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Finding velocity taking the pulley into consideration

  1. Nov 13, 2006 #1
    The picture for this problem is at http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1423590496

    It is problem number 9. I also have my work so far there.

    Two blocks are connected as shown. Friction between M1 and the horizontal surface is characterized by mu_k = .25 and mu_s=.5. The pulley has I = .16kgm^2 about its center and a radius of .2m. Find the speed of the masses after they moved 2m. Assume that M1 is still on the horizontal surafece after those 2m so that the friction force will act over the entire 2m. Use M1 = 2kg and M2 = 6kg.

    I did my work and ended up with accerleration being 3.26 however, I can't get the velocity. I don't know how to get the velocity, and I am not sure if I am overlooking some kinds of relationship between the variables.
     
  2. jcsd
  3. Nov 13, 2006 #2

    Doc Al

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    Staff: Mentor

    There's a simple kinematic formula that relates velocity and distance for a constant acceleration. You can either look it up or derive it for yourself. (Consider the definitions of acceleration and average speed.)
     
  4. Nov 13, 2006 #3
    I know that d = (1/2)at^2 since v_o and x_o are both zero.

    2 = (1/2) at^2
    4 = at^2
    4/3.26 = t^2

    t = 1.107

    v = v_o + at
    v = 3.26(1.10)
    v = 3.61

    However, the correct answer should be 4.24. What am I doing wrong?
     
  5. Nov 13, 2006 #4

    Doc Al

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    Nothing wrong with your kinematics. So better check that you calculated the acceleration correctly.
     
  6. Nov 13, 2006 #5

    Doc Al

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    That's the problem: Your calculation of acceleration is incorrect.
     
  7. Nov 13, 2006 #6
    Are my free body diagrams correct? For M1, I have normal force pointing up, Tension 1 point right, static friction pointing left, and gravity pointing down. For M2, I have tension2 pointing up and gravity pointing down. For the pulley, I have Tension 1 pointing left and Tension 2 pointing down. I think that I might have made some mistakes here.

    M1

    x) T1- f = M1*a
    y) N - M1*g = 0

    M2

    T2-M2*g = -M2*a

    Pulley

    -T1*R -T2*R = I (-alpha)
    T1*R + T2*R = I*(alpha)

    So I have

    R(T1+T2) = I (a/R)
    R^2(T1+T2) = Ia
    Ia = R^2(M1*a + mu_s*M1*g + M2*g - M2*a)
    Ia -R^2*M1*a + R^2*M2*a = R^2*M2*g + R^2 * mu_s*M1*g
    a = [R^2*M2*g + R^2*mu_s*M1*g]/(I+R^2*M2 -R^2*M1)


    I plugged in all of my values from the problem and got 8.575 for a, which is wrong when I try to find velocity.
     
  8. Nov 13, 2006 #7

    Doc Al

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    Good.

    Realize that T1 and T2 exert opposite torques on the pulley and thus must have opposite signs.
     
  9. Nov 13, 2006 #8
    Since T1 is trying to cause the pulley to rotate counterclockwise, while T2 is causing it to rotate clockwise, would it work if I have

    T1*R-T2*R = I (-alpha)
     
  10. Nov 13, 2006 #9

    Doc Al

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    That looks good to me.
     
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