1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tension in string

  1. Jan 25, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-1-26_11-19-48.png

    A string is wound around a uniform disk of radius ##10cm## and mass ##300g##. The
    disk is released from rest when the string is vertical and its top end is tied to a
    fixed bar. Determine the tension in the string. [ ##g = 10 ms^{-2} ##]



    2. Relevant equations


    3. The attempt at a solution
    I think it is just ##300g \cdot 10 ms^{-2}## ;
     
  2. jcsd
  3. Jan 25, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If so, what is the net vertical force on the disk? What acceleration would result?
     
  4. Jan 25, 2015 #3
    The accelaration of the disc is ##g## .
    upload_2015-1-26_11-37-19.png
    But, the disc will rotate, I think. And there should be a torque. I shown in the picture, the tension of the string ## T## will create the torque.
     
  5. Jan 25, 2015 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The acceleration of the disk would be g if you took the string away. If there's any tension in the string the acceleration must be less.
    Write the ##\Sigma F = ma## equation for the vertical direction.
     
  6. Jan 25, 2015 #5
    upload_2015-1-26_11-51-40.png
    I shown in the picture, there is tension in the string, but the net force in the vertical direction is still ##mg## .
     
  7. Jan 26, 2015 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    OK, I see where you are confused. There are different ways of treating how the portion of the string wrapped around the disk comes into it.
    The least confusing is to assume it barely wraps around at all, i.e. as though it is attached to the disk at the point where they meet.
    If you want to consider a longer wrap, it gets complicated because the least friction will soon result in its being effectively attached. If we say there's no friction at all then you have to follow it around to where it is ultimately attached, at some arbitrary point on the rim. T will apply tangentially there, but nowhere else. At all other parts of the string touching the disk, the force exerted by the string is radial. We can go down that path if you like, or you can trust me that it will all come out the same in the end... as though the string is attached to the disk at the point where they first meet.
     
  8. Jan 26, 2015 #7
    But why?
     
  9. Jan 26, 2015 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If there's no friction how can the string apply a tangential force all along its length? It has no grip. Can you open a screw-top jar with greasy hands?
     
  10. Jan 26, 2015 #9
    I got your point. So, in this problem, the tension is simply 0.
     
  11. Jan 26, 2015 #10

    Stephen Tashi

    User Avatar
    Science Advisor

    I suggest you follow the standard procedure and draw a correct free body diagram of the disk instead of the diagram you drew. Are you familiar with the concept of a free body diagram?
     
  12. Jan 26, 2015 #11
    Yes.
    I think it will be as I drawn below:
    upload_2015-1-26_13-11-36.png
    And, ##F_{net} = ma## ; where, ##a## is the accelaration of the disc.
    Will it be, ##a = g## ?
     
  13. Jan 26, 2015 #12
    And I think there will be a torque acting on the disc, which is equal to ##Tr## ; where ##r## is the radius of the disc.
    Again, torque = ##I \alpha## ; where, ##I## is moment of inertia and ##\alpha## is the angular accelaration;
    So, ##Tr = \frac{1}{2} mr^2 \alpha##; because, moment of inertia of a disc with radius ##r##, is ##\frac {1}{2}mr^2##
    ##\alpha = \frac {2T}{mr}##
    Again, ##a = r \alpha##
    ##a = r \frac {2T}{mr}##
    ##a = \frac {2T}{m} ##
    And so,
    ##mg - T = ma = m \frac{2T}{m}##
    ##mg -T = 2T##
    ##T = \frac{1}{3} mg##
     
  14. Jan 26, 2015 #13
    That looks correct.:)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Tension in string
  1. Tension in the String (Replies: 2)

  2. Tension of a String (Replies: 2)

  3. String Tension? (Replies: 2)

  4. Tension in strings (Replies: 18)

Loading...