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B Conservation of Energy for Unwinding Mass

  1. Aug 6, 2017 #1
    As a spinning mass is unwound, angular momentum is conserved, meaning ##Iω## remains constant. However, since rotational energy is proportional to the square of the angular velocity, how is it possible for energy to be conserved as well?
     
  2. jcsd
  3. Aug 6, 2017 #2

    Merlin3189

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    Is energy conserved?
    If you think of the spinning ballerina/skater scenario, or child on a roundabout, then as the mass is moved to the centre, I decreases and conservation of angular momentum speeds up rotation (ω). To achieve this, they must do work pulling in their arms. Similarly, when they lean outwards or move arms/ legs outwards, I increases and conservation of angular momentum slows down rotation. Work must be done on them to stretch their muscles as they reduce the centripetal force.

    You don't say how your reel unwinds. Perhaps work is done stretching the rope.

    Edit 2: By, is energy conserved, I meant rotational KE. Obviously total E is conserved, but I suggest work being done or conversion to spring PE.
     
    Last edited: Aug 6, 2017
  4. Aug 6, 2017 #3
    Thanks for the response—it seems reasonable enough, but I still have some confusion for an example such as this:

    A mass is attached to a piece of string and the string is wound around a rod. The string is latched on to the rod and the rod is rotating at a constant angular velocity. The string is then unlatched, allowing the mass to move outward, increasing the moment of inertia, decreasing the angular velocity, and decreasing the rotational energy.

    How could work be done on the string or mass in this example?
     
  5. Aug 6, 2017 #4

    Merlin3189

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    Right. I'll need to start doing some sums ! But for now, maybe:
    CircularKE&Momentum.png
    M can be a mass on the end of the rope, or just the CoM of the extended rope maybe.
    Now due to the tangential tension, there is a component of force accelerating it in the v direction, as well as the radial centripetal force. This force accelerates the mass as it moves outward, so that v increases to keep up with the constant ω as r increases. (v = rω)
    So whatever is turning the drum is doing work on the mass to increase its KE.

    Edit: Whoops ! I forgot ω isn't constant. So I definitely need to look at the sums first.
    It's getting complicated with I increasing, ω decreasing, r increasing. I don't think v can stay constant, but there's the opportunity for work to be done either way through the rope tension. But once the drum is being forced to rotate, you are changing the rotational momentum as well as KE. Maybe if the drum is light and free to rotate how it will, when the rope is freed, the mass flies off at a tangent with constant KE and decreasing angular velocity to compensate for the increasing moment of inertia about the axis.
    Edit: I think we'have to specify the situation more clearly.
     
    Last edited: Aug 6, 2017
  6. Aug 6, 2017 #5

    Dale

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    The string could be used to do work here, so the KE does decrease (unless the mass just flies off in a straight-line tangent at constant speed)
     
  7. Aug 6, 2017 #6
    The system contains a rod (or drum) of radius ##r##, a string, and a mass of mass ##m##. The mass of the rod and string is ignored. Friction is ignored. No external forces are applied to the system.

    The string is initially wound around the rod and the entire system is rotated with angular velocity ##ω_i##. The mass is allowed to unwind.

    The energy in the system would be ##½mr^2ω_i^2##.

    Is everything correct so far?
     
  8. Aug 6, 2017 #7

    jbriggs444

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    Under these conditions, the mass at the end of the string moves in a straight-line trajectory, unaffected by the massless string wrapped around the massless drum.

    What is the real setup?
     
  9. Aug 7, 2017 #8

    Merlin3189

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    Ok.. ## \frac 1 2 mr^2ω^2 \ \ \ or\ equally \ \frac 1 2 mv^2 ## ie. it has so much KE whether we call it linear or rotational.
    And ## angular momentum \ = Iω \ \ = mr^2ω \ =\ mvr ##

    So now it flies off on its tangent with no restriction from the light frictionless string and rod. So neither the velocity, KE nor momentum change.
    With no force / torque acting, linear and angular momentum are conserved, but one can look at it innocently and see:
    I always need a picture, so
    CircularKE&Momentum3.png
    R (the distance from the axis) increases, causing changes in some angular elements.
    I is increasing with R2, ω is decreasing as the component of velocity perpendicular to the radius decreases and R increases.
    The velocity, instead of being perpendicular to the line from the mass to the axis, is at an angle α to that line,
    ## R = \frac {r} { cos(α)} ##
    perpendicular component of velocity ## v_P = v \cos (α) ##
    radial component of velocity ## v_R = v \sin (α) ##
    ## ω = \frac {v_P} {R} = \frac {v \cos^2 (α)} {r } ##
    ## I = mR^2 = \frac {mr^2} {\cos^2 (α)} ##
    ## L =Iω = \frac {mr^2} {\cos^2 (α)} \frac {v \cos^2 (α) } {r } \ \ = m r v ##
    So the angular momentum is still the same if we calculate it in the new position, as expected.

    If we now change the spec, as jb suggests, and allow the string to exert some force on the mass, this is a torque about the axis and can change the angular momentum. If the rod is massive, its angular momentum is part of the sum, so it could gain while the mass could lose angular momentum, or vice versa.
     
  10. Aug 7, 2017 #9
    Yes, that was a dumb mistake on my part.

    Let's instead imagine the rod is secured to the ground and stationary, which would therefore cause the earth to apply force on the system. All the constants and values I wrote before would remain the same. I would still imagine that angular momentum and energy would be conserved, however.
     
  11. Aug 7, 2017 #10

    jbriggs444

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    Angular momentum of the earth plus mass on end of string? Yes.
    Energy of earth plus mass on end of string? Yes.

    To a good approximation, the work done on mass by string and on earth by string will be zero -- the mass will continue in a trajectory at constant velocity and increasing angular momentum. But why would you expect energy to be given by ##\frac{1}{2}I \omega^2##? We are not dealing with a rigid assembly rotating around a fixed axis.
     
  12. Aug 11, 2017 #11
    Would it be possible to approximate the scenario by imagining the earth had an "infinite" mass and therefore didn't move at all? I can't figure out how to solve the problem while taking earth's motion into account.
     
  13. Aug 11, 2017 #12

    jbriggs444

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    Sure. Do that. Assume the earth is motionless. In this case, the direction of the force of the string is always at what angle to the current motion of the mass?
     
  14. Aug 11, 2017 #13
    Since the string applies a force radially inward, the angle would be 90 degrees, right?
     
  15. Aug 11, 2017 #14

    jbriggs444

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    The angle is indeed 90 degrees. But what makes you say that the string applies a force radially inward? Certainly it is not a force directed toward the center of the drum/spool.
     
  16. Aug 11, 2017 #15
    Right—it isn't exactly radially inward. That would only be the case if the mass were rotating around in circles and the string was connected to the center. In this case, the force would be applied off-center, tangential to the center rod, I believe. This force, which is the force of tension, would always be applied in the same direction as the string.
     
  17. Aug 11, 2017 #16

    jbriggs444

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    Good. Now, because the force is not exactly radially inward, we do not have conservation of angular momentum. The angular momentum of the mass will be increasing due to the net torque from the string. But let us not worry about that just yet.

    Now, if we agree that the force of the string is at right angles to the motion of the mass, can we also agree that the speed of the mass will be constant?
     
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