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Homework Help: Two difficult translational and rotational motion problems

  1. Oct 29, 2008 #1
    At least, they're difficult for me. I find them very hard to set up, in particular the interaction of the rotation and translation is hard for me to grasp. I've worked on both of these very hard, with only some headway, so be liberal in your help.
    (For some reason, all my greek letters are superscripted - ignore that, they should not be)

    The first
    1. The problem statement, all variables and given/known data
    Hanging from one side of an ideal pulley is a mass m. On the other side, there is a cylinder with the pulley-rope wrapped around it. The cylinder has mass M and radius R, and we can assume it to be uniform (though it shouldn't matter to the problem - just say it has moment of inertia I if you like). Both objects are acted on by the force of gravity, (gravitational acceleration = g), as well as the tension of the pulley rope. Mass m accelerates downwards at rate a, while cylinder M accelerates at rate A, with the rope around it unwinding at angular acceleration a [tex]\alpha[/tex].

    From this, determine the values of a, A and [tex]\alpha[/tex] in terms of m, M, R and g.

    2. Relevant equations
    General equations, the letters don't stand for the quantities of the problems.
    a = g -T/m
    Ek = 1/2mv^2
    Er = 1/2I[tex]\omega[/tex]^2
    Eg = mgh
    [tex]\alpha[/tex] = [tex]\omega[/tex]t
    a = [tex]\alpha[/tex]R
    I = 1/2mR^2
    [tex]\tau[/tex] = FXR = [tex]\alpha[/tex]I
    Some important constraints (the letters represent the quantities of the problem)
    a + A = [tex]\alpha[/tex]*R, due the constraint of the rope length
    -g <= a, A <= g
    0 <= [tex]\alpha[/tex]*R <= 2g

    3. The attempt at a solution
    I tried two different ways, both of which ended in dead ends. The first was to set up the following (though they may be incorrect):
    Note: T = force of tension in the rope;
    1) a + A = [tex]\alpha[/tex]*R
    2) a = g - T/m
    3) A = g - T/M + [tex]\alpha[/tex]*R
    4) [tex]\alpha[/tex]R = [tex]\tau[/tex]R/I = T*R^2/I = T/M*(M*R^2/I)

    From this, I get:
    2g - T(m+M)/(mM) + [tex]\alpha[/tex]R = [tex]\alpha[/tex]R
    2g - T(m+M)/(mM) = 0
    Giving T = 2g(mM)/(m+M)
    a = g - T/m = g - 2g(M)/(m+M) = g(m+M)/(m+M) - 2gM/(m+M) = g(m-M)/(m+M)
    [tex]\alpha[/tex]R = T/M*(M*R^2/I) = 2g(m)/(m+M)*(M*R^2/I)
    divide by R to get [tex]\alpha[/tex]
    A = [tex]\alpha[/tex]R - a = 2g(m)/(m+M)*(M*R^2/I) - g(m-M)/(m+M)

    This looks fine and dandy at first, but for one problem:
    I = 1/2(M*R^2)
    (M*R^2)/I = 2
    Therefore, if m>>M,
    [tex]\alpha[/tex]R ~= 2g*1/1*2 = 4g
    THIS DEFIES OUR EARLIER CONSTRAINT! It's impossible, in other words, which is a good indication that something is wrong. I suspect I made a false assumption in one of the equations, probably #3. With a good formula for A, or for T, I'm sure I could solve this problem easily. As it is, I'm having a tough time.

    I also tried another way. Using work-energy, I derived a quadratic equation for a in terms of A (which, strangely enough, was identical to A in terms of a). Said equation is so complicated however, that I'm practically certain I either messed it up (and I'm pretty sure that's not so, since I did it twice), or energy simply isn't a fruitful path to this solution.

    The Second problem:

    1. The problem statement, all variables and given/known data
    This one is a good deal more straightforward.
    A uniform disk of mass m and radius R lies on a surface of frictionless ice in an x-y plane. A boy of mass m (same m) is skating towards it at Vi, directly parallel to the x axis. The boy jumps and lands on the disk outer edge of the disk (at coordinate (0,-R) in other words), sticking to it completely, at t = 0. Give equations for the boy's subsequent path the x coordinate and the y coordinate. Also, give the amount of energy dissipated in the boy's jump.

    2. Relevant equations
    L = I[tex]\omega[/tex] + mv X r(vector, not radius)
    Ek = 1/2mv^2
    Er = 1/2I[tex]\omega[/tex]^2
    I = 1/2mR^2

    3. The attempt at a solution
    I figured that angular momentum was conserved. After the boy's jump, the disk is going to have a translational velocity Vf and an angular velocity [tex]\omega[/tex].
    1) Li = mViR
    2) Lf = 2mVfR + (mR^2 + 1/2mR^2)[tex]\omega[/tex] = 2mVfR + 3/2mR^2[tex]\omega[/tex]
    I'm not completely sure about #2
    3) Li = Lf; simplifies to
    4) Vi = 2Vf + 3/2R[tex]\omega[/tex]

    The problem is, I don't know what to do from there - I have no clue how the momentum distributes between linear and rotational motion. I feel like it's 2/3rds rotational, 1/3rd linear, due to I relative to mR^2, but I don't know why or how to show it if so.
    The lack of constraints or relations leaves me floundering - I'm also clueless on the second part, about energy dissipation; I'd assume 0, and that energy would be a constraint that would help solve the motion part of the problem, but clearly this isn't so.

    To whomever can help me, THANK YOU.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 29, 2008 #2
    The superscript greek letters seems to be just a problem of using latex next to normal text.

    First Problem:

    I assume that A refers to the acceleration of the centre of mass of the cylinder downwards. In that case, I can get equation 2: a = g - T/m. I get a different equation for 3 and 4, however. Equation 3 becomes A = g - T/M - a. The 2 other equations for α can be obtained by taking the centre of the cylinder and the point where the rope contacts it as the axis of rotation, and equating τ = Iω.

    You then have enough equations to solve the problem. There is an additional equation αR = a+A that you may use, although it is not necessary given the others above.

    Second problem:

    You can try working in the frame of the centre of mass, then use conservation of angular momentum to find the angular velocity of the disk with the boy on it. From there you can get a set of parametric equations for the boy's position, as well as calculate energy dissipated.
  4. Oct 29, 2008 #3
    Yes, A is the acceleration downwards of the center of mass of the cylinder.
    On the first, how do you get your equations? If A = g - T/M - a (not alpha), then A + a = 2g - T(m+M)/(mM) = [tex]\alpha[/tex]R, which can't be the case either, as it doesn't factor in moment of inertia at all, and gives you a non-sensical answer at m = M of [tex]\alpha[/tex] = 0. Also, neither T nor Tau = I[tex]\omega[/tex]

    Similarly, I'm still confused by the 2nd problem. If I look at it from an inertial frame at the center, I get:
    Li = mViR, of course. But then I don't know how to decide how much of this momentum tranlates into angular momentum for the disk-boy system, and how much is translational momentum? For instance, if the boy jumps on at a coordinate closer to the center of mass, I know that it translates more, rotates less, while if he jumps on at the outer edge it rotates more, translates less. I have NO IDEA how to decide and calculate this though - can I assume that it all becomes rotational? I don't think so.
  5. Oct 30, 2008 #4
    The cylinder is acted on by 2 forces: the tension T and its weight Mg. Its acceleration in the frame of the rope is g - T/M. Since the rope is accelerating in the opposite direction at a, the linear acceleration of the cylinder's centre of mass in the frame of the observer is A = g - T/M - a.

    (I was a little confused about this part for a while. If you arrived at A = g - T/M, as I initially had, it turns out to be inconsistent with αR = a + A. It seems that g - T/M refers to the acceleration of the cylinder in the frame of the rope, so A = g - T/M - a instead. I admit that I am still not fully satisfied with visualising it in this way, but this is the only way to get consistent equations. Alternatively you may give up on this equation entirely and just use αR = a + A, which gets you the same answer.)

    If A = g - T/M - a, then A + a = g - T/M = αR which is reasonable.

    The other equations you get is (about the centre of mass of the cylinder) TR = (.5MR^2)α, and (about the point that it first contacts the rope) MgR = (1.5MR^2)α. You can first solve for α using the last equation, then subsequently the remaining unknowns. If you generalise the moment of inertia of the cylinder by using k as the coefficient for MR^2, you should eventually arrive at a solution that reduces to a 2 point mass system in the limit where k->inf.

    For the second problem, you may notice that the centre of mass moves at .5V in the negative x direction, with its vertical position equidistant to the centre of the disk and the boy. Hence the initial angular momentum in the frame of the centre of mass is Li = m(.5V)(.5R) [the boy] + m(.5V)(.5R) [the disk] = .5mVR. The final angular momentum is in the form of a pure rotation of the disk with the boy at the edge, so you can solve for the position of the boy first in the frame of the COM, then simply add the latter's velocity.
  6. Oct 31, 2008 #5
    I was able to figure out the second problem from your advice, but the first problem still eluded me. I probably should have just brute-forced it with energy, but it would have looked ridiculous. Ah well.

  7. Nov 1, 2008 #6
    My apologies. After pondering about this problem for a while, I realised that this is wrong. The linear acceleration A of the cylinder is indeed just g - T/M. and the problem of inconsistency arose from another equation, namely one of the 2 for the rotation. I will present the solution which I now think is correct.

    The accelerations of both masses are
    a = g - T/m
    A = g - T/M

    About the centre of the cylinder,
    TR = (.5MR^2)α

    About the point of contact with the rope,
    M(g+a) = (1.5MR^2)α

    This was where the problem was. I had forgotten to consider the rotation in the (accelerated) frame of the rope. In that frame, the inertial force on the cylinder has to be included, hence the (g+a) in the last equation.

    This set of equations should lead to
    α = 4mg / (3m+M)R
    A = (m+M)g / (3m+M)
    a = (3m-M)g / (3m+M)

    which turns out to satisfy αR = a + A.

    If I generalised the moment of inertia of the cylinder by using k as described in my previous post, the solutions I got were
    α = 2mg / ((k+1)m+kM)R
    A = ((1-k)m+kM)g / ((k+1)m+kM)
    a = ((k+1)m-kM)g / ((k+1)m+kM)

    As k->inf, they do reduce to α = 0, A = (M-m)g / (m+M), and a = -A respectively, as in the case of nonrotating masses.
    Last edited: Nov 1, 2008
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