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When to apply conservation of angular momentum?

  1. Jul 1, 2015 #1
    So my question is when to apply conservation of angular momentum?When there is no external force ,right?But in the case below
    A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached to a string which passes through a smooth hole in the plane as shown.
    The tension in the string is increased gradually and finally m moves in a circle of radius R0/2 . The final value of the kinetic energy is
    AIPMT_2015_11_alagu_3515_22_56_3_Ouput.png
    A) mv20
    B) 1/4 mv20
    C) 2 mv20
    D) 1/2 mv20
    My teacher applied conservation of angular momentum.But I don't know why?As tension force is external force the angular momentum should not be conserved.
    Note:I am not posting this question in homework sectio because I know it's solution .I don't want method to solve rather I am having a conceptual doubt.
     
  2. jcsd
  3. Jul 1, 2015 #2

    Dale

    Staff: Mentor

    You can apply conservation of angular momentum whenever there is no external torque. The tension produces no torque about the axis.
     
  4. Jul 1, 2015 #3
    Because tension force is the only force,and for torque we require a pair of force separated by perpendicular distance,right?
     
  5. Jul 1, 2015 #4
    right?Sir,please answer.
     
  6. Jul 1, 2015 #5
    Torque requires you to choose an origin

    Nop.

    In magnitude the torque of a given force about certain point is given by
    ##
    \mathbf{\tau}=rFsin\theta##

    where r measured to point in question. In this case choose to measure r about the center of the circle, I mean choose the center of the circle as the origin of the coordinates.

    In this case r is the radius of the circle, note that the tension and r are parallel, so the angle is 0 so the sin(theta)=0. Then, the torque associated to the tension is null, therefore the angular momentum is conserved.
     
  7. Jul 1, 2015 #6

    Drakkith

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    Staff: Mentor

    The force of tension in the string always points perpendicular to the direction the object is moving, so its component in the tangential direction is zero, and thus it does no torque. In order for a force to exert torque on an object, the component of the force in the tangential direction has to be non-zero.
     
  8. Jul 1, 2015 #7
    This is not true, in this example the motion have a tangential and a radial component.

    This is true and is the reason there is no torque.
     
  9. Jul 1, 2015 #8

    Drakkith

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    Staff: Mentor

    Yes, but the force from tension only has a radial component, right?
     
  10. Jul 1, 2015 #9

    Nathanael

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    Homework Helper

    Yes this is correct. But the object also has a radial component of speed (it is not moving in a circle, it is spiraling inward) and so the tension is not perpendicular to the motion.
    This is why the energy is larger at the end of the problem: the tension did some work.
     
  11. Jul 1, 2015 #10

    Drakkith

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    Staff: Mentor

    Roger.
     
  12. Jul 2, 2015 #11
    This pair of force is called "couple",I think it is required in calculation of moment.
    So,torque doesn't need two forces.It just needs a force and some angle theta with r which is distance between point of application of force and axis of rotation.
     
  13. Jul 2, 2015 #12

    Dale

    Staff: Mentor

    A torque requires a force that has a component perpendicular to the axis of rotation some distance from the axis of rotation. In this case, the tension force is through the axis of rotation and therefore it's torque is 0.
     
  14. Jul 10, 2015 #13
    The important information given in the question is that tension is increased gradually . The increasing tension causes the block to move closer to axis of rotation.

    The block then moves towards the center in a spiral path , such that tension has both radial and tangential components , of which the radial component can produce a torque making conservation of angular momentum invalid .

    But because tension increases gradually the block moves towards the center in an approximately circular path and tension will have only a tangential component . Tension doesn't produce torque .

    You can now conserve angular momentum .
     
  15. Jul 10, 2015 #14

    A.T.

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    Science Advisor
    Gold Member

    The tension in the string is always purely radial, towards the center. You might mean normal and tangential components, perpendicular and parallel to the blocks velocity.

    The string tension cannot produce any torque around the center, no matter how you decompose it into components. The torques from the two components will simply cancel.

    Now you seem to confuse tangential with radial.

    Yes, because it is only radial. No matter how quickly you increase it.
     
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