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B Need a bit more clarification on W=F.ds

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  1. Jun 22, 2017 #1
    I was a bit confused about the work energy theorem. I perfectly understood it's applications for point sized particles but I'm a bit confused about its application on extended bodies both rigid and non rigid.
    Case 1 {for rigid bodies} :-
    Consider a rod of definite mass hinged at the top. It's initially vertical. Now you apply a horizontal force F to the lowest point in horizontal direction of constant magnitude. The magnitude is low enough so that the rod doesn't gain any significant kinetic energy. However it gains potential energy as it rotates. The hinge doesn't do any work because there is no displacement of the point of contact. The applied force F is always horizontal. So F.ds in vertical direction is always zero. My question is how does it gain potential energy then?
    OR
    Consider a square block kept on a rough horizontal surface (friction is sufficient enough to prevent slipping/sliding). You apply a horizontal force F on the topmost point ,constant in magnitude and direction. The magnitude is adjusted so that the block doesn't gain significant kinetic energy. Due to rotation about the axis passing through the line of contact of the ground and block, the potential energy of the block increases as it loses contact with the surface. How does it gain potential energy?
    Case 2 {non-rigid body}:-
    Consider 2 blocks each of mass M, connected by a massless spring. One of the blocks is in contact with a vertical wall. You now compress the spring and let the motion begin after you have left the other block. The normal force doesn't do any work because there is no displacement of the point of contact. So How does the system gain kinetic energy?

    On the whole I want to understand what is actually meant by the elemental displacement term in the work energy equation. I assume that in case of a rigid body it's displacement of centre of mass and in case of non rigid body it's displacement of point of contact. Am I correct assuming this?
     
  2. jcsd
  3. Jun 22, 2017 #2

    A.T.

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    Work is a scalar. It doesn't have a direction.

    From internally stored potential energy.
     
  4. Jun 22, 2017 #3
    Can you please write the corresponding equation for case1 in work energy.. for both sub cases?
     
  5. Jun 22, 2017 #4
    I also need more clarification on the meaning of ds. Is it displacement of centre of mass or point of contact?
     
  6. Jun 22, 2017 #5

    jbriggs444

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    Point of contact.

    If you want to get picky, it is the displacement of the material in the object on which the force is being exerted at the point where the force is currently being applied that is relevant. This nit-picky point becomes crucial when discussing, for instance, the work done on a tire by the road. The "point of contact" is moving. But the "material at the point of contact" is not. So the road does zero work on the tire.

    Now, before @Doc Al jumps in to point it out, there is a concept of "center of mass" work in which it is the displacement of the center of mass that counts. So that the road is seen to do center of mass work on the car.

    For a rigid, non-rotating body, "center of mass work" is equal to plain old "work" and there is no distinction to be made.
     
  7. Jun 22, 2017 #6

    CWatters

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    I think the answer is F must have a vertical component. If there is no friction then the force acting on the rod must be normal to the rod, and at all angles (other than vertical) that has a vertical component.
     
  8. Jun 22, 2017 #7
    I was surfing on the internet and I came across this theorem
    https://en.m.wikipedia.org/wiki/Chasles'_theorem_(kinematics)
    If I apply this for a rigid body I think displacement of centre of mass would do( by treating it as a particle). But I'm still confused about non rigid bodies like ropes and bodies with many point of contact.
     
    Last edited: Jun 22, 2017
  9. Jun 22, 2017 #8
    I think it's not necessary. You can even perform this experiment at home using a pencil held in your hands. The vertical component of hinge reaction may exceed mg by infinitesimal amount to lift the centre of mass.
     
  10. Jun 22, 2017 #9
    But we can use F.ds in any direction we choose.
     
  11. Jun 22, 2017 #10

    jbriggs444

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    F has a direction. ds has a direction. Their dot product does not.

    If F is purely horizontal then clearly it has a non-zero horizontal component and a zero vertical component. If the pendulum is moving in a narrow region around bottom dead center then clearly ds has a non-zero horizontal component. So their dot product will certainly be non-zero.
     
  12. Jun 22, 2017 #11
    What I'm trying to quote is that the horizontal force F doesn't have any vertical components hence the work done by F is zero. But hinge reaction has a vertical component. Hence maybe hinge is doing work by displacing centre of mass. I know energy is scalar.
     
  13. Jun 23, 2017 #12
  14. Jun 23, 2017 #13
    In the above case I have tried to proceed from centre of mass energy to conservation of energy for a rigid body. Can anybody help me out on how does this work for non rigid bodies like ropes.
     
  15. Jun 23, 2017 #14
    For example a rope falling from a desk{with a hole} I can write the centre of mass energy but I can't proceed further.
     
  16. Jun 23, 2017 #15

    A.T.

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    This is not how work is defined.

    A static frictionless hinge does no work.
     
  17. Jun 23, 2017 #16
    So How would you explain the situation
    ?
     
  18. Jun 23, 2017 #17

    A.T.

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    See post #10.
     
  19. Jun 23, 2017 #18
    I'm talking about centre of mass energy above.
     
  20. Jun 23, 2017 #19

    jbriggs444

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    While there is a concept of center-of-mass work, the term "center of mass energy" is undefined (to my knowledge). The concept of center-of-mass work is poorly suited to a conservation of energy problem since it ignores some of the energy transfer due to the force.
     
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