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Rotation physics help

  1. Nov 30, 2005 #1

    Päällikkö

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    This should be an easy one, but I can't fully figure it out.

    If a ball is given some initial speed v0, but no angular speed as it's thrown on a surface with a coefficent of friction [itex]\mu[/itex], how long will it take for the ball to execute "perfect roll" ie. roll without slipping?

    I did it through conservation of energy, and got
    [tex]t = \frac{v_0}{\mu g}[/tex]

    Next, I tried with Newton's laws.
    I got
    [tex]t = \frac{v_0}{\mu g\left(\frac{mr^2}{I} + 1\right)}[/tex]

    I suppose the latter's wrong because I cannot simply assume the ball would only rotate about the center of mass -axis (that's what I did).
    How do I take this into account?
     
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  3. Nov 30, 2005 #2

    siddharth

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    Paallikko,
    I don't see anything wrong with what you have done using Newton's law.
    I don't understand why you think it is incorrect to assume the ball will rotate about it's center. Perhaps you could explain that?

    Also, how exactly did you use conservation of energy?
     
    Last edited: Nov 30, 2005
  4. Nov 30, 2005 #3

    Päällikkö

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    This, once again, isn't a textbook problem, so I don't know the correct answer, but I think (my brain's been completely jammed this week, I can't think straight) the conservation of energy is the correct way to go here.

    From the initial conditions we get:
    [tex]\frac{1}{2}mv_0^2 + W = \frac{1}{2}I \omega ^2 + \frac{1}{2}mv^2[/tex]
    The force doing work is the friction force, [itex]W = - \mu mgx[/itex]. Divide by m and 1/2, and [itex]\omega = \frac{v}{r}[/itex]
    [tex]v_0^2 - 2 \mu g x = \left(\frac{I}{mr^2} + 1\right) v^2[/tex]
    Newton's second is now used, but it wasn't the core idea here:
    [tex]v = v_0 - \mu gt[/tex]
    [tex]x = \left(v_0 -\frac{1}{2}\mu gt\right)t[/tex]

    [tex]\Rightarrow v_0^2 - 2 \mu g \left(v_0 -\frac{1}{2}\mu gt\right)t = \left(\frac{I}{mr^2} + 1\right) (v_0 - \mu gt)^2[/tex]

    Now I put maxima to solve it and get:
    [tex]t = \frac{v_0}{\mu g}[/tex]

    (I did later manually make sure that the I does in fact disappear)


    EDIT: I think the part I fail to do it with dynamics is the part torque part (which is a derivation of F = ma, sort of), not the F = ma "itself". I hope you understood what I meant by this :smile:.
     
    Last edited: Nov 30, 2005
  5. Nov 30, 2005 #4

    siddharth

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    Ok. But the friction force also has a torque. So while it does work in slowing down the translational velocity of the ball, it also does work in increasing the angular velocity. So I think you have to take that into account as well.

    Sorry, I don't understand that
     
    Last edited: Nov 30, 2005
  6. Nov 30, 2005 #5

    Päällikkö

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    It is taken into account: rotational energy increases. Or how do you mean? Can you please elaborate?
     
  7. Nov 30, 2005 #6

    Päällikkö

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    The way I did it with the laws of dynamics:
    [tex]v = v_0 - \mu gt[/tex]

    [tex]M = I \alpha = r \mu mg[/tex]
    [tex]\omega = \alpha t = \frac{r \mu mg}{I} = \frac{v_t}{r}[/tex]
    The above only applies when alpha is constant, which is it.

    When the tangential speed equals the speed of the center of mass, the ball is in perfect roll:
    [tex]v_0 - \mu gt = \frac{r^2 \mu mg}{I}t[/tex]

    [tex]\Rightarrow t = \frac{v_0}{\mu g\left(\frac{mr^2}{I} + 1\right)}[/tex]
     
  8. Nov 30, 2005 #7

    siddharth

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    What I think is that the force due to friction (say [itex] f [/itex]) exerts a torque [tex] rf [/tex] on the ball. It is this torque which speeds up the ball's angular velocity.
    So the work done due to this frictional force in changing the Rotational Kinetic Energy will be [itex]\int rf d \theta [/tex] much the same way as the Work done in changing the Kinetic energy is [itex] \int fdx [/itex].
    I'm not too sure of this myself so I may be wrong.
    In any case, I think the answer you obtained by the newton's method is right.
     
  9. Nov 30, 2005 #8

    arildno

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    Hi, Paaliko:
    The reason why your answers differ is that you use an incorrect expression for the work done:

    You cannot use the distance traversed by the centre of mass in this manner. Your work is given by [itex]W=-\int_{0}^{t}\mu{g}v_{c.p}d\tau[/itex], where [itex]v_{c.p}[/itex] is the velocity of the point on the ball instantaneously in contact with the surface, not the center of mass velocity, as you have basically assumed.

    (These will be equal in the case of SLIDING, in which the phenomena of "rolling" and "falling to rest" become the same, and the time is, indeed, given by [itex]t=\frac{v_{0}}{\mu{g}}[/itex]
     
    Last edited: Nov 30, 2005
  10. Nov 30, 2005 #9

    siddharth

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    arildno, I didn't understand the last bit.
    In the case of sliding, as in this case, [tex] v_{com} > r \omega [/tex]
    So [tex] v_{cp} = v_{com} - r \omega [/tex]
    And the work done will consequently be
    [tex] \int -\mu mg v_{cp} dt [/tex]
    which will be

    [tex] \int -\mu mg v_{com}dt + \int \mu mg r \omega dt [/tex]

    which is
    [tex] \int -\mu mg dx + \int \mu mgr d\theta [/tex]
    So the answer will not be [itex]t=\frac{v_{0}}{\mu{g}}[/itex]
     
    Last edited: Nov 30, 2005
  11. Nov 30, 2005 #10

    arildno

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    1. The case of sliding is the case where the object does not rotate about its own axis, i.e, [itex]\omega=0[/itex]. This is, of course, impossible for a ball (due to the frictional force's production of a net torque), but is quite possible for, say, a box sliding along the ground.

    2.
    I have written the work done as an integral over time; the tau is merely a dummy variable.

    3. Yes, your dynamics stuff is correct.
     
  12. Nov 30, 2005 #11

    Päällikkö

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    Thanks both, it all makes sense now.

    I suppose the case's closed.
     
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