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Advanced 2-D Motion Problem - Bead on Wire

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    "Consider a bead of mass m that is free to move around a horizontal, circular ring of wire (the wire passes through a hole in the bead). You may neglect gravity in this problem (assume the experiment is being done in space, far away from anything else). The radius of the ring of wire is r. The bead is given an initial speed v_0 and it slides with a coefficient of friction mu_k. In the subsequent steps we will investigate the motion at later times. You should begin by drawing a free-body diagram at some instant of time. Note that there will be a radial acceleration, a_R, and a tangential acceleration, a_T, in this problem.

    1.1 Write Newton's 2nd law for the radial and tangential directions.

    1.2 Combine the above equations to write a differential equation for dv/dt, where v is the speed at time t.

    1.3 Solve the above differential equation to determine v(t). The solution has the form v = c1/(1+c2*t) - find c1 and c2. Hint: if v_0 = 3 m/s, mu_k = 0.1, r = 10 cm, and t = 3 s, v(3) = 0.3 m/s. "

    Thats just the entire problem for reference.

    2. Relevant equations

    F_c=mv^2/r, F_f=u_k N, F = ma, a = a_R^2 + a_T^2

    3. The attempt at a solution

    Ok so this is what I tried so far. For the first part, 1.1, there are only 2 forces acting on the object, the normal force of the wire, and kinetic friction. Now, I made the assumption that the normal force is the centripital force, which I believe is a safe assumption. Therefore, for the radial direction, it becomes:

    F = m a_R
    Normal = m a_R
    mv^2 / r = m a_R --- > v^2/r = a_R

    For the second equation, I assumed the force of friction was mu_k*N. Therefore, it became mu_k * mv^2 / r. Using this information, the net force in the tangential direction became:

    F = m a_T
    mu_k *mv^2/r = m a_t ---> mu_k *v^2/r = a_T

    Now things get dicey for part 2. I am not sure how to combine these equations. I'm fairly certain a= root(a_R^2 + a_T^2) (pythagorean theorem), and from there you use a = dv/dt and solve that, but it gets messy. If I try that out, it becomes something like this:

    a = root(v^4/r^2 + mu_k^2 * v^4/r^2) <factor our v^4/r^2 out of square root)
    dv/dt = v^2/r * root (1 + mu_k^2)

    For some reason that doesn't sit well with me. Because for part 1.3, when I'm thus solving for the constants c1 and c2, it simply doesnt work. What am I doing wrong? If I integrate that, the answer I got above, this is what I get:

    dv/dt = v^2/r * root (1 + mu_k^2) <--- r = .1 m, mu_k = .1
    dv/v^2 = 10 root (1.01) dt
    -1/v = 10 root (1.01) t + C

    This doesnt work with the initial conditions. What am I doing here?
  2. jcsd
  3. Feb 27, 2008 #2
    Don't banish me to page 2 :(
  4. Feb 27, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

    Have respect for vectors!

    No, you've treated acceleration (a) and velocity (v) as scalars (numbers without direction). They're vectors (numbers with direction), and you must respect that!

    You must work out the acceleration and velocity in one direction at a time.

    You did it for the radial direction, and correctly used it to calculate the "effective weight" and therefore the friction force.

    Now use that value of the friction force to calculate the acceleration in the tangential direction (as a function of the speed in that direction)! :smile:
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