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Torque and tension?

  1. Mar 13, 2005 #1
    A string is wrapped several times around the rim of a small hoop with radius r and mass m. The free end of the string is held in place and the hoop is released from rest.. (picture attached~)

    So I tried using moment of inertia as mass*radius^2, so torque would be MR^2*alpha, and it would also be tension*R.

    Solving for tension would give me mass*acceleration as tension.

    Then net force would be tension-mass*gravity.


    Now that doesn't make any sense.

    Can anyone help?

    Attached Files:

  2. jcsd
  3. Mar 14, 2005 #2


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    Staff Emeritus
    Science Advisor

    You're right- it doesn't make sense. T is NOT equal to m*a- for one thing, that doesn't take into account the increasing angular momentum.
  4. Mar 14, 2005 #3
    The moment of inertia you used is for a hoop, with the axis of rotation at the centre of the hoop. You need to use the moment of inertia with the axis of rotation at the outter radius.
    In this case:

    [tex] I = \frac{mR^2}{2} [/tex]


    Last edited: Mar 14, 2005
  5. Mar 14, 2005 #4

    Doc Al

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    Staff: Mentor

    Nothing wrong with that. You have applied the acceleration constraint that connects [itex]\alpha[/itex] with a: [itex]a = \alpha R[/itex].

    Right. T = ma (where "a" is the magnitude of the acceleration).

    Right. The net force on the hoop is T - mg, taking up as positive. Applying Newton's 2nd law gives you: T - mg = - ma. (note the minus sign; the acceleration acts downward thus is negative)

    It doesn't make sense because you mixed up the signs. Combine these equations and it will make sense:
    T - mg = - ma
    T = ma
  6. Mar 14, 2005 #5
    heres what happened when I did it:
    [tex] RT = I\alpha [/tex]
    [tex] RT = I\frac{a}{R} [/tex]

    since the moment of inertia is [tex] \frac{mR^2}{2} [/tex]


    [tex] RT = \frac{mR^2}{2} \frac{a}{R} [/tex]
    [tex] T = \frac{ma}{2} [/tex]

    now we can sub this in into our free body diagram equation:

    You should be able to do the rest.


  7. Mar 14, 2005 #6

    Doc Al

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    Staff: Mentor

    There's nothing wrong with taking a point on the edge of the hoop as your axis, but that's not the correct moment of inertia.
  8. Mar 14, 2005 #7
    Doc Al was right. (though maybe your way Nenad could also produce the "correct" answer)

    [tex]T - mg = -ma[/tex] because acceleration is downward.

    [tex]2T = mg[/tex]

    [tex]T = (mg)/2[/tex]

    Darn little technicalities.

    Thanks guys! You helped me finish my homework on time :biggrin: .
    Last edited by a moderator: Mar 14, 2005
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