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Homework Help: Torque and rotational dynamics

  1. Nov 29, 2006 #1
    1. The problem statement, all variables and given/known data
    A wheel is turning initially at 1200 turns/min and stops in 4 min because of friction. If we add a supplementary torque of 300 N*m, the wheel stops in 1 min.

    a) What is the moment of inertia of the wheel?
    b) What is the torque of friction?


    2. Relevant equations
    Torque = alpha * I
    time = w/alpha
    f = ucF
    Torque = -fR

    3. The attempt at a solution

    I figured i'd start by converting the 1200 turns/min into rad/sec

    (1200 *2pi)/60 = 126 rad/sec (this gives me angular speed - w)

    Then I figured i'd find the angular acceleration using the angular speed that I just found with the time it takes to stop (in this case its 4min or 240seconds).

    alpha = 126/240
    = 0.525

    Now at first I thought I could just substitute these in the equation Torque = alpha * I but it doesn't work out.
    I then realised that they are saying that there is a SUPPLEMENTARY torque and it then stops in 1min. I thought it would then be something like "initial torque + 300 = something...". Does that make any sense?

    Any help to clear this up would be appreciated!
     
  2. jcsd
  3. Nov 29, 2006 #2

    radou

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    Homework Helper

    You could think of it this way: [tex]M=\alpha_{1} I[/tex], and [tex]M+300=\alpha_{2} I[/tex]. The additional torque creates a new net torque, which creates a new angular deceleration. Try to work something out from here on. (P.S. Note that the moment of inertia is, of course, the same.)
     
  4. Nov 29, 2006 #3

    Doc Al

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    Staff: Mentor

    You are given two (related) situations with two different accelerations. Find both accelerations, then write two "Torque = alpha * I" equations. Solve both equations together and you'll get your answers.

    (Oops... radou beat me to it!)​
     
  5. Nov 29, 2006 #4

    AlephZero

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    Homework Helper

    You are nearly there. Yes, Torque = alpha * I is a good equation to use.

    You have got two cases. In the first one you have the unknown friction torque (call it T) and the wheel stops in 4 min. You worked out the acceleration alpha correctly.

    In the second case you have an additional torque so (as you said) the torque is "initial torque + 300 = something" ... well, the "something" is just "T+300". This time the wheel stops in 1 min, so the value of alpha is different.

    Using Torque = alpha * I for the two cases will give you 2 equations in the 2 unknowns (T and I) which you can solve.
     
  6. Nov 29, 2006 #5
    Thanks a bunch! I got it :)
     
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