Circular motion vector question

[joker_900]

Homework Statement

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.

Homework 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

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?!?

 

[kdv]

Homework Statement

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.

Homework 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

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?!?

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

 

[joker_900]

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

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?

 

[kdv]

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?

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

 

[joker_900]

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

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

 

[kdv]

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

Yes, I meant -v^2/r.

using a= +v^2/r as yopu had posted I got your final answer of 0.5 mg.
when I redid it with the correct acceleration of a=-v^2/r, I get the correct 1.5 mg.
You are saying that you got 0.5 mg using a=-v^2/r? I am confused. Please check your calculation.