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Tension in the strings after string is cut

  1. Mar 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Q .Find the tensions in the strings (1), (2) and (3) and the acceleration of the mass ‘m’ just after :
    (a) string (1) is cut
    (b) string (2) is cut
    (c) string (3) is cut


    2. Relevant equations



    3. The attempt at a solution

    I feel the answer should be

    (a) T1 = 0 ; T2 = mg ; T3 = 2 mg ; a = g
    (b) T1 = mg ; T2 = 0 ; T3 = 0 ;a = 0
    (c) T1 = mg ; T2 = 0 ; T3 = 0 ;a = 0

    I would be grateful if somebody could check my understanding of the problem . Have I answered correctly ?
     

    Attached Files:

  2. jcsd
  3. Mar 16, 2014 #2

    NascentOxygen

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    When string 2 is cut, why do you say the mass won't start falling?

    What properties are you attributing to the spring?
     
  4. Mar 16, 2014 #3
    The extension in the spring will not change instantaneously .Just after the string 2 is cut,the mass will still have upwards force T1 and mg downwards ,with kx=T1=mg ,which we had obtained from the equilibrium state (just before the string is cut) .
     
    Last edited: Mar 16, 2014
  5. Mar 16, 2014 #4

    NascentOxygen

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    There are some arbitrary assumptions that can be made here, and, as always, it's prudent that you fall in line with the view of whoever will be marking your work. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]

    It appears that you are attributing to the spring a significant mass, i.e., the spring has mass which is not insignificant under dynamic conditions. Accordingly, I agree with your answers for (a) and (b).

    Are we likewise to assume the pulley has mass which is not insignificant under dynamic conditions? If not, why not? If so, then I don't agree with your answer (c).

    Your answer for (c) would accord with the pulley being massless, without inertia.

    I'm hoping someone else will confirm or refute my answer, as I'm not confident answering questions which seem to begin, "First, flip a coin ...".
     
    Last edited by a moderator: May 6, 2017
  6. Mar 17, 2014 #5
    I think the pulley and the spring are assumed massless in this problem.Do you mean a) and b) are correct if spring has mass and c) is correct if pulley is massless ?

    Could you please explain how the mass of spring affects our results ?
     
    Last edited by a moderator: May 6, 2017
  7. Mar 17, 2014 #6

    ehild

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    Imagine you fasten the spring to a wall, the other end to a small ball. Pull out the ball and then release on a horizontal surface, what happens? The ball will oscillate.
    The same happens if you fasten balls at both ends. Pull the balls apart, then release them. They will oscillate about the common centre of mass.

    Choose one mass smaller and smaller. The centre of mass gets closer and closer to the centre of the other ball, which will oscillate with less and less amplitude. At the limit, the amplitude of the oscillation gets zero, while the frequency stays finite. The acceleration of the ball becomes zero.

    Here the spring was considered massless, but you get similar result if you replace the other mass with a massive spring. There is initially tension along the spring. That tension acts on the mass at the end. Releasing the other end, the spring-ball system start to do SHM. The spring contracts, but the ball feels the initial tension so accelerates inward, toward the CM of the system.

    If you release the free end of the spring, the tension at the free end becomes zero and a wave is initiated. The speed of the wave depends on the mass of unit length of the spring. If that mass is zero, the speed is infinite. The disturbance reaches the other end in no time.

    ehild
     
  8. Mar 17, 2014 #7
    Thanks ehild :smile:

    Is it correct to say that conclusions(answers) drawn in post#1 are correct under the assumption that spring is not massless and pulley is massless ?
     
    Last edited: Mar 17, 2014
  9. Mar 17, 2014 #8

    NascentOxygen

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    Why not? Ask yourself "why can't the spring instantly revert to being unstretched?" The only answer can be "because the coils of the spring have mass, therefore inertia", and it's this that slows down the spring's movement.

    The only basis on which the spring is able to exert any force on the mass below it once string 2 is cut must be inertia due to the spring coils themselves having mass. When string 2 is cut, the two ends of the stretched spring both tend to retract towards the middle of the spring, and it's this upward pull by the lower end of the spring which explains how tension persists in string 1 after string 2 is cut. If the spring were massless, tension in string 1 would instantly fall to zero simultaneously with the cutting of string 2.

    I'm undecided on for how long that tension in string 1 will equal mg. It would make for an interesting classroom demonstration to verify the initial magnitude of this tension and its variation--if any--with time.
     
    Last edited: Mar 17, 2014
  10. Mar 17, 2014 #9
    Have I concluded correctly in post#7 ?
     
  11. Mar 17, 2014 #10

    NascentOxygen

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    Yes, that's my conclusion.
     
  12. Mar 17, 2014 #11
    Wonderful...Thank you very much Nascent :smile:

    Is it correct to say that the massless spring is equivalent to a massless string as far as transmission of tension is concerned across its length ?
     
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