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Homework Help: Sphere rolling with slipping on a movable platform

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A sphere (of radius r and mass m) rotating with angular velocity ω0 is lowered onto the edge of a floating platform of length L and mass M. The platform can move freely on water. The platform is rough and the sphere rolls all the way from one edge to the other edge of the platform with slipping except the very end of the other edge.

    The question requires to find the final speed of the platform w.r.t. the water and also to show that the time of contact of the sphere and the platform is
    [itex]\triangle t = \frac{L(7M+2m)}{(M+m)r\omega_0}[/itex]

    2. Relevant equations
    Basically Newton's law of motion

    3. The attempt at a solution
    I want to know where did i get it wrong and I've done the following to find the time [itex] \triangle t [/itex] but with the [itex] 2m [/itex] term missing:

    I first find the speed [itex] v_s [/itex] of the sphere relative to the water:
    Kinetic friction [itex] f_k = \mu mg = ma [/itex]
    Torque [itex] -f_k r = I \alpha [/itex]
    [itex] v_s = v_0 + at = at[/itex]
    angular speed [itex] \omega_s = w_0 + \alpha t [/itex]
    Denote the final speeds and final angular momentum of the sphere (at the other edge of the platform) as
    [itex] v_f ,\omega_f[/itex] respectively
    [itex]\Rightarrow \omega_f = \omega_0 - \frac{mr}{I} v_f =\frac {v_f}{r}[/itex] [itex] I [/itex] is the moment of inertia of sphere
    [itex]v_f = \frac{\omega_0}{\frac{1}{r}+\frac{mr}{I}}=\frac{2}{7}\omega_0[/itex]
    By Newton's third law,
    [itex] mv_s = Mv_p [/itex] where [itex] v_p [/itex] is the velocity of the platform
    [itex]\rightarrow v_p =\frac{m}{M} \frac{2}{7} \omega_0[/itex]
    The time will then be the distance divided by the average (since acceleration is constant) relative velocity of the sphere and the platform:
    [itex] \triangle t = \frac{L}{\frac{1}{2}(1+\frac{m}{M})\frac{2}{7}\omega_0}=\frac{7ML}{(M+m)r\omega_0}[/itex]
    But i am missing the 2m term and I cannot figure out where i was wrong
  2. jcsd
  3. Apr 7, 2016 #2


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    Science Advisor
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    Gold Member

    Sorry, but I don't understand how you get that equation. I assume this is from conservation of angular momentum, and your reference point is the centre of the sphere (or some fixed point in its path). If so, I would expect to see an M term but not an m term.
    (One should always state the reference point/axis for considerations of angular momentum etc.)

    That doesn't look right either. What about the linear motion of the platform?
    By the way, I get a different answer from the one given.
    Last edited: Apr 7, 2016
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