Circular motion vector question

In summary: 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.
  • #1
joker_900
64
0

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?!?
 
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  • #2
joker_900 said:

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
 
  • #3
kdv said:
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?
 
  • #4
joker_900 said:
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 !
 
  • #5
kdv said:
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
 
  • #6
joker_900 said:
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.
 

Related to Circular motion vector question

1. What is circular motion vector?

Circular motion vector is a vector quantity that represents the direction and magnitude of an object's motion in a circular path. It is typically represented by an arrow pointing towards the center of the circle and its length represents the speed of the object.

2. How is circular motion vector calculated?

Circular motion vector is calculated by using the formula: v = ωr, where v is the magnitude of the velocity vector, ω is the angular velocity, and r is the radius of the circular path.

3. What is the difference between tangential velocity and circular motion vector?

Tangential velocity is the component of velocity that is perpendicular to the radius of the circle, while circular motion vector represents the overall velocity of an object in circular motion. Tangential velocity changes as the object moves along the circular path, while circular motion vector remains constant.

4. How does the direction of circular motion vector change?

The direction of circular motion vector changes continuously as the object moves along the circular path. It always points towards the center of the circle and changes in the direction of the object's motion.

5. What factors affect circular motion vector?

The magnitude of circular motion vector is affected by the speed of the object and the radius of the circular path. The direction of the vector is affected by the direction and rate of change of the object's motion along the circular path.

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