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Thread: [SOLVED] Difficult rotational mechanics question?

  1. #1
    Super Member fardeen_gen's Avatar
    Jun 2008

    [SOLVED] Difficult rotational mechanics question?

    A uniform circular cylinder of mass m and radius r is given an initial angular velocity \omega_{0} and no initial translational velocity. It is placed in contact with a plane inclined at an angle \alpha to the horizontal. If there is a coefficient of friction \mu for sliding between the cylinder and plane, find the distance the cylinder moves up before sliding stops. Also, calculate the maximum distance it travels up the plane. Assume \mu > \tan\alpha

    d_{1} = \frac{r^2\omega_{0}^2(\mu \cos \alpha - \sin \alpha)}{2g(3\mu\cos \alpha - \sin \alpha)^2}
    d_{max} = \frac{r^2\omega_{0}^2(\mu \cos \alpha - \sin \alpha)}{4g\sin \alpha(3\mu\cos \alpha - \sin \alpha)}

    Please help!
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  2. #2
    May 2009
    This is actually quite simple to solve and here is how you do it:


    To begin with here's a diagram of the initial state of the cylinder:

    For the duration of time when there is still sliding taking place then the friction force is at maximum and therefore the acceleration is a constant and so is the torque on the cylinder. This means the following equations of motion hold:

    \omega = \omega_0 + \ddot{\theta}t and v = a t

    where v and \omega are the velocity and angular velocity at the point when sliding stops respectively. Also
    v = r \omega.
    Hence, v can be found in terms of \ddot{\theta} and a as
    v = \frac{a}{a-r \ddot{\theta}} r \omega_0 .

    This equation for the velocity can then be used with the equation of motion relating velocity to distance to find d_1, i.e.
    v^2 = u^2 +2aS so then
    \color[rgb]{0,0,1}d_1 = \frac{a}{(a-r\ddot{\theta})^2} \frac{r^2 \omega_{0}^{2}}{2}.

    Now all you need to do is find \ddot{\theta} and a and substitute in the above equation. These two can now be found by resolving forces in the diagram.
    Friction force is given by

    F_\mu = \mu R

    Resolving perpendicular to the plane we have:

    R = m g \cos \alpha.

    Resolving along the plane

    F_\mu  - mg \sin \alpha = m a .

    From these three equations we find that

    \color[rgb]{0,0,1} a = (\mu \cos \alpha - \sin \alpha) g.

    Finally the torque is given by:

    I \ddot{\theta} = -r F_\mu where \color[rgb]{0,0,1}I = \frac{1}{2} m r^2.


    \color[rgb]{0,0,1} \ddot{\theta} = - \frac{2 \mu g \cos \alpha}{r} .

    So that gives you d_1 easily enough!

    For the second part you can just use the conservation of energy to work out the further distance travelled, d_2, before stopping. So equating kinetic energy lost to the potential energy gained gives:
    \color[rgb]{0,0,1}mg \, d_2 \, \sin \alpha = \frac{1}{2} I \omega^2 +\frac{1}{2} m v^2


    d_{max} = d_1 +d_2 .
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