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Rotational Energy Pulley Problem

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data

    The pulley in Fig. 9-25 has radius 0.150 m and moment of inertia 0.440 kg·m2. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00 kg block just before it strikes the floor.

    Figure: http://www.webassign.net/yf9/9-25.gif

    2. Relevant equations

    Ug = mgh
    Krotational = .5I

    3. The attempt at a solution

    I tried using torques and kinematics but that didn't work. I've tried doing mgh = .5mv^2 + .5mv^2 + .5mv^2 (each m being each block and the pulley) but that didn't work either.

    Help please!

    Thank you!
     
  2. jcsd
  3. Dec 6, 2009 #2
    Ok this is a basic Atwood machine problem. There are many ways to solve it. I'll do it using the Lagrangian formulation since it's asking for energy methods, but you can do it through standard Newtonian mechanics as well (the two formalisms are physically equivalent). I'll assume that the mass on the left (the larger one) is m(1) and the mass on the right is m(2).

    The kinetic energy of the system can be written as:

    T = (1/2)*m(1)v^2 + (1/2)*m(2)v^2 + 1/2Iw^2

    where w = v/a and w^2 = v^2/a^2.

    so T = (1/2)*m(1)v^2 + (1/2)*m(2)v^2 + 1/2I*(v^2/a^2)

    The potential energy of the system can be written as:

    V = -m(1)gx - m(2)g(L - pi*a - x)

    where L is the length of the rope, pi* a is half the circumference of the pulley, and x represents the distance of the first mass to a straight line going through the center of the pulley. So the expression L - pi*a - x represents the distance of mass m(2) to that same axis (since we want to find the potential energy associated with m(2)).

    The Lagrangian is L = T - V (kinetic minus potential). Put it into the http://nerdwisdom.files.wordpress.com/2007/10/lagrange001.jpg" [Broken] (where x represents q and v represents velocity, or q "dot") and you get the following equations of motion:

    (m(1) + m(2) + I/a^2)*acceleration = (m(1) - m(2))*g

    so the acceleration, which is constant, is:

    acceleration = (m(1) - m(2))*g / (m(1) + m(2) - I/a^2)

    plug in everything we know to get:

    acceleration = 0.77 meters per second squared.

    From here it's simple kinematics:

    v(final)^2 = v(initial)^2 + 2ax

    Initial velocity is 0, so that term goes away. You're left with:

    v(final) = sqrt(2ax), where a is the acceleration (0.77 m/s^2) and x is the distance the 4 kg block has to travel, which is 5 meters.

    Plug everything in and you should get:

    v(final) = 2.77 m/s

    Hope this helps.
     
    Last edited by a moderator: May 4, 2017
  4. Dec 6, 2009 #3
    Thank you so much!!
     
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