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Wrapped Cylinder Kinetics

  1. Dec 22, 2009 #1
    Well, I had a couple problems on my final I was hoping to go over- hope nobody minds. Here's the first.

    1. The problem statement, all variables and given/known data
    A uniform cylinder of radius r and mass m, wrapped around by an unstretchable and massless string, is suspended from a point. The cylinder comes down unwrapping the string and oscillating around the horizontal axis passing through the point under the action of gravity. Let l be the distance from the support point to the contact point of the string with the cylinder and phi be the angle the string forms with the vertical axis through the support point.
    a.) The magnitude of the linear momentum of the cylinder.
    b.) The angular momentum of the cylinder.
    c.) The kinetic energy of the cylinder in terms of l, phi, dl/dt (written as l'), and d(phi)/dt (written as phi').

    (Sorry, not sure how you do the fancier presentation codes.)
    2. Relevant equations
    Linear Momentum
    P = (m/2)*(x'^2+y'^2)

    Angular Momentum
    M = I*omega

    Moment of Inertia for a Uniform Cylinder
    I = (mr^2)/2

    Kinetic Energy
    T = (P^2)/2m + (M^2)/2mr^2

    3. The attempt at a solution
    First, was part A.

    P = (m/2)*(x'^2+y'^2)

    The trick being finding x'^2 and y'^2. Preferably in terms of l, l', phi, and phi'.

    Let's call the angle from the contact point on the cylinder to the center of the cylinder theta. Distance between these two points is always r- which is a constant in this problem. Woot!

    So, x = l*sin(phi) + r*cos(theta) and y = l*cos(phi) + r*sin(theta).

    Now, admittedly my weakest assumption, I assumed that the angle formed by the string to the contact point to the center of mass of the cylinder was usually around 90 degrees. Especially if the string wasn't close to being completely unwound. So, by method of similar triangles phi is about equal to theta.

    Thus, roughly, x = l*sin(phi) + r*cos(phi) and y = l*cos(phi) + r*sin(phi).

    Therefore x' = d(l*sin(phi) + r*cos(phi))/dt = l'*sin(phi) + l*phi'*cos(phi) - r*phi'*sin(phi)
    Thus x'^2 = (l'*sin(phi) + l*phi'*cos(phi) - r*phi'*sin(phi))^2

    And by similar arguments y'^2 = (l'*cos(phi) - l*phi'*sin(phi) + r*phi'*cos(phi))^2

    So P = (m/2)*(x'^2 + y'^2)

    P = (m/2)*((l'*sin(phi) + l*phi'*cos(phi) - r*phi'*sin(phi))^2 + (l'*cos(phi) - l*phi'*sin(phi) + r*phi'*cos(phi))^2)

    Which expands into something that on the surface looks rather messy, but a few terms add out and other terms are simplified by the good old cos^2(theta) + sin^2(theta) = 1 identity.

    So P = (m/2)*(l'^2 + l^2*phi'^2 + r^2*phi'^2 + 2*r*l'*phi'(cos^2(phi) - sin^2(phi)) - 4*r*l*phi'^2*sin(phi)*cos(phi))



    For a uniform cylinder I = (mr^2)/2. Also known is omega = (v/r) = (P/mr)

    Thus M = P*(r/2)


    T = (P^2)/2m + (M^2)/2mr^2

    Which when you plug in P and M, do a little multiplication by constants, and you get.

    T = (5m/32)*(l'^2 + l^2*phi'^2 + r^2*phi'^2 + 2*r*l'*phi'(cos^2(phi) - sin^2(phi)) - 4*r*l*phi'^2*sin(phi)*cos(phi))^2
  2. jcsd
  3. Dec 28, 2009 #2
    No bites, eh?

    Well, if it helps I don't need to be taken by the hand. I just want to see how the professor got these answers.

    A) P = m*sprt((l'-r*phi')^2 + l^2*phi'^2)

    B.) M = (3/2)*m*r*(l' - r*phi')

    C.) T = (3/4)*m*(l'-r*phi')^2 + (m/2)*l^2*phi'^2
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