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Homework Help: Velocity Of pulley

  1. Nov 30, 2011 #1
    Question: Mass one is 15.0 kg and mass 2 is 9.0 kg. They are connected by a massless rope which passes over a frictionless pulley. The pulley has a shape of a solid disk. They pulley mass is 8.0 kg and the pulley radius (R) is 0.20 meter.
    What is the speed of the falling body?

    Solution:

    I think my issue is determining if the forces in my equations are either positive or negative...
    In my diagrams I forgot to label the the direction of my force vectors...
    Body 1 and Body 2: Both T1 and T2 point up towards top of page and w1 and w2 point down towards bottom of page.

    Equations:

    1. Ʃ (torque) = T1(R) - T2(R) = (I)(a/R) = (MR^2/2)(a/R) = M*a*R/2

    2. Ʃ(F1) = T1 - w1 = (m1)(a) .... T1 = m1*a + w1

    3. Ʃ(F2) = w2 - T2 = (m2)(a) .... -T2 = m2*a - w2

    4. T1 - T2 = m1*a + w1 + m2*a - w2

    * Substitute eq. 4. into eq. 1.

    (m1*a + w1 + m2*a - w2) = M*a*/2

    a = (w1 - w2)/(M/2 - m1 - m2) = -2.94 m/s^2

    V = (2*-2*-2.94)^(1/2) = 3.42 m/s

    The answer is 2.9 m/s
     

    Attached Files:

    Last edited: Nov 30, 2011
  2. jcsd
  3. Nov 30, 2011 #2

    ehild

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    Homework Helper

    Check equations 2 and 3. Object 1 accelerates downward and object 2 upward.
    What was the question about speed? Do you need to give it at a certain time instant or after falling a given length?

    ehild
     
    Last edited: Nov 30, 2011
  4. Dec 1, 2011 #3
    After it falls 2.0 meters


    2. Ʃ(F1) = -T1 + w1 = (m1)(a)

    3. Ʃ(F2) = -w2 + T2 = (m2)(a)

    Is this correct?
     
  5. Dec 1, 2011 #4

    ehild

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    Correct.

    ehild
     
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