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Pully with mass and weights

  1. Sep 28, 2011 #1
    Hi guys,

    I have this question below but I don't have the answer to it and I'm not sure if what I am doing is right. I attempted the solution by balancing the forces as I am not as clear on how to use the energy method yet.

    Is my solution/answer correct? (:

    Thanks guys!

    The problem statement, all variables and given/known data

    Consider the system shown in Figure 10.42 with m1 = 20.0kg, m2 = 12.5kg, R(radius of pully)=0.2m and mass of uniform pulley M=5.00kg. Object m2 is resting on the floor and m1 is 4.00 meters above the floor when it is released from rest. The pully axis is frictionless. Calculate the time for m1 to hit the floor.

    Relevant equations and attempt at solution
    T[itex]_{1}[/itex] = m[itex]_{1}[/itex]g - m[itex]_{1}[/itex]a
    T[itex]_{1}[/itex] = 196.2 - 20a

    T[itex]_{2}[/itex] = m[itex]_{2}[/itex]g + m[itex]_{2}[/itex]a
    T[itex]_{2}[/itex] = 122.625 + 12.5a

    T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = I[itex]\alpha[/itex]
    T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = (0.5)(MR[itex]^{2}[/itex])(a/R)
    T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = 0.5(5.00)(2.00)a

    Subbing in T[itex]_{1}[/itex] and T[itex]_{2}[/itex] I get a = 2.23m/s[itex]^{2}[/itex]

    Then I use Xf = Xi + Ut +0.5at[itex]^{2}[/itex] to get t and get 1.89s

    Is this right?
    Last edited: Sep 28, 2011
  2. jcsd
  3. Sep 28, 2011 #2


    User Avatar

    Staff: Mentor

    Remember that torque is F x L, where F is the applied force and L is the "lever arm" or radius vector to the center of rotation. So check your Pulley section to make sure that you're equating torques with torques.
  4. Sep 28, 2011 #3
    Thanks for your reply! (:

    T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = I[itex]\alpha[/itex]

    Should be:
    T[itex]_{1}[/itex]R - T[itex]_{2}[/itex]R = I[itex]\alpha[/itex]

    Right? (:
  5. Sep 28, 2011 #4


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    Staff: Mentor

  6. Sep 28, 2011 #5
    Just one more question. If the cylinder was a hollow one I understand that:
    I = (0.5)M(R[itex]_{1}[/itex][itex]^{2}[/itex] - R[itex]_{2}[/itex][itex]^{2}[/itex])

    Does my [itex]\alpha[/itex] = a/(R[itex]_{1}[/itex] - R[itex]_{2}[/itex]) as well?

    What about my [itex]\tau[/itex]? Does it become:
    (R[itex]_{1}[/itex]-R[itex]_{2}[/itex])(T[itex]_{1}[/itex]- T[itex]_{2}[/itex]) as well?

    Thanks! (:
  7. Sep 28, 2011 #6


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    Staff: Mentor

    You don't need to change anything but the moment of inertia unless the radius at which the torque is applied changes too.

    The moment of inertia is the rotational analog of mass. In a linear system, if you had a hollow sphere (spherical shell) rather than a solid one you'd have to do something similar in order to calculate the mass of the object, but after that you'd just use the calculated mass and get on with things. :smile:
  8. Sep 28, 2011 #7
    So the radius I use for the calculation of [itex]\tau[/itex] and [itex]\alpha[/itex] will be that of the full radius (i.e. from att the way out to the centre) and the only thing is actually affects is the calculation of the inertia of the disk? :D
  9. Sep 28, 2011 #8


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    Staff: Mentor

    Yes. The torque calculation always involves the actual distance from the center of rotation to where the force is applied.
  10. Sep 28, 2011 #9
    Thanks! You're a LIFESAVER! (:
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