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Circular Motion Without Gravity- Calc Based

  1. Oct 9, 2008 #1
    Circular Motion Without Gravity- Calc Based URGENT!!

    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.

    We will continue our analysis of Problem 1 with the bead.

    2.1 Given your solution for v(t), calculate the radial and tangential components of the acceleration, a_R(t) and a_T(t), respectively. From these calculate the total acceleration a_tot(t).

    2.2 Given your solution for v(t), write the differential equation involving ds/dt, where s is the position of the bead around the circumference of the ring. Solve this equation for s(t).

    2.3 Graph s(t), v(t), a_R(t), and a_T(t) for v_0 = 3 m/s, mu_k = 0.1, r = 10 cm and let t vary from 0 to 10 s.

    2.4 What is the speed, v_1, of the bead after 1 revolution of the ring (using the parameter values given in part 2.3)? What is the speed v_2 after 2 revolutions of the ring? (On your own you may wish to calculate these two speeds for the same bead with the same parameters, except moving along a straight wire.)

    3. The attempt at a solution

    Okay, there's a thread on here that answers part of the first half, which I understood. Since there's no gravity, the frictional force is mv^2/r * mu_k, and that is equal to the tangential force, dv/dt. The masses cancel out, and that's the acceleration formula.

    I set the differential equation as this:

    mu_k * v^2/r = dv/dt

    dt* mu_k/r = dv/ v^2

    After I integrate that...

    (mu_k)(t)/r + C1 = (-1/v) + C2

    The solution is supposed to be in the form c1/(1+c2*t), which I can't comprehend being possible. So I tried to solve it using the equation above.

    Using these initial conditions:
    v_0 = 3 m/s, mu_k = 0.1, r = 10 cm, and t = 3 s, v(3) = 0.3 m/s.

    I found C1-C2 to be -6.333. And solving for v(t), I found

    v(t)= -1/ ((mu_k)t/r-6.333)

    I don't feel like that's correct, and then using that formula to find the acceleration components is a nightmare.

    If anyone can find where I went wrong, it would be much appreciated!!!!
    Last edited: Oct 9, 2008
  2. jcsd
  3. Oct 9, 2008 #2
    Nothing...? :-(. I probably should have posted this earlier than the day before it's due.
  4. Oct 9, 2008 #3
    Re: Circular Motion Without Gravity- Calc Based URGENT!!

    Let's start from here.
    You don't need two constants when you integrate. C1 and C2 in the problem are not two integration constants. Maybe this thing is confusing you.
    The solution is
    (mu_k)(t)/r = (-1/v) + C

    Find C from the initial condition (v=vo at t=0)
    Then solve the equation for v.
    You'll find something that look exactly like they said. You just have to identify the constants.

    I hope it helps. If not, let me know.
  5. Oct 10, 2008 #4
    Yes, that fixed everything for me. I had thought that my professor wanted me to integrate both sides and leave the constants separate, but it was just coincidental that they were defined as c1 and c2. Thanks a lot!
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