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Force, Energy, and a Rolling ball

  1. Apr 17, 2006 #1
    I was engaging in a thought exercise and got into an argument with a friend the other day.

    Suppose you have a sphere on a frictionless plane (or in space or whatever).

    If you apply a force directly in normal to the surface, it obviously accelerates at a=F/m. This gives (after some delta t) an energy of 1/2mv^2.

    Now, if instead you apply a force tangent to the ball at the same point for the same delta t, the ball instead starts rotating, and translating. But this gives an energy of .5*i0*w^2 + .5mv^2 (due to rotational kinetic energy and translational energy).

    Even though the force is the same in both cases, the energies are different. Is this correct?

    My friend on the other hand believes that because the force is applied tangentially, there is only rotation and no translation of the object. In this case, perhaps the energies are equal?

    I've included a diagram of what I mean.
    http://separatereality.hackorp.com/files/pictures/sphere.jpg
     
  2. jcsd
  3. Apr 17, 2006 #2

    arildno

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    Sure it is correct.
    The rate of energy transfer to the object equals the power, that is the force multiplied with the velocity of the object point upon which the force acts.

    So, if the contact point where the force acts has a higher velocity than the C.M (as in your case), the energy gain for the object will be greater.
     
    Last edited: Apr 17, 2006
  4. Apr 17, 2006 #3

    Hootenanny

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    Just to add to what arildno said, the only way to prevent the ball from translating is to fix it's axis. A good rule of thumb is that if you apply an unbalanced force the centre of mass of an object will move.

    Regards,
    ~Hoot
     
  5. Apr 18, 2006 #4
    But why would the body rotate? The axis is not fixed. Nor is there an opposing force (e.g frictional force in rotation with translation).
    In other words, suppose u applied a force into a face of a rectangular block but away from the centre of mass, would the block rotate? The block is free from other forces (friction, air resistance etc)
     
  6. Apr 18, 2006 #5

    Hootenanny

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    Yes, it would rotatate and translate.
     
  7. May 13, 2006 #6
    Suppose first that a force [itex]F[/itex] is applied to the center of mass of a rigid sphere of mass [itex]M[/itex] and radius [itex]R[/itex].

    Translation is described by

    [tex]MA_{CM} = F[/tex]

    In the case being discussed, the surface is frictionless. The net torque about the mass center of the sphere is zero and hence the angular acceleration [itex]\alpha = 0[/itex]. This does NOT mean that the sphere cannot rotate on the frictionless surface. It simply means that there is no tendency for the angular velocity to change during the course of motion of the sphere. So if you placed an already spinning sphere, all this means is that it will continue to spin with the same angular velocity as the force being exerted on the sphere is unable to give a torque.

    If I instead exert the force at the topmost point of the sphere, summing torques about the mass center gives

    [tex]I_{CM}\alpha = RF[/tex]

    In this case, an [itex]\alpha[/itex] does develop and so the angular velocity does change, depending on F. In this case there is no friction yet the force being applied is causing rotation (if you started out with zero angular speed, then the acceleration would give rise to an increasing angular speed for time [itex]t \geq 0^{+}[/itex]).

    As for energy, the kinetic energy of sphere can be written as

    [tex]T = \frac{1}{2}I_{CM}\omega^2 + \frac{1}{2}MV_{CM}^2[/itex]

    It is obvious that the torque [itex]RF[/itex] increases the first term and the force [itex]F[/itex] increases the second term so there is on overall pumping of kinetic energy into the sphere. The power is simply [itex]dT/dt[/itex].

    As for the axis, well this is a case of parallel translation of axes and hence there is no ambiguity as far as these equations are concerned. The axis of the sphere does translate (with an acceleration [itex]A_{CM}[/itex]) but it always remains parallel to itself hence the directions of torque, angular velocity and acceleration remain the same though each of these quantities as vectors translate parallel to each to other in space (in a plane parallel to the linear surface).

    As for the block, even on a frictionless surface there will be three forces acting: the external force F, the weight of the block (W) and the normal reaction (N). When you apply F so that its line of action does not pass about the mass center, there are 2 forces which give torques about the CM: F and N. Since you have no friction to balance F (to keep the block from translating), there will be translation. The line of action of N will adjust according to the tendency to topple (that is according to the point of application of F). It will not in general pass through the CM. If the torques about the center of mass balance, then there will be no rotation but only translation. However, if they do not balance, then there will be a translation + rotation.
     
    Last edited: May 13, 2006
  8. May 13, 2006 #7
    The forces N and Mg act in one plane and the force F in another - both the planes being perpendicular to each other. In such a case, why would their torques add up or cancel? I'm considering say a book kept on a frictionless surface and being pushed by a force parallel to the surface i.e perpendicular to N and Mg.


    What if the block was in space i.e free from gravity and all forces, a single force applied away from the center of mass would be able to cause rotation would it? This is what i actually wanted to ask.
     
  9. May 13, 2006 #8
    okay, i got the cause of confusion. I believe maverick280857 has considered an axis parallel to the surface. I was visualizing a book lying on a surface - the way we usually place books, and so an axis perpendicular to the surface.
     
  10. May 13, 2006 #9

    rcgldr

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    Energy is related to force times distance, not force times elapsed time. In order to apply a tangental force, there has to be friction with no normal force, so let the sphere's diameter be like a gear, and the force applied via a flat plane that has geared teeth on it.

    Using this example of applying the tangential force via a flat plane with an infinite coefficient of friction, the flat plane travels further while applying the force than the normal force would in the same amount of time.

    If the forces were limited so that the distances involved were the same, then the net increase in energy would be the same.
     
  11. May 14, 2006 #10

    arildno

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    Eeh? Whoever has stated that?
     
  12. May 14, 2006 #11

    rcgldr

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    It's a simplified way to describe work, as in Newton-meters, pound-feet, which is related to a change in energy. The point was that it's not related to time (Newton - seconds?).

    The techinal definition for work is the (line) integral of the component of force in the direction traveled, over the distance traveled.

    [tex]\int _{s0} ^{s1}\ F(s)\ cos(\theta)\ ds[/tex]
     
    Last edited: May 14, 2006
  13. May 14, 2006 #12

    Hootenanny

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    I think arildno was refering to this bit;
    ~H
     
  14. May 14, 2006 #13

    rcgldr

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    It's in the starting post for this thread:

    as opposed to "for the same delta s".
     
    Last edited: May 14, 2006
  15. May 14, 2006 #14

    arildno

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    Well, an arc length parameter definition of work is entirely equivalent to defining work as the net sum ("integral over") of power times time, since the arc length parameter is in bijective correspondence to the time parameter, so you can perfectly well define work in terms of power over elapsed time.
    This is slightly more useful for dissipative systems in which the "distance" concept is rather abstract. For energetically closed (non-dissipative) systems, as in, say, classical thermodynamics, where intermediate velocities between different states is of secondary interest, the distance definition is more useful and natural.
     
  16. May 14, 2006 #15

    Hootenanny

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    Arildno was refering to the equation P = Fv;

    [tex]\frac{dE}{dt} = Fv \Rightarrow dE = Fv dt[/tex]

    Energy is equal to the product of force, velocity and time. RE post # 2.

    ~H

    Edit sorry arildno, didn't see your post
     
  17. May 14, 2006 #16

    rcgldr

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    Note my response was to the original post, not to the other replies. The original post mentions force and time, not power nor velocity. This is the issue for the difference in work done in the two cases, force and time are the same, but not the distance or velocity as previously posted, which is why the change in energy for the two cases is different. I mentioned distance because force times distance is the classical definition for work.

    I left out mentioning that a longer distance in the same time implied a higher average velocity, but this was already mentioned in another post, and I was responding to the original post.
     
  18. May 14, 2006 #17

    arildno

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    Okay, but OP does not anywhere state that energy is gained by multiplying force with time.
    He roughly said that over some time interval, the object will eventually gain energy.. . This is both correct and meaningful to say.
     
  19. May 14, 2006 #18

    rcgldr

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    But the OP implies an apparent dilemma for his friend where the the same force and time are input, and the resulting increase in energy is different. It also seems to imply that the point of application of the force is what is causing the difference in energy increase, as opposed to the fact the the distance that the force is applied is different in both cases.

    If the OP were changed to stating that the same force and distance were applied in the two cases, then the resultant gain in energy would be the same. Instead, the OP mentions the same "delta t", instead of the same distance. This is what I was trying to point out.
     
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