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K&K Question 3.22 - Mass, String, and Ring

  1. Apr 10, 2014 #1

    Radarithm

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

    A mass m whirls around on a string which passes through a ring, as shown. Neglect gravity. Initially the mass is distance r0 from the center and is revolving at angular velocity ω0. The string is pulled with constant velocity V starting at t = 0 so that the radial distance to the mass decreases. Draw a force diagram and obtain a differential equation for ω. This equation is quite simple and can be solved either by inspection or by formal integration.

    2. Relevant equations

    Image: http://gyazo.com/15430c4d669103e00a52b49dd533be0c
    [tex]r=Vdt[/tex]
    [tex]\frac{V^2}{r}=Vd\omega[/tex]
    [tex]T=\frac{mV^2}{r}=mr\omega^2=mV^2dtd\omega[/tex]
    [tex]N=m\ddot{y}[/tex]

    3. The attempt at a solution

    I have absolutely no idea on how to start this problem. I cannot obtain a valid differential equation from the tension, and I do not know how to relate the normal force to everything. Even if I write:
    [tex]\frac{dv}{dt}=V^2dtd\omega[/tex]
    It is still impossible to get a valid differential equation. I don't even think the relation [itex]\theta=s/r[/itex] would help.
     
    Last edited: Apr 10, 2014
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  3. Apr 10, 2014 #2

    BvU

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    Hello Rad,
    First a general comment: you can't have a differential on one side and not on the other side of an equation. So instead of [tex]r=Vdt[/tex] you should write [itex]dr=Vdt[/itex] etc. etc. ([edit] actually: [itex]dr=-Vdt[/itex])

    But there's no need for that one: you already are given ##r(t) = r_0 - V\, t##, right ?

    In your second equation you start mixing up V's. tangential speed ##v(t) = \omega(t) \, r(t)## alright, but big V comes in differently.

    It is time to think physics: what exactly is happening, what is conserved, what work is done (by what on what) and such.
    If too complicated, explore the time before t=0 assuming steady state. What pulling force is needed to have equilibrium ?

    Did you draw the diagram ?
     
    Last edited: Apr 10, 2014
  4. Apr 10, 2014 #3

    Radarithm

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    I did draw the diagram, and as far as I can tell the only forces acting on the mass are the normal force from the ring and the tension from the string that's being used to decrease the radius. I don't see how the system is still in equilibrium; gravity is neglected so there is nothing to oppose the normal force. The "centrifugal" force doesn't exist and the tension is perpendicular to the normal force. The linear velocity is proportional to the angular one through the radius, but the radius decreases as the angular velocity increases. This means that:
    [tex]\frac{dv}{dr}=d\omega[/tex] The fact that there is angular velocity in the x direction shows that the system is not in equilibrium, but it doesn't make sense as to why the normal force is the net force in the y direction and yet the mass doesn't go recoiling upwards.
     
  5. Apr 10, 2014 #4

    Radarithm

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    I realize that the string is being pulled down; what I don't understand is how the tension contributes to that. It is parallel the plane, but would it be correct to say that it is at an angle with the mass (eg. from the string below the ring to the mass an angle is created but I doubt this assumption is correct).
     
  6. Apr 10, 2014 #5

    BvU

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    Ah, Ok, let's take a step back and really look at the V = 0 case: can you agree the ring only has the function of a kind of pulley: changing the direction of the tension ? A wire can only exercise a force lengthwise. And tensions at both ends have the same magnitude, but (thanks to the pulley / ring) different directions.

    That way the pulling force is delivering the centripetal force required for the circular motion, and we can write it down for t = 0: ## F = m\, \omega_0^2\, r_0##. Or, indeed, for t in general: ## F = m\, \omega^2\, r##, where now all three of ## F, \omega, r## vary with time !

    Again: you can't have a differential quotient on one side of an equation and a differential on the other ! So [itex]\frac{dv}{dr}=d\omega[/itex] can't be right. At best it should be something like [itex]dv=d(\omega\, r)[/itex].

    [edit] Oh, and by the way: the ring doesn't exercise a force on the mass, at least not directly: the only thing that is connected to the mass is the wire !
     
  7. Apr 10, 2014 #6

    Radarithm

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    Would using polar coordinates simplify the problem in any way? I am currently stuck; I don't even know how to approach this problem. There doesn't seem to be a starting point. I now know the conditions at t = 0, and the only force is acting in the radial direction. There is something I'm not seeing here. We don't know what F is. Is there a way to relate it to the velocity of the string being pulled? I'm going to try and set the tangential equations of motion to zero and solve for the angular velocity that way.
     
  8. Apr 10, 2014 #7

    Radarithm

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    And apparently that method wont work. If I had to define "stumped" it would be this problem.
     
  9. Apr 10, 2014 #8

    BvU

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    What's wrong with ## F = m\, \omega^2\, r## ? I wonder if we really need it, though.

    Remember I was asking for physics thinking? Why don't we play teacher and student and I am the student:

    student: wrt to the origin, the wire can not exercise a torque on the mass because force and r are aligned. With torque ##= 0 = \tau = I\omega## :redface:

    [edit] Should be: torque ##= 0 = \tau = {\bf d\over dt}(I\omega)##

    and ## I = mr^2## I get ## m\omega r^2 = ## constant. So ## \omega r^2 = \omega_0 r_0^2##. Since we know how r changes with time, this leaves as a solution for ##\omega## the (non-differential) expression $$ \omega = \omega_0 \, {r_0^2 \over (r_0-Vt)^2} $$

    So what's wrong with this reasoning ?

    In the mean time I brushed up my jargon and discovered that K&K stands for BPC1 (which I don't have or ever have had), and these guys have a solid brick reputation. So if they want a differential equation, who am I with no reputation worth mentioning to come up with something else ? You be the judge (or anyone else listening in) and show me wrong !

    A good check might be an energy balance: The hand definitely does work, and the only place it can go is in kinetic energy. I'll chew on that for a while.
     
    Last edited: Apr 10, 2014
  10. Apr 10, 2014 #9

    Radarithm

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    I have to leave right now so I'll have to come back to this question at a later time. I don't see anything wrong with the way you answered this question, mainly because it obeys the hint and is dimensionally consistent
     
  11. Apr 10, 2014 #10

    BvU

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    It obeys the hint in the sense that ## {d\over dt}(I\omega) = 0 ## can be considered a differential equation in ##\omega##, yes. Perhaps I am digging too deep.

    Worked out the energy balance also. Want a try at it ?

    Time to go for me too. Nice exercise (internet only shows much duller equilibrium cases), BPC stature grown even higher in my book.
     
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