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Ball rolling up a incline

  1. Aug 10, 2014 #1
    A hollow sphere with mass $$M$$ , radius $$R$$ , and moment of inertia $$I=2/3MR^2$$ about its center rolls without slipping with a initial center of mass speed $$v$$ towards a fixed ramp. It then rolls without slipping up the ramp. The ramp forms an angle $$\theta$$ with the horizontal. The coefficient of static friction between the ramp and ball is$$\mu_s$$ , and the coefficient of kinetic friction is $$\mu_k$$ .
    How long will it take for the ball to reach its maximum height?

    I can find the height at which the ball stops using energy equation. But not really sure how to calculate time it takes to get there. I always get that $$t=v/g \sin\theta$$, but I think it is wrong
     
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  3. Aug 10, 2014 #2

    olivermsun

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    One thing you can do is write out the energy equation and notice that you have a relationship between velocity and height, from which you can get velocity or height vs time.
     
  4. Aug 10, 2014 #3
    Ok. So here is the energy equation: $$1/2Mv^2+1/3Mv^2=Mgh$$
    from which I get this relationship; $$h=\frac{5}{6g}v^2$$
    And now, I supose, I have to express the distance traveled on the incline.$$d=h/\sin\theta$$
    Now I could find the acceleration $$a=v^2/2d=v^2sin\theta/2h$$
    and from here I can get the time that it takes to slow down $$t=v/a=v\frac{2h}{v^2sin\theta}$$
    is this correct?
     
    Last edited: Aug 10, 2014
  5. Aug 10, 2014 #4

    Nathanael

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    Yes that is correct [itex]t=\frac{5v}{3gsin(\theta )}[/itex]

    I did it a much different way, though. (Involving the frictional force.)

    Your way was easier, but you assumed constant acceleration.


    Edit:
    If you assume constant acceleration, then the time is the distance (d) divided by the average velocity (in this case v/2)

    So you get [itex]t=\frac{2d}{v}=\frac{2h}{vsin(\theta )}=\frac{5v}{3gsin(\theta )}[/itex]
     
    Last edited: Aug 10, 2014
  6. Aug 10, 2014 #5

    haruspex

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    Taken literally, there is a sudden change from a level flat surface to a slope. If so, there's a sudden impulse when the ball hits the slope. Work will not be conserved. You can use a different conservation law to figure out what happens in that instant.
     
  7. Aug 10, 2014 #6
    Could you show how did you do it?
     
  8. Aug 10, 2014 #7

    Nathanael

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    Sure.



    The net force on the object as it rolls up the hill is [itex]mgsin(θ)[/itex] (gravity) minus [itex]F_f[/itex] (friction)
    (The reason the friction is in the opposite direction becomes clear if you imagine what the direction of the torque is)
    So you have this equation for the acceleration:
    [itex]a=gsin(\theta )-\frac{F_f}{M}[/itex]

    I will call this the "first equation"

    ...

    We know that the angular acceleration is the linear acceleration of the CoM (what I've called "a") divided by the radius of the ball
    (this is because it rolls without slipping)
    [itex]\alpha = \frac{a}{R}[/itex] ......... (Where [itex]\alpha[/itex] is the angular acceleration)
    We know the frictional force is what is causing the torque.
    [itex]RF_f=\tau[/itex]
    We also know that the torque is the angular acceleration multiplied by the rotational inertia.
    [itex]\tau =I{\alpha}[/itex]

    So, this all leads to the following equation:
    [itex]RF_f=\frac{2}{3}MR^2\frac{a}{R}[/itex]
    Which becomes:

    [itex]F_f=\frac{2}{3}Ma[/itex]



    Plugging that in to the "first equation" we finally get:
    [itex]a=gsin(\theta )-\frac{2}{3}a[/itex]

    Solving for a we get:

    [itex]a=\frac{gsin(\theta )}{1+\frac{2}{3}}=\frac{3gsin(\theta )}{5}[/itex]

    And of course, [itex]\frac{v}{a}=t[/itex] so we get

    [itex]t=\frac{5v}{3gsin(θ)}[/itex]
     
  9. Aug 10, 2014 #8

    Nathanael

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    Now that seems like a difficult way to do the problem, but it took a lot more effort to explain it than to actually do it :tongue:

    The other method ([itex]t=\frac{distance}{avrg.vel}=\frac{2d}{v}[/itex]) is much faster, but the only thing is, (if you didn't have an answer sheet) you would have to justify your assumption that the acceleration is constant.
     
  10. Aug 10, 2014 #9

    olivermsun

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    It's an energy equation with kinetic energy proportional to distance traveled. Would it be fair to infer constant acceleration from that alone?
     
  11. Aug 10, 2014 #10

    haruspex

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    Two things...
    First, did you notice my post #5? I can't be sure the question intends you to take this into account, but I feel you should.
    Secondly, once you get past that, you can use conservation of energy without having to worry about forces or constancy of acceleration. Just use that the loss in KE in coming to a stop will equal the gain in PE. (Nathanael, you're making it much harder than it needs to be.)
     
  12. Aug 10, 2014 #11

    olivermsun

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    I seriously doubt the question is concerned with the sudden onset of the inclined plane. Otherwise maybe you'd also have to take into account the finite radius of the sphere and how that interacts with the "corner." :wink:

    I was just talking about justifying the constant acceleration to get the travel time (in response to Nathanael's post).
     
  13. Aug 10, 2014 #12

    Nathanael

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    Sorry, but how does that give you the time?

    All of my posts had to do with the time aspect of the problem (because conservation of energy disregards time)


    Edit:
    I think so.

    Can I prove it? Nope. :smile:
     
    Last edited: Aug 11, 2014
  14. Aug 10, 2014 #13

    olivermsun

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    Ah.

    I was thinking along these lines:
    $$\begin{align}
    E_\mathrm{total} &= RE + KE + mgh & \\
    &= \dfrac{5}{3}KE + mgh \\
    KE &= \dfrac{3}{5} E_\mathrm{total} - \dfrac{3}{5} mgh = \mathrm{const.} - \dfrac{3}{5} mgh.
    \end{align}$$
    This is just the usual KE equation for falling under constant acceleration, except here it's as if the gravitational force is ##\dfrac{3}{5}g##.

    To be more rigorous, you can take derivatives with respect to time on both sides to get
    $$\begin{align}
    v \dfrac{dv}{dt} &= -\dfrac{3}{5} g v\sin\theta \\
    \dfrac{dv}{dt} &= -\dfrac{3}{5} g \sin\theta = \mathrm{const.}
    \end{align}$$.
     
  15. Aug 10, 2014 #14

    haruspex

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    No I'm sorry, you're right - I misremembered what the question was asking for.
    But maybe one could simply observe that the forces and torques are constant, therefore the acceleration is constant? Is that slightly easier than solving the force equation?

    I don't see why it's so unreasonable. You are given all the information to do it.
    That's not an also - it's exactly what I'm referring to. It's not particularly hard, and the effect is quite significant.
     
  16. Aug 10, 2014 #15

    olivermsun

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    Because it's an extra complication that doesn't really have anything to do with the main concept, which is
    $$RE + KE + PE = \mathrm{const.}$$
    You can always make the effect of the corner negligible by specifying that the radius is "small" or the sphere has always been on the incline, etc., i.e., the usual idealizations in this kind of problem.
     
  17. Aug 11, 2014 #16

    Nathanael

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    Yes, I completely agree that the constant-acceleration-method is far far superior.

    I did not mean to imply that it would be difficult to show that acceleration is constant, I just wanted to point out that the assumption was made, and so the OP should recognize that it's been made.


    (The only reason I made that long post involving force equations is because the OP requested my alternate solution.)

    (And the only reason that I solved it that way is that I'm imperfect and I overlooked the simplest solution.)
     
  18. Aug 11, 2014 #17

    haruspex

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    No, that doesn't fix it. The fraction of energy lost depends on the angle of the slope, not on the radius of the sphere.
     
  19. Aug 11, 2014 #18

    olivermsun

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    It fixes the problem that the sphere encounters the corner before the center is above the incline.
    Other than that, what is the problem you perceive with respect to the energy equation?
     
  20. Aug 11, 2014 #19
    Ty Nathanael for your solution. I ,realy think, it is also a good and easy way to solve this. Actualy I think this is the way the problem was intended to be solved in the test I took. Thankyou again. What regard the asumption of constant acceleration, constant force implies constant acceleration. So better call it "consequence" rather than asumption. :)
     
    Last edited: Aug 11, 2014
  21. Aug 11, 2014 #20

    haruspex

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    Only if you assume the transition from horizontal to incline has some smoothness. If it is an abrupt change, it makes no difference how small the radius is.
    For the present problem (hollow sphere), the fraction of energy lost looks like this:
    30o: 15%
    45o: 32%
    60o: 51%
    90o: 84%

    So in general it is an important consideration.
    Now, it's quite possible that the problem setter also overlooked its significance. If so, I regard it as a poorly specified question.
     
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