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Speed of objects rolling down slopes

  1. Oct 17, 2009 #1
    I'm a bit confused about the speed of objects rolling down slopes.

    In my textbook, it says
    "Neither the mass nor the size of the object will affect its speed when rolling downhill." And that solid balls of different masses/sizes will all reach the bottom of the slope together.

    And then it goes on to say that because a hollow cylinder has mass far from the centre, it has a large rotational intertia, so gains a larger proportion of rotational Ek, so a smaller proportion of linear Ek, so it will have a slower speed when rolling downhill. Hence a solid ball (smaller I) will reach the bottom of the slope before the hollow cylinder.

    My ques is - if rotational inertia is what determines the speed of objects rolling down slopes, won't a larger solid ball have a larger rotational inertia than a smaller solid ball? So a larger solid ball should (by the reasoning above) have a slower speed, so will reach the bottom of the slope AFTER (not at the same time as) a smaller solid ball?

    Yet my textbook says "Large or small, light or heavy, all of these solid balls will reach the bottom of the slope together".

    Can someone help me clear up my confusion?
     
  2. jcsd
  3. Oct 17, 2009 #2
    Let's just do the equation and see what we get
    Conservation of energy states
    [tex]\frac12 J\omega^2+\frac12 mv^2+mgh=\text{const}[/tex]
    where the energies are rotational energy, kinetic linear energy and potential energy. [itex]J[/itex] is the moment of inertia around the rolling axis. With [itex]v=\omega r[/itex] this simplifies to
    [tex]\frac12\left(\frac{J}{mr^2}+1\right)v^2+gh=\text{const}[/tex]
    So the velocity only depends on height and [itex]\frac{J}{mr^2}[/itex]. For bodies with the same shape this expression is the same.
     
  4. Oct 17, 2009 #3
    Thanks Gerenuk!!!

    But do you mind explaining this:

    Why's it the same for bodies of the same shape?
     
  5. Oct 17, 2009 #4
    That's actually not quite a general statement.
    At least for object with some degree of symmetry (maybe cylindrical; like spheres or cylinders) it turns out that
    [tex]J=amr^2[/tex] where a is a constant (consider a scaling argument for [itex]J=\int r^2\mathrm{d}m[/itex]). This case is easy.
    Basically if an object with cylindrical symmetry (and constant radius) is scaled up or its mass density is changed, the expression [itex]J/(mr^2)[/itex] is invariant.

    I'm not sure how much one can generalize this statement.

    For rolling ellipsoids one actually has to check all preconditions again. I think all equations are valid again, however r would be the distance from ground to the center of mass which changes as the ellipsoid rotates and also the velocity in [itex]v=\omega r[/itex] is not the velocity of the center of mass projected along the inclination of the ground. And I guess [itex]J=amr_\text{avg}^2[/itex]. So I could image that this equal shape theorem doesn't apply for ellipsoids, but I haven't checked all details.

    Maybe someone else can clarify which class of objects reaches the end of the track at the same time.
    I wouldn't be surprised if someone is able to prove that a general upscaling of the physical situation gives same time for rolling. But that would also mean that for comparing rolling at one type of slope only, on needs a self-similar slope, i.e. a flat slope.
     
    Last edited: Oct 17, 2009
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