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Work done by tension

  1. Aug 29, 2015 #1

    Titan97

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    When a particle is attached to a rope and rotated, then tension acts perpendicular to displacement and work done is zero. But I am thinking about this case (top view):
    Untitled-1.png
    Here, the rope is attached to a vertical cylinder and the particle is given a velocity perpendicular to the rope. The length of rope that's in motion decreases and the particle comes closer and closer to the cylinder. Hence, can I say that the particle moves towards the point P which is along the direction of T. Hence Tension does work. Am I correct?
    Another problem is gravity. The particle can go downwards (into the plane of webpage) due to gravity. Let me assume that there is no acceleration due to gravity.
    (I have not taken this from any book. I used MS paint to draw this)
     
  2. jcsd
  3. Aug 29, 2015 #2
    Yes.
     
  4. Aug 30, 2015 #3

    Titan97

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    But here is the problem:
    Let theta be the angle between the line joining C and bottomost point of cylinder and the line CP.
    Let C be origin. And the particle be N
    The vector along CP is ##\vec r_{CP}=R(\sin\theta \hat i - \cos\theta \hat j)##
    The vector along T is ##\vec r_{PN}=(l-R\theta)(cos\theta \hat i + \sin\theta \hat j)##
    The vector ##r_{CN}=\vec r_{PN}+\vec r_{CP}##
    So the position vector of the particle is ##r = r_{CP}+r_{T}##. The tangent vector to the path of the particle is
    $$\frac{dr}{d\theta}=(L-R\theta)(-\sin\theta,\cos\theta)$$
    which is perpendicular to the direction of tension.
    But since I haven't learnt vector differentiation yet, I am getting doubts on what I did o0).
    Is the particle supposed to have velocity along tangent?
     
  5. Aug 30, 2015 #4

    Titan97

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    Can you help me find the equation of trajectory of the particle? I thought It might be useful to find the equation for further analysis.
    I think its easier to find the equation in polar coordinates.
     
  6. Aug 30, 2015 #5

    OldEngr63

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    There is no work done by the the string tension. Here is a solution, in the attached PDF. As for the equation of the trajectory in the plane, the particle travels on an involute of the circle.
     

    Attached Files:

  7. Aug 30, 2015 #6

    Titan97

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    What about @Chestermiller 's post?
    Also, can you show me how you found dr/dθ?
    Can I say that the centripetal force increases as radius decreases and hence tension, which provides centripetal force also increases and velocity of particle increases?
     
    Last edited: Aug 30, 2015
  8. Aug 30, 2015 #7

    OldEngr63

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    Regarding Chester's post, what can I say except that Chester is mistaken in this case?

    In my PDF, I showed you the full development of dr/dθ. What else do you need?

    As the particle approaches the fixed cylinder, the particle velocity will increase. The acceleration in the direction of the cord is v^2/R where R is progressively getting smaller.

    Without any math at all, simply observe:
    1) The tension in the cord must act along the cord;
    2) As long as the cord remains taut, the only possible motion of the particle is perpendicular to the cord;
    Since the force and the motion are at right angles, there is no work done.
     
  9. Aug 30, 2015 #8

    Titan97

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    Can you help me find the equation of the particles trajectory? I thought it would be useful.
    The trajectory is like a spiral. So polar coordinates might be better.
     
  10. Aug 30, 2015 #9
    Yes. I was too hasty in saying what I said. Your analyses both show that the tension is normal to the velocity, and no work is done by the cord. Sorry for any confusion I may have caused.

    Chet
     
  11. Aug 30, 2015 #10

    Titan97

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    That's OK. But can you help me find the equation of trajectory?
    I am stuck here.
    WP_20150831_00_22_05_Pro.jpg
     
  12. Aug 30, 2015 #11
    From your original post, the vector velocity is given by:

    $$\vec{v}=(L-Rθ)[-sinθ\vec{i}_x+cosθ\vec{i}_y]\frac{dθ}{dt}$$

    Since no work is being done, the KE should be constant.
    $$KE=\frac{1}{2}(L-Rθ)^2\left(\frac{dθ}{dt}\right)^2$$
    So, for constant kinetic energy,
    $$(L-Rθ)\frac{dθ}{dt}=v_0$$
    where v0 is the initial velocity. So,

    $$Lθ-R\frac{θ^2}{2}=v_0t\tag{*}$$

    When θ=L/R, the cord is fully wound. This occurs when ##t=L^2/(2Rv_0)##

    Chet
     
    Last edited: Aug 30, 2015
  13. Aug 30, 2015 #12

    Titan97

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    I found the time it takes to hit the cylinder. I was trying to make the situation interesting. For example, someone could take this situation and ask "find the equation of trajectory". I want to find the equation of trajectory to make things clear. I am stuck in the steps shown in post #10. The trajectory is like a spiral.
     
  14. Aug 30, 2015 #13
    The equation tagged with a * in post #10 can be used to solve for θ at any arbitrary value of t by applying the quadratic formula. Once you know θ vs t, you know the entire trajectory (since you know ##\vec{r}(θ)##).

    Chet
     
  15. Aug 30, 2015 #14

    Titan97

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  16. Aug 30, 2015 #15

    OldEngr63

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    As I mentioned in #5, the trajectory is an involute of the circle. You can look this up on-line for more details about the involute.
     
  17. Aug 30, 2015 #16

    rcgldr

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    On a side note, in this case it would appear that angular momentum isn't conserved, until you take into account that the tension on the post creates a torque on the post and whatever the post is attached to, such as the earth, and only by considering the entire system, particle, rope (if it has mass), post, and earth, is angular momentum conserved.
     
  18. Aug 30, 2015 #17

    Titan97

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    ##L=m(v\times r)##
    v is constant while r changes and the angle between v and r also changes. So the angular momentum is not conserved right?
     
  19. Aug 30, 2015 #18

    Titan97

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    In post 10, I got the equation for x in terms of ##\phi##. But how do I bring it to the form of ##x=r(\sin\theta - \theta\cos\theta)##?
     
  20. Aug 30, 2015 #19
    $$r^2=x^2+y^2$$
    $$\tanθ=\frac{y}{x}$$
    $$x = r\cosθ$$
    $$y = r \sin θ$$

    Chet
     
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