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Homework Help: Tension in a string swinging a rock.

  1. Oct 31, 2007 #1
    A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane.
    A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
    B) what is the tension in teh string at the bottom of the swing?
    C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal?

    I got the answer for part a) and part c). I am not sure about Part b).
    Here is what i did..:
    A) At A: Net force= -T-mg
    = -(T+mg)
    But weight at A should be zero.
    Hence, Net force= -T = -mg= -0.2*9.8= 1.96 N (Down)
    F_c= mv^2/r= 0.2*v^2/0.75
    v^2= 1.96*0.75/0.2
    v = 2.71m/s
    At B: Net force= T-mg
    F_C= mv^2/r
    = 0.2*2.71*2.71/0.75
    = 1.958 N
    Net Force= T-mg
    T= Net Force+mg
    T= 1.958+0.2*9.8
    T= 1.958+1.96
    T= 3.918N
  2. jcsd
  3. Oct 31, 2007 #2


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    The first question does not make sense. The weight is mg and that's fixed. Do they mean that the tension must be zero?
  4. Oct 31, 2007 #3
    no they mean that the weight is zero. I know that dosent make sense, but i guess the answer i got is the right one. But I seem to have trouble with part b..Firse i used the conservation of energy formula, but i found out the there is not change in h.
  5. Oct 31, 2007 #4
    I guess, you have already discussed this problem in some other thread.

    weight at A should be zero => T = 0. So, net force = -m*g. [I guess, you mis-typed. If not, plz take note of this.]
  6. Oct 31, 2007 #5
    what does your F_c stand for... centripetal force or centrifugal force??

    i am asking because, in either one of the part, you have taken wrong sign for it.
  7. Oct 31, 2007 #6
    I did type net force=-mg..:S
  8. Oct 31, 2007 #7
    centripetal force
  9. Oct 31, 2007 #8


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    I know that you get the right answer but the reasoning is illogical.

    How do you get from net force = - (T+mg) to syaing that -T = -mg ???

    No prof would mark this as correct even if the final answer is right.

    The only thing that makes any sense is to say that the tension is zero there fore the net force is -mg. You set this to m a_y with a_y given by -mv^2/R. This tehn gives the same answer but is now logically consistent.
  10. Oct 31, 2007 #9
    Furthermore, as now the string (read, rock) is at the bottom-most point... how come you are using same speed... 2.71 m/s?? This was speed of the rock when it was at A, isnt it?
  11. Oct 31, 2007 #10
    Ok..Net force= -T-mg or i can write this as -(T+mg) ok..?
    Then since the weight is Zero, the total net force will be -(T+0)= -T
    T=mg Hence -T=-mg..this is how i got it..Any help for part b)..is it right?
  12. Oct 31, 2007 #11
    Okay, if centripetal..

    Fc would be in downward direction at A.. thus.. Fc = -mv^2/r= -0.2*v^2/0.75
    [Note the -ve sign]
  13. Oct 31, 2007 #12
    umm..yes thts what i thought saket. But when I used the conservation of energy to find out the velocity of the rock at the botton, and checked it with one of my teachers in school, she said that this was wrong..:S:S. She asked me to use..Net force= T+mg..But how can mg b positive..it should be negative..:S
  14. Oct 31, 2007 #13
    No, that's a wrong perception! This is what I was trying to explain.

    Okay, what is weight? Mass*gravitational acceleration?? Ya, fine.. but that is when you are weighing yourself on a weighing machine.. then your acceleration is zero!

    If, you are accelerating upwards with an acceleration "a", your weight will be m*(g + a).
    If, you are accelerating downwards with an acceleration "a", your weight will be m*(g - a), where a is not greater than g. In this only, if the lift is accelerating downwards with acceleration "g", your weight is zero! (since, a = g) ... This is what they call, a situation of "weightlessness"! haven't you heard about it?
  15. Oct 31, 2007 #14


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    I will look at part b in a second.

    But realize that what you wrote above is wrong. You first say mg=0 and in the next line you say T=mg...therefore T should be zero according to what you just wrote!!
  16. Oct 31, 2007 #15

    Shooting Star

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    The v at the top is different from the v at the bottom. Find v at bottom by using conservation of total energy. Then find the tension in the usual way.

    C) Again, the total energy is conserved.

    It's not so tough!
  17. Oct 31, 2007 #16
    nrged..Ok so i got the net force as -T.
    Now I have to find T
    the formula for finding T is T=mg
    So-T=-mg...Isin't that right..?..:S
  18. Oct 31, 2007 #17
    So, in this problem, weight becomes zero when T = 0 and not when m*g = 0!!
  19. Oct 31, 2007 #18
    pleeeeeeeeeezzeee... "nrqed" and "pinkyjoshi65" .. read my posts! you two are discussing something which is fundamentally wrong.
  20. Oct 31, 2007 #19
    Okay, a more well-known example is that of an artificial satellite orbitting around Earth. An astronaut in it feels "weightlessness", because he is falling towards Earth with an acceleration "g" at any point of the path! (i.e. a = g)
  21. Oct 31, 2007 #20


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    I won't get into a debate here but I find that pedagogically speaking, it's confusing to talk about "gravitational acceleration" as if there are different types of accelerations. I find that's it's better to distinguish "apparent wight" from true weight which is simply the force of gravity on an object and is always equal to mg. And apparent weight involves the notion of non-inertial frames which is a subtle point.

    This problem can be done without ever talking about non-inertial frames, apparent weight or "gravitational acceleration". It's simply a simple application of Newton's second law!
  22. Oct 31, 2007 #21
    I am sorry saket, yes i get what you are saying. mg cannot be zero. Thanks Nrqed and Saket for correcting me.
  23. Oct 31, 2007 #22
    I am not sure what your teacher meant! [Or, maybe you took her wrong.. don't take it personally. It happens with students. It used to happen/happens with me as well.]
    But, yes, if I were to solve this problem, I would certainly use energy conservation to get speed at B.
  24. Oct 31, 2007 #23


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    The question b is ambiguous. . Assuming that the speed is kept constant during the entire motion then your answer b is correct. using conservation of energy won't work because there is some other force doing work on the mass in this case.

    However if there is only the tension in teh rope an dthe motion is purely around a fixed center, then the speed cannot be constant and the problem must be done using conservation of energy
    Last edited: Oct 31, 2007
  25. Oct 31, 2007 #24
    I am not taking physics now. I am just refreshing the concepts on my own since I am starting university in January. Sometimes I go to my old school and ask a few questions to a physics teacher.
  26. Oct 31, 2007 #25
    Thank You, Sorry for not getting what you were trying to say..!..:)
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