1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Circular motion vector question

  1. Feb 1, 2008 #1
    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


    However the given answer is 0.5mg - where have I gone wrong?!?
  2. jcsd
  3. Feb 1, 2008 #2


    User Avatar

    The radial acceleration is toward the center of the circle so th eequation should be
    ma = - mgcosb + R
  4. Feb 1, 2008 #3
    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?
  5. Feb 1, 2008 #4


    User Avatar

    Ok, but then your acceleration will be a_r = - v^2/R !!!
  6. Feb 1, 2008 #5
    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
  7. Feb 1, 2008 #6


    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook