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Centripetal Force: Rotating mass

  1. Feb 3, 2010 #1
    Mass on a string undergoing uniform circular motion by hand:

    I was taught that the centripetal force F = mV^2/R is a resultant force, and in this case was due to the tension force by the wire on the mass. I am trying to understand what is causing the tension and what causes the string to break at high speeds. Looking from above the plane of rotation, in order to keep the mass rotating, I appear to be moving my hand linearly back and forth to continue the circular motion of the mass. There has to be an applied torque to counteract the torque from friction (hand on string) trying to decrease the angular speed. I was wondering, is the torque that is causing the rotation a component of the normal force between the string and my hand? Is this normal force changing as my hand moves back and forth, or is it constant with respect to the center of mass of the system?
    What force causes the string to break/ over comes the tension force? It can't be the centripetal force because the centripetal force is a resultant force. Does this have to do with the reactive centrifugal force, and what would the reactive centrifugal force be in this case (tension force by hand on wire)? Should I be thinking about this in terms of angular momentum and impulse?

    Any help will be greatly appreciated
    Last edited: Feb 3, 2010
  2. jcsd
  3. Feb 3, 2010 #2


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    First, consider linear force. For example, the mass just hangs straight down on the end of the string. You hold the other end of the string. Is there tension in the string? (Yes, there is.) What causes it? Is there a support force on the hanging mass? (Yes, there is.) This support force is something like the resultant centripetal force, except that, instead of resulting in the centripetal acceleration, it results in (counteracting) the weight.

    Another example is to rest the mass on a slippery table top, connected to the string, and then pull the string in a straight line across the slippery table top. Is there tension in the string? (Yes, there is.) What causes it? The resultant force in this case would simply be the F=ma of the mass, where the F is equal to the tension in the string. (I made the table slippery so that I could neglect friction.) If you yank the string hard enough across the table, it will break.

    Now, back to the circular motion. You have already recognized that you must oscillate your hand back and forth in order to maintain the circular motion. This basically just amounts to pulling tension into the string, very similarly to pulling the string across the table top in the previous example. Except, in circular motion, you must persistently change the direction that you pull the string, thus the oscillation. It is the same basic kind of force that you apply to the string in the first two examples. The only difference is the oscillation, and this is needed because the motion is changing direction. The tension results from the fact that you pull on one end of the string, and the inertia of the mass resists the change in motion.
  4. Feb 3, 2010 #3
    The thread "Is centrifugal force real" might be helpful to you, as far as where the force comes from I would think it is because when you move your hand forward, your hand and the ball get linear velocity relative to each other in opposite directions(assuming the motion of your hand is perpendicular to the ball, if it's not you will either pull it behind your hand or run into it, non-rotation in either case.) but since we're talking in opposite directions, the centripetal force of the wire causes the linear accelerations to be angular.
  5. Feb 3, 2010 #4
    Agh, beat me to it, well at least we kind of answered in different scopes...
  6. Feb 3, 2010 #5
    I'm confused by the statement "the F=ma of the mass". Is this the reactive force to the string pulling on the mass (mass pulling on string)? So mass, in resisting acceleration (inertia), sets up a reactive force on the other object. So this reactive force should be constant, regardless of velocity, unless tension is a restorative force and increases with the radius of rotation. Is F=ma ever unequal to tension? Is the string breaking because the components of the string can accelerate faster than the mass?

    Could you elaborate on this? I was wondering how this is derived. Also, when I'm rotating the mass, aren't I in an inertia frame of reference? So the tugging force I'm feeling on my hand is a reactive force by the mass on the string resisting the centripetal acceleration? Is this why the centrifugal force is called and inertial force, because the reactive force to tension is due to the inertia of the object?

    Thank you for you replies.
  7. Feb 3, 2010 #6


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    Yes. Another name is "resultant force" or "net force".

    Yes. This is sometimes referred to as inertial force. However, the typically preferred way to deal with these problems is to consider the free body diagram of the mass, with tension as an applied force, and centripetal acceleration as the acceleration.

    If the string tension is the only force acting on the mass, then F=ma is always equal to the tension. That's just Newton's 2nd Law. Of course, there is gravity in the case of whirling a mass around your head on a string. And, in the case of the mass sliding on the table top, there is friction. However, you can still sometimes neglect these other forces, for instance, if the spinning mass on the string becomes almost horizontal.

    I guess that's one way to put it.

    The string breaks because it cannot handle more than a certain amount of tension (see tensile strength or breaking strength). So, if F=ma exceeds this amount, then the string will break, for the same reason that it will break if you hang a really heavy stationary mass from it vertically, whose weight exceeds the breaking strength.

    In the case of whirling the mass around your head, this implies that there is a maximum speed at which you can whirl the mass before the string breaks, because the a in F=ma is given by a=v2/r. However, the oscillatory motion that you must apply in order to provide the tension will most likely result in fluxuations of the tension over the course of a cycle. I suppose that you can train yourself to minimize this.
    Last edited: Feb 4, 2010
  8. Feb 4, 2010 #7
    the centripetal force of the wire causes the linear accelerations to be angular.

    the Ftension = Fcentripetal,


    neglecting the movement of you hand making it and ellipse it would work like that,

    it would go v, and 1/2ac every t and ac is always pointed toward the center, so it "tries" to moves linearly but the centripetal force constantly curves the path toward the center so if they are both constant they should have a constant curve, and end in the same place......not sure exactly how it was derived..seems pretty crazy that if you draw right triangles with the straight sides the lengths v and ac with the v side aligned with the hypotenuse of the last one it ends up making a circle with a radius equal to v2/ac.
  9. Feb 5, 2010 #8


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    I'm not sure what you're getting at here. A "resultant" force is normally defined as the net force produced from a sum of forces. There's no reason that a centripetal force has to be the sum of multiple sub-forces.

    Your hand is generating a pulling force at one end of the string, and the reaction force of the mass being accelerated is generating a pulling force at the other end of the string, so you have tension. It doesn't matter if this acceleration is linear or circular.

    In order to overcome the energy losses in this system, you're adding kinetic energy back into the system with linear acceleration of the mass. As you move your hand a back and forth, the actual path of the mass is spiral like and a component of tension in the string is in the direction of travel of the mass, which increases it's linear speed, or at least maintains it's speed against the friction forces opposing the speed.

    The torque is related to changes in the angular momentum of the twirling mass, and the increase in linear speed from your hand movment is also maintaining the angular momentum against the opposing torques due to friction. The string is too flexible to be able to generate a signifcan torque on this system, all it can do is generate linear forces.
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