the circle with axes along the orthogonal unit vectors r^ and b^ where b is the angle between the radius vector r and the vertical line that passes through the origin.

Assuming the particla is released from rest at the top of the semi-circle, use conservation of energy and Newton's Second Law in the r^ direction to calculate the reaction force exerted by the surface at b=60 degrees.

2. Relevant equations

The first part of this equation led to finding the formula for acceleration:

[/b]a[/b] = r*b''b^ - (v*v/r)r^ where b'' is the second derivative of the angle

3. The attempt at a solution

For conservation of energy I had

2grcosb = v*v, at b=60 => gr = v*v

And for Newton in the r^ direction I had

ma = mgcosb - R , at b=60 => ma = 0.5mg - R where R is the reaction force

From the equation for acceleration in the section above, I took the acceleration in the r^ direction to be

v*v/r and I substituted this into the equation above, and then substituted the energy conservation equation to get

R=1.5mg

However the given answer is 0.5mg - where have I gone wrong?!?

2. Feb 1, 2008

### kdv

The radial acceleration is toward the center of the circle so th eequation should be
ma = - mgcosb + R

3. Feb 1, 2008

### joker_900

I've defined r^ as going from the centre to the circumference though, so surely the mgcosb bit is positive as it acts outwards, and as R acts in the opposite direction it should be negative?

4. Feb 1, 2008

### kdv

Ok, but then your acceleration will be a_r = - v^2/R !!!

5. Feb 1, 2008

### joker_900

Oh sorry (you mean v^2/r ?) I did do that but I mistyped it. I took the acceleration in the r^ direction as -v*v/r

6. Feb 1, 2008

### kdv

Yes, I meant -v^2/r.