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Tensions and centripetal acceleration

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data

    An object of mass m1 of 4.00kg is tied to an object of mass m2 of 3.00kg with a string of length 0.5m. The combination is then swung in a verticular circular path on a second string of length 4.00m. The two strings are always colinear (they are on the same line). At the top of the motion, m2 has a speed of 4.00 m/s. Now, answer the following:
    a. What is the tension in the short string (the 0.5m string) at this moment?
    b. What is the tension in the long string (the 4.00m string) at this moment?
    c. If you rotate the system faster and faster, which string (assuming they have identical strengths) breaks first?

    2. Relevant equations

    -


    3. The attempt at a solution

    I have no idea how to approach this problem. I keep getting a negative tension while doing this exercise: I've talked to a friend and we keep getting the same result. Our FBD for the first mass (the farthest one) is just one tension and its weight. On the other hand, for the second mass (the closest one) is the tension upwards, the other downwards and the weight. These two forces provide the centripetal force which makes them go in a circle.
     

    Attached Files:

  2. jcsd
  3. Dec 9, 2013 #2


    who will count the centrifugal force out side??? mv2/r
     
  4. Dec 9, 2013 #3
    You probably made a sign error. Can you show the equation you get and specify which of the direction you consider positive?
     
  5. Dec 9, 2013 #4
    no man........ just see above....... he has not counted the centrifugal force for the second and the first mass
     
  6. Dec 9, 2013 #5
    no man.....just look at his post.....he has not said he is working in a rotating frame.
     
  7. Dec 9, 2013 #6


    i think this means generally that the system is rotating............
     
  8. Dec 9, 2013 #7
    Both masses are rotating. So the FBD must take into account not just the forces, but the accelerations as well.

    Or you could use the non-inertial co-rotating frame, in which case you will need to take into account fictitious forces.
     
  9. Dec 9, 2013 #8
    see.... a PF Patron has justified it... Pranav.........
     
  10. Dec 9, 2013 #9
    Anyone can become a PF Patron. Don't attach too much meaning to that.

    Pranav, however, was right. You mentioned centrifugal force, but it is only present in a rotating frame. A rotating mass does not mean a rotating frame.
     
  11. Dec 9, 2013 #10
    but i think a rotating object as above will experience a outward force which will create a tension in the upper string
     
  12. Dec 9, 2013 #11
    Negative tension means the string is slack and therefore the objects will not follow a circular path - and indeed tangential velocity of 4 m/s is nowhere near enough to make this system work as described on the Earth's surface.

    The question is incomplete: it does not state what gravity is to be taken into account (and if it is gravity at the Earth's surface it does not work), and it does not state whether m1 or m2 is attached to the long string.
     
  13. Dec 9, 2013 #12
    That is not correct. The outward force is fictitious, i.e., not real, and exists only in a non-inertial (co-rotating) reference frame.

    In an inertial frame, the only force is the force of tension. It arises because the rotating object, in accordance with Newton's first law, tries to fly in a straight line, which would increase its distance from the center of rotation, and the rope opposes that.
     
  14. Dec 9, 2013 #13
    Stop it. Learn about centripetal force.
     
  15. Dec 9, 2013 #14
    Note that the minimum tangential velocity to maintain tension at the top of the swing and therefore follow a circular path is given by ## \sqrt{rg} ## which is approximately 6.27 m/s at 4 m radius or 6.65 m/s at 4.5 m radius.
     
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