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Changing moment of inertia

  1. Aug 31, 2006 #1

    daniel_i_l

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    If you are spinning around with your arms flung-out to the sides and then pull them in it makes you spin faster. I think that the conventional explanation for this is that pulling your arms closer to your body makes your MOI get smaller and in order to conserve angular momentum you have to spin with a greater angular velocity.
    My question is - if the force applied to pull your hands to your body was always prependicular to the velocity (the force was applied along the radius of the circle that your hads trace as you spin around) how can it affect the angular velocity, in other words, how can a force that applies no torque create angular acceleration?
    I think that the answer could be related to the centripital force - that force is in the same diretion as the force pulling your arms together (along the radius) and it is the force that causes circular motion. could it be that increasing this force increases the angular accelaration?
    Any explanation of this would be very helpfull.
    Thanks.
     
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  3. Aug 31, 2006 #2

    Galileo

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    It's nowhere stated that the angular velocity can't increase if there is no torque. The torque is equal to the change in angular momentum, no torque means that the angular momentum is constant. If the MOI decreases, the angular velocity has to increase to keep the product constant.

    There IS an increase in kinetic energy, though. And that's because you do work when bringing your arms closer to the axis of rotation.
     
  4. Aug 31, 2006 #3

    daniel_i_l

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    So what force is making you spin faster?
    Thanks
     
  5. Aug 31, 2006 #4
    Why do you say that Centripetal force can't apply torque? I know it is true for the circular motion with fixed radius. But in this case, you're arm is really going spiral motion by changing it's radius. The tangential velocity is no longer perpendicular to the raidual vector.
     
    Last edited: Aug 31, 2006
  6. Aug 31, 2006 #5

    ZapperZ

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    Torque is defined as r x F, where r is the radial position of the force being applied, and F is the applied force. If you look at it carefully, both the r vector and F are parallel to each other when F is a centripetal force. So what is the cross product of 2 vectors when they are parallel?

    A radial force does not add an external torque to the circular motion. That is why angular momentum is conserved. If there is an external torque, then the system's angular momentum will no longer be conserved the radius changes.

    Zz.
     
  7. Sep 7, 2006 #6

    daniel_i_l

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    Ok, so i still don't understand, which FORCE makes you spin faster? or do you not need a force - how can something accelerate with no force?
     
  8. Sep 7, 2006 #7
    when you pull your arms, at first the arms still have the same velocity as before. but then your arms will move too fast for their radius, so they will attemp "escaping" the circle. but since you are a solid(not exacly, but it isnt biology here), and avoid letting the arms go, you will use force to prevent the arms from excaping the circle(a tangental force which reduces the velocity of the arms to the one fitting their radius), by the third law, there would be a reaction force, which will accelerate the rest of the body.

    but thats just my guess, i didnt read it in a book or anything...
     
  9. Sep 11, 2006 #8
    When your hands are flung-out the path they move along is a large circle, lets say they make one rotation in a second, when you move your hands towards the centre of the circle the path they will move along is now a smaller circle so in order for the hands to move a similar distance per second they will now have to travel more than one rotation in one second.

    You also do make the hands move faster when you pull them inwards as they now have the tangential component Vt and a radial component Vr, so the velocity is sqrt(Vt^2 + Vr^2).
     
  10. Sep 11, 2006 #9

    rcgldr

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    Although there's no outside force, there are internal angular forces accelerating your body. Since the component of energy of your arms and hands is conserved, their speed tends to remain constant, regardless of the path they follow. If you pull your arms and hands inwards, the initial reaction is to retain their current speed, conserve angular momentum, and rotate faster. However, since your hand are attached to your arms, which are attached to your body, there are internal angular forces generated, forwards by your arms to your body, and backwards from your body to your arms (equal and opposing). The net result is your body's rate of rotation increases, while your arms and hands rate of rotation increase less than they would if they were free to rotate independently of your body. Your body perimeter ends up with more speed, and your arms and hands ends up with less speed, an internal redistribution of the kinetic energy. The overall angular momentum is conserved, as well as the overall energy.

    To make this simpler, imagine a ball on a string threaded through a tube. For the first case, imagine that there is no friction between the string and the tube, and the weight of the string is negligble. In this case, the speed of the ball remains constant regardless of it's path, and pulling the string inwards increases the rate of rotation, but doesn't change the speed. For the second case, imaging that the string is threaded through a frictionless slot in the tube, and that the tube has significant angular momentum. As the string is pulled inwards, the rate of rotation of the ball, and string increase, and the string generates an angular force on the tube at the slot, and vice versa, increasing the speed of the perimeter of the tube, and decreasing the speed of the ball. The system as a whole retains it's energy and angular momentum, but there is a redistribution of the angular momentum and speed in the components of this system.

    Here's a short video of a guy doing a quadruple back flip off "swinging rings". During the swinging phase, the person shifts his weight outwards at the end of the swing where there is little g-force, and inwards (upwards) through the center part of the swing where there is a lot of g-force. Work is being done during this phase. In just 4 half cycles, the swing angle increases from about 45 degrees each way to just over 90 (center of gravity slightly above the support bar). In this case, the person starts tucking before release, but the rate of rotation is much greater in the tucked position versus open where instead of 4 flips, he only does 2 (plus a double twist) in the later part of this short clip, viewed from above (I climbed one of the chains, as the support poles were too slippery) in the second part.

    quad.wmv
     
    Last edited: Sep 11, 2006
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