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Problem with rotational torque and angular acceleration

  1. Oct 15, 2013 #1
    I've attached a screenshot of this particular question. I was able to figure out part A but I'm having some difficulty with part B.

    Relevant equations:

    T = I*alpha
    I = .5mr^2 (for a solid cylinder like the grinding wheel)
    F = ma

    Attempt at solution:

    I've gotten as far as setting up the equation and solving for Tf.

    (T1-T2)(r) - Tf = (.5)(m)(r)(a)

    So Tf = (T1-T2)(r) - (.5)(m)(r)(a)

    Which gives me .01878 Nm, which rounds to .019 Nm.

    For the acceleration, I've been using .28/6.7 = .04179 m/s^2
     

    Attached Files:

  2. jcsd
  3. Oct 15, 2013 #2

    TSny

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    Consider a free body diagram for each of the hanging masses to get the tensions T1 and T2.
     
  4. Oct 15, 2013 #3

    haruspex

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    How are you calculating the tensions in part B? They will not be the same as in part A, because the acceleration is different. Pls post all your working for part B.
     
  5. Oct 15, 2013 #4
    Ok, I think my problem was caused by using the wrong acceleration to calculate the tensions. I get that the system should have the same acceleration, so using a = .28/6.7 universally:

    (T1-T2)(r) - Tf = (.5)(m)(r)(a)
    (1.2g-6.5a)(r) - Tf = (.5)(m)(r)(a)
    (11.76 - .27)(.049) - Tf = (.5)(.8)(.049)(.042)
    .56 - Tf = .00082
    Tf = .56 Nm
     
  6. Oct 15, 2013 #5

    TSny

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    You might want to check if you plugged in the correct sign for your acceleration.
     
  7. Oct 15, 2013 #6
    Since the the acceleration is negative, that should mean that Tf is negative as well?
     
  8. Oct 15, 2013 #7

    TSny

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    I think you already took into account the negative direction of the friction torque when you wrote a minus sign in front of the friction torque in your first equation in post #4.
     
  9. Oct 15, 2013 #8
    How do you know that the acceleration should have a negative sign convention?
     
  10. Oct 15, 2013 #9

    TSny

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    The choice of positive direction is up to you and should be made explicit at the very beginning of setting up the equations. I had to guess what direction you are taking as positive by examining your equations. From the left hand side of the first equation in post #4 it appears to me that you are taking positive direction to be the way the heavier mass tries to accelerate the system. But you are given that when you give the system an initial velocity in this positive direction, it slows down rather than speeds up.
     
  11. Oct 15, 2013 #10
    So it's pretty much keeping the sign convention consistent. Thanks!
     
  12. Oct 16, 2013 #11

    TSny

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    Yes. Good work.
     
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