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Falling Rods Conceptual Question

  1. Mar 30, 2009 #1
    The problem statement, all variables and given/known data

    The three thin rods shown below are initially at the same angle, with one rod twice as long as each of the others two.
    rods.jpg
    Rod X is made of brass (density = 8.6 g/cm3), the others are made of aluminum (density = 2.7 g/cm3). All rods are released from rest at the same time. Select an answer for each statement below.

    A. Time for Z to reach the ground is ... for Y. (greater than, less than, or equal to)
    B. The initial linear acceleration of the top end of Y is ... that of Z. (greater than, less than, or equal to)
    C. Just before landing the kinetic energy of X is ... that of Y. (greater than, less than, or equal to)

    The attempt at a solution

    Here is my reasoning. For part A, Z would take longer, since the center of mass of the rod has a greater distance to fall. For part B, the linear accelerations would be the same, since gravity acts the same on both of them. For part C, the kinetic energy of X would be greater than Y since X will have a greater mass and K=(1/2)mv^2. One or more of my answers are wrong, though, so can anyone see where I'm going astray?

    Thank you for your help.
     
  2. jcsd
  3. Mar 30, 2009 #2

    LowlyPion

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    Welcome to PF.

    Tell me a little more about the linear acceleration at the end of y and z.
     
  4. Mar 30, 2009 #3
    Thanks. The way I figure it, to find the linear acceleration for any of the rods, I would use the formula for torque:

    T=I(a/r)
    Fr=I(a/r)
    mgcos(theta)r=(2/3)(mr^2)(a/r)

    After canceling terms, I get: a=(3/2)gcos(theta). I took that to mean that the linear acceleration does not depend on the mass or the radius of the rod, so they would be equal. Does that make sense, or am I going about it wrong?
     
    Last edited: Mar 30, 2009
  5. Mar 30, 2009 #4

    LowlyPion

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    Isn't it

    m*g*cosθ = 1/3m*(L/2)2*(a/L)

    Looks to me like you end with a dependency in L for linear acceleration a at L.
     
  6. Mar 30, 2009 #5
    You're right about MI being (1/3)mr^2 for a rod; sorry about that.

    But shouldn't there be an L on the left side of your equation, since torque is Fxr? (I accidentally left it out when I typed up my equation too.) If there is, then the Ls cancel out completely.
     
  7. Mar 30, 2009 #6

    LowlyPion

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    Right you are. I copied yours, but thought you had canceled out the L2 with an L. Sorry I should have caught that.

    You get different α 's but at the end that translates to the same a.
     
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