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Return of the Inclined Plane Double Pulley: Part II: Energy

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data

    My last inclined plane problem has a second part to it: There are some energy problems I need to solve.

    A double pulley consists of two separate pieces that are welded together; a hoop of radius 10.0 cm and a disk of radius 15.0 cm. The moment of inertia of the pulley is 0.160 kg-m2. A 15.0 kg block (m1), on a 35.0° incline, is attached to the outer pulley by a massless cable and a 24.0 kg block (m2) is hanging from the inner pulley by a massless cable. The incline has a kinetic coefficient of friction of 0.250. At the instant shown in the diagram, the two blocks are at a height of 1.50 m and are moving in the appropriate direction with m2 having a speed of 0.240 m/s. The motion is considered complete when either m 1 reaches the bottom of the incline or just before m2 hits the ground.

    2vi4mmr.jpg

    Use energy methods to answer the following questions.
    Assign the zero point reference for potential energy to be the ground for both blocks!

    a) What is the speed of block m 1 when the motion is complete?
    b) What is the speed of block m 2 when the motion is complete?
    c) What is the total energy and the distribution of the energy of the system at the initial instant described?
    d) What is the final distribution of the energy of the system just before the motion is complete?



    2. Relevant equations

    (Possibly these are relevant?)
    PE = KE
    mgh = 1/2 mv^2

    3. The attempt at a solution

    I managed to answer everything from Part I of the problem. I was able to use forces to find that m2 moves downward, that m1 moves up the ramp, that m2 travels distance h (1.50m) and m1 is pulled 2.25m. The angular acceleration of the pulley should be 8.63. Acceleration of m1 is 1.29, m2 is 0.863, tension in cable 1 is 133.95 and cable 2 is 214.73. Great.

    I'm trying to set up some equation to find the velocities of these blocks. Unfortunately, in all the examples I'm finding there are only regular pulleys, so in this answer they solve for a single velocity.

    Using the above example I've made a really crude attempt in setting up the problem (which I have little to no confidence in):

    [tex]E_1=m_2*g*h_2-m_2*(v_2)^2/2[/tex]
    [tex]E_2=m_1*g*2.25*\sinθ+m_1*(v_1)^2/2[/tex]
    [tex]E_3=μ*m_1*g*2.25*\cosθ[/tex]
    [tex]E_1=E_2+E_3[/tex]

    Not even sure if using 2.25 like that would be the right way to go; I was just trying to get the x and y components of the distance m1 travels, which I calculated in the first part, which seems relevant. But I wouldn't be able to use step 4 to set up an equation because I'd have two unknowns, v1 and v2, and that would be no good. And I bet the double pulley should be playing into this equation too but I'm still really confused about how that thing works.

    Since the radii in my problem are different I have a very strong feeling the velocities will be different because the smaller inner pulley is going to turn faster. However I also need to factor in friction because m1 is encountering it and is (?) going to slow down m2.

    I'd like to apply "PE = KE" which I found here, but that problem is so radically different than this one that I'm not sure how.
     
  2. jcsd
  3. May 20, 2014 #2

    Doc Al

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    Staff: Mentor

    The key is that the inner and outer pulleys have the same angular speed and acceleration as they are welded together. Use that fact to relate the linear speeds of the two masses.
     
  4. May 20, 2014 #3

    Doc Al

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    Staff: Mentor

    Also, when tracking energy changes, don't forget to include the energy of the pulley.
     
  5. May 21, 2014 #4

    ehild

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    Homework Helper
    Gold Member

    Wrong. The potential energy is not equal to the kinetic energy in general.
    Conservation of energy states that the sum of the kinetic energy and the potential energy is constant during the motion of a system, but it is not true when there is friction.


    What are those E-s?????

    Why do you keep on copying the solution of a completely different problem??? You feel it will not work, and it does not work. There are different accelerations, velocities and distances, all differing by the same factor R1/R2.

    There is friction, you can not ignore it. In case of friction, is the mechanical energy conserved?

    Anyway, you know the acceleration and the distance travelled for both blocks. It is uniformly accelerating motion. Given the path taken and the initial velocity of m2. Can you find the final velocity then?

    ehild
     
    Last edited: May 21, 2014
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