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Rotating frame + heat

  1. Jul 29, 2010 #1
    http://img14.imageshack.us/img14/1399/blahblahr.jpg [Broken]

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    1. The problem statement, all variables and given/known data

    These stairs are rotating as a whole while a dog is climbing on them with a velocity v relative to them, as shown.
    The angle that these stairs make is 60 degrees, and the radius is R.

    1. find the radial component of the friction force between the dog's feet and the stairs.
    2. after a while, the dog stop's moving completely relative to the stairs (but the stairs keep on rotating). find the amount of heat produced as a result of the friction which stopped the dog from moving.


    1. The attempt at a solution

    1. What i've is:

    [tex]m \vec{a}_{rel} = \vec{F} -2m \vec{\omega} \times {\vec{v}}_{rel} -m\vec{\omega}\times(\vec{\omega}\times\vec{r}})[/tex]

    where

    [tex] \vec{v}_{rel} = vcos(60) \hat{\theta} + vsin(60) \hat{z} [/tex]
    [tex] \vec{r} = R\hat{r} [/tex]
    [tex] \vec{a}_{rel} = -\frac{(vcos(60))^{2}}{R} \hat{r} [/tex]

    SO,

    [tex] \vec{F}=-\left ( \frac{v^{2}}{4R}+\omega{v}+\omega^{2}R \right )m\hat{r} [/tex]

    Does this seem reasonable?

    2.

    Here i'm a little puzzled as I know that the work done by friction equals the difference of the mechanical energies, where the final K.E. vanishes. Now, how can I treat potential energies?

    If I take the K.E. difference relative to the stairs, I get:

    [tex] K_{2}-K_{1} = -\frac{1}{2}mv^{2} [/tex]

    However, doing this with respect to a fixed frame, we get:

    [tex]K_{2}-K_{1}=\frac{1}{2}m \left ( (\omega R)^{2}-\left ( \frac{1}{2}v+\omega R \right )^{2}-\frac{3}{4}v^{2} \right )[/tex]
    [tex]K_{2}-K_{1}=-\frac{1}{2}m\left ( v^{2}+\omega v R \right )[/tex]

    which is higher than the previous result, WHY?!?!? and what is the amout of heat from all of these?! does the rotating of the stairs contribute anything to the amount of heat produced?!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 30, 2010 #2
    help would be appreciated, guys =/
     
  4. Jul 30, 2010 #3
    60 degrees? Wow, the dog must be jumping :biggrin:

    1/ That's correct. I think you would find this interesting when rewriting your final result a bit:

    [tex]\vec{F}=-m\frac{(\omega R+vcos60)^2}{R}\hat{r}[/tex]

    It looks like the centripetal force on a mass in constant circular motion at some speed that you may easily see :smile: From that, you can get another solution (surprised! :biggrin:)

    2/ Friction is exerted on both the dog and the stair, and heat = total work done by friction. Recalculate it and see if you can get the same result in both reference frames.
     
  5. Jul 30, 2010 #4
    recalculate what? I mean, I can't do the integral for the work done by friction because i'll get zero: [tex]\int \vec{F}\cdot d\vec{l}=\int F\hat{r} \cdot dl \hat{\theta}[/tex], so I have to find this work (heat) by the difference of mechanical energies. Maybe i was somewhat unclear with my questions, so i'll try to restate them differently:

    - Is potential energy considered? If so, then how? (i'm not given anything about it)

    - What is the difference between the two results that i got previously:

    [tex]
    K_{2}-K_{1} = -\frac{1}{2}mv^{2}
    [/tex]

    [tex]
    K_{2}-K_{1}=-\frac{1}{2}m\left ( v^{2}+\omega v R \right )
    [/tex]

    and which one is correct?!


    By the way, the answer to this is :[tex]K_{2}-K_{1} = -\frac{1}{2}mv^{2} [/tex] but I don't understand why, in addition, why are the two results of the difference in kinetic energies different?!
     
  6. Jul 30, 2010 #5
    No, you won't get it zero. You should show how you got zero.

    It's not that I don't understand your question or your problem. I am only allowed to give you guidance, and the rest is your effort. So now I restate my guidance again: The heat generated by friction = work done by friction on the dog + work done by friction on the stair (we have to add them both, and that's the only thing we know; we cannot point out how much heat is transfer to each one. To explain why this is the case, we must go further, and at that point, I have no idea). So there is nothing to do with potential energy or work done by any other force. You should apply directly the formula dW=Fdr. If you calculate it via the change in kinetic energy, then you have to calculate the change in potential energy and the energy supplied to maintain the angular speed of the stair, which makes things more complicated. But first, if you want to apply dW=Fdr, you must obtain the other component of the frictional force. However, I don't see anything related to the friction here. I guess it is either that the author wants us to solve this in a tricky way, or that something is missing.

    Oh yes, I forgot to tell you that your friction force's formula is a bit vague, though it's correct. The [tex]\vec{F}[/tex] term in the 1st equation should be the total "real" force on the dog (normal force, friction, weight), but the term [tex]\vec{F}[/tex] in the 2nd equation is just the radial component of the friction force, not the friction force.

    EDIT: About why the changes in kinetic energy are different (which is irrelevant in this problem), I will say it is because kinetic energy depends quadratically on the speed, but velocity (and speed) depends on the reference frame. Because of the quadratic rule, the change in kinetic energy doesn't have to remain the same when we switch the reference frame. The total energy, however, is still always conserved.
     
    Last edited: Jul 30, 2010
  7. Jul 31, 2010 #6
    you're right that this force vector corresponds to the total real forces on that dog, but it doesn't contradict with what i've written at all as the rest force components have no acceleration on the LHS to be equated with. So, the friction force vector, which corresponds to the [tex] \hat r [/tex] & [tex] \hat \theta [/tex] directions has a component only in the radial direction as obtained previously, while in the [tex] \hat z [/tex] direction there is: [tex] N-mg=0 \to N=mg[/tex]. At least, this is how I understand this.

    I don't think this is true as the velocity vector may be expressed in a whole lot of different ways but its magnitude (or magnitude squared as we have in the KE expression) should always yield the same result.

    Anyway, the final answer of the second part of the problem I posted on the first message is [tex] -\frac{1}{2}mv^{2} [/tex]. Why is that? :tongue2: ... I don't think that this result can be obtained unless the work-energy equation is used (without performing the integral over the frictional force which has an expression, now, different than that obtained in the first part of the problem ... I think).
     
  8. Jul 31, 2010 #7
    First, let me make clear one point which is about the way we perceive the problem. Until now, we have assumed that the dog is like a point mass, running up the stairs which are like an inclined road which circles around a cone. This is why we can have heat produced by friction: friction here is kinetic friction. The real case is that the dog runs up the stairs, the dog is not a point mass and the stairs are not a smooth road, and friction is in fact static friction. Static friction does not produce heat. But let's leave that fact aside.

    Please think again about my words you cited. Rearrange your 1st equation:

    [tex]\vec{F} = m \vec{a}_{rel} + 2m \vec{\omega} \times {\vec{v}}_{rel} + m\vec{\omega}\times(\vec{\omega}\times\vec{r}})[/tex]

    Where [tex]\vec{F}=\vec{N}+m\vec{g}+\vec{F}_{friction-radial}+\vec{F}_{friction-other}[/tex] is the net force.

    It's easy to see that all the terms in the right side of the 1st equation only have radial component. In addition, the normal force N and weight mg don't have radial component. Therfore:

    [tex]\vec{F}_{friction-radial}= m \vec{a}_{rel} + 2m \vec{\omega} \times {\vec{v}}_{rel} + m\vec{\omega}\times(\vec{\omega}\times\vec{r}})[/tex]

    [tex]\vec{N}+m\vec{g}+\vec{F}_{friction-other}=0[/tex]

    See?

    Think carefully. If I see you run at 200m/s :biggrin: then we have [tex]v^2=40000m^2/s^2[/tex]. But your friend who is running side by side with you will see you running at 0m/s, so now we get [tex]v^2=0[/tex]. Are they the same?

    Then I don't think it's correct. The problem is way more complicated and requires more information.
     
  9. Jul 31, 2010 #8
    Ok, got ya! Thanks man :thumb up: (someone should really update the smilies over here)

    By the way, thanks for the compliment about my running speed ... I've really blushed the hell up :biggrin:
     
  10. Aug 4, 2010 #9
    come to think about it once more, can't I say that the rotation of the stairs doesn't contribute anything to the heat produced on the feet of the dog?! ... since if it were standing still while the stairs are rotating, the amount of heat would be zero, however, because this dog is moving along those stairs, then its velocity relative to the stairs is what determines the amount of heat produced as a result of the friction than made it stop. SO,
    [tex] Q=-\frac{1}{2}mv^{2} [/tex]
     
  11. Aug 4, 2010 #10
    The rotation (or motion in general) of the stairs does somehow "contribute" to the heat. It is because the heat is determined by the relative motion of the dog and the stairs, as you reasoned.

    If we simplify the problem like this: the dog is a point mass given an initial speed v relative to the stairs and gravity's effect is negligible, then Q=0.5mv^2 obviously. If you want a rigorous solution in that simplified case, here it is:

    [tex]F_{friction}=-mdv/dt[/tex]

    [tex]dQ=dW_{friction-on-dog}+dW_{friction-on-stairs}=F_{friction}\times (v+\omega Rcos(60))dt+(-F_{friction})\times \omega Rcos(60)dt[/tex]

    Therefore: [tex]dQ=-mvdv[/tex] or [tex]Q=\int _{v_o}^{0}-mvdv = 0.5mv_o^2[/tex]

    However it is simplified unrealistically. First, effect of gravity must be ignored, while this isn't stated in the problem. Second, this model (i.e. "point-mass" dog, kinetic friction, etc) doesn't agree with reality at all.
     
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