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Pulley, Ramp with Moment of Inertia

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data

    http://imgur.com/ZkjXv

    The given answers are (b) .309 m/s^2 (c) T1= 7.67N, T2=9.22N

    2. Relevant equations

    I_disc = .5(mass_disc)r^2
    Sum of Torque = I α
    a = r α

    3. The attempt at a solution

    We will define the system moving right and down as positive to get a positive acceleration.

    The tension between mass 1 (2kg) and the pulley is T_1
    The tension between mass 2 (6kg) and the pulley is T_2

    Summation of Torque = I α = rT_2 - rT_1

    plug in: α = a/r and I_disc = .5(mass_disc)r^2

    into

    I α = rT_2 - rT_1

    to get

    a(.5)(mass_disc) = T_2 - T_1

    The summation of the forces in the x-direction for m1 is: F_x = m1(a) = T_1 - μ(m1)g

    T_1 = (m1)a + μmg

    How do you define the summation of the forces for m2 and get an expression for T_2?

    My first attempt:

    Sum of Forces for M2 in x direction:

    F_x = (m2)a = T_2 - μ(m2)gcos30

    T_2 = (m2)a + μ(m2)gcos30

    is incorrect. This was done through rotating the system 30 degrees but the summation for mass 1 was not rotated therefore the equations are not consistent and apparently, wrong. What is the best way to do this?
     
    Last edited: Apr 2, 2012
  2. jcsd
  3. Apr 2, 2012 #2
    [itex]T_1-μm_1g=m_1a .......1[/itex]

    [itex]m_2Sinθ-μm_2Cosθ-T_2=m_2a .......2[/itex]

    [itex]r(T_2-T_1)=Iα........3[/itex]

    3 equations with 3 unknowns- a,T1 and T2
    Rotating the system make it more complicated.
    You should be consistent with direction of acceleration.
     
    Last edited: Apr 2, 2012
  4. Apr 2, 2012 #3
    The equation you gave for T_2 seems to be incorrect. Plugging in .309 for a yields a T_2 of -.725 which is incorrect
     
  5. Apr 2, 2012 #4
    Bump? Can anyone give some insight?
     
  6. Apr 2, 2012 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi kc0ldeah! :smile:

    (try using the X2 button just above the Reply box :wink:)
    that's only tension and friction …

    what about gravity? :wink:
     
  7. Apr 2, 2012 #6
    Thousand apologies.
    The second equation should be....missing g

    [itex]m_2gSinθ-μm_2gCosθ-T_2=m_2a .......2[/itex]
     
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