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Homework Help: Atwood Machine and block acceleration

  1. Nov 23, 2007 #1


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    [SOLVED] Atwood Machine

    1. The problem statement, all variables and given/known data

    A pulley which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. One block has mass M = 500 g and the other has mass m = 460 g. When released from rest, the more massive block falls 75 cm in 5.00 s (without the cord slipping on the pulley.)

    (a) What is the magnitude of the blocks’ acceleration? What is the tension in the cord that supports

    (b) the more massive block and

    (c) the less massive block?

    (d) What are the magnitude and direction of the pulley’s angular acceleration?

    (e) What is its rotational inertia?

    (f) Find the rotational kinetic energy after 5.00 s.

    http://img264.imageshack.us/img264/6785/atwoodue6.th.jpg [Broken]

    2. Relevant equations

    [tex] K_i + U_i = K_f + U_f [/tex]

    [tex] v= R\omega [/tex]

    [tex]a_t= r\alpha[/tex]

    3. The attempt at a solution

    know that:

    r= 5.00cm=> 0.05m

    M= 500g=> 0.5kg

    m= 460g = 0.46g

    h= 75cm=> 0.75m

    I'm not sure exactly how to get this at all.
    notation confuses me somewhat for rotation motion
    I was just reading the chapter today since we just learned this the other day.

    a) What is magnitude of block's acceleration?

    I was thinking about using Newton's second law with the forces that are on the blocks.

    [tex]\sum F_{y}= m_2g - T= m_2a_y [/tex]

    [tex]\sum F_{y}= T- m_1g = m_1ay [/tex]

    m_2g - m_1g = a_y(m_1 + m_2)

    a_y= g [(m_2 - m_1)/ (m_1 + m_2)]

    However I don't think I'm supposed to get the acceleration this way....since they give the radius and the time...I don't know how to incorperate the time and the radius if I need to.

    Help please.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 23, 2007 #2
    As I know should be




  4. Nov 23, 2007 #3


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    so tension isn't the same for both then..

    but why is it [tex](T_1-T_2)r=I\frac{a}{r}[/tex]

    Is the [tex]T_2 [/tex] negative since that direction is clockwise?
  5. Nov 23, 2007 #4
    [tex]T_1[/tex] is trying to rotate pulley in clockwise direction.

    [tex]T_2[/tex] is trying to rotate pulley in counterclockwise directon.If there were no M,it would rotate in counterclockwise direction.Here difference of them contributes to the rotation of pulley.

    So torque by M is clockwise [tex]T_1r[/tex] and torque by [tex]T_2[/tex] is counterclockwise [tex]T_2r[/tex]
  6. Nov 23, 2007 #5


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    I was wondering how would I find the acceleration?

    I expanded the last equation but ran into some trouble.


    I don't have (a) but I did this...

    since I= mr^2
    and a= r[tex]\alpha[/tex]

    [tex]r(Mg- Ma - mg - ma) = (mr^2)(a/r)[/tex]

    r(Mg-Ma - mg -ma) = m*r*a

    however I'm not quite sure what m are they refering to when I replace the m in the I for moment of inertia.

    Is I= (M+m)(r^2) ?

    Help..I'm still not sure how to find the acceleration b/c I found that the accelerations would just cancel out if I'm not wrong..so how would I isolate the accleration?

    And I'm not sure what I = ? since generally I= mr^2 but here there are 2 masses...so is mass = M+ m?

    Help please:frown:
  7. Nov 23, 2007 #6
    When you look at pullley,you'll have two tensions and pulley.We need to take just moment of inertia of pulley.
  8. Nov 24, 2007 #7


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    I don't get what your saying.....I was speaking of the Inertia of the pulley but what if I don't have the mass of the pulley?

    can anybody help me out?
  9. Nov 24, 2007 #8

    Doc Al

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    You don't need the mass of the pulley. You'll deduce its rotational inertia, not calculate it directly.

    Start by finding the acceleration of each mass using kinematics. Then use that to find the tensions. Then use the tensions to figure out the rotational inertia of the pulley.
  10. Nov 24, 2007 #9


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    a) Magnitude of block's acceleration

    when you say kinematics do you mean rotational or translational equations?

    I tried to use the
    distance equation..

    [tex]y-y_o= v_{iy}(t) + .5a_y(t)^2 [/tex]

    -.075m = 0(5.00s) + .5(a_y)(5.00s)^2

    [tex] a_y= -.75m/12.5s^2 [/tex]

    [tex] a_y= 0.06m/s [/tex] ===> this looks odd to me...but is it right?

    I was thinking that the accelerations for both blocks has to be the same since they are both connected by one cord, However you said to...

    this has confused me somewhat.

    so If they ask for the "magnitude of the blocks acceleration"

    I assume that this would be what they want right?


    Tension in cord that supports

    b) more massive block

    c) less massive block

    Assuming the acceleration I calculated before was correct.

    [tex]a_y= 0.06m/s^2[/tex]

    [tex]\sum F_{y}= Mg - T_1= Ma [/tex]

    T1= Mg- Ma

    [tex] T_1= .5kg(9.81m/s^2)- .5kg ( 0.06m/s^2)[/tex]

    [tex] T_1 = 4.875N [/tex]

    [tex]\sum F_{y}= T_2 - mg= ma [/tex]

    [tex]T_2 = ma + mg [/tex]

    [tex] T_2 = (0.460kg)(0.06m/s^2) + (0.460kg)(9.81m/s) [/tex]

    [tex] T_2 = 4.5402 N [/tex]

    [tex]\sum F_y = T_2 - mg = ma [/tex]

    Um..for the rotational inertia

    (T1-T2) r = [tex]I \alpha[/tex]

    [tex]\alpha = a/ r[/tex] ==> you said to use these to find the rotational inertia but what is the a in the [tex]\alpha eqzn[/tex] equal to? Is this tangial acceleration the same as the acceleration I found for the blocks?

    And why would I need the rotational inertia if they didn't ask for it?

    d) Magnitude and direction of angular accleration

    Not sure but do I use this equation


    or since [tex]\alpha = a/ r[/tex] can I plug in acceleration as the one I found before for

    the blocks ([tex] a= \alpha/r [/tex]) since I have the radius?

    Last edited: Nov 24, 2007
  11. Nov 24, 2007 #10

    Doc Al

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    Looks good to me.
    Yes, the blocks have the same magnitude of acceleration. (So you only have to calculate it once.)
    Yes, [itex]a = \alpha r[/itex] with the same acceleration (a) as the blocks (and rope) have.
    They ask for it in part e.
    You'll use this for the next part (e).
    That's the one you want. (But that last equation should be: [itex]\alpha = a/r[/itex].)
  12. Nov 24, 2007 #11


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    just like I thought

    oh what do you know..they do ask for it..

    d) What is magnitude and direction of pulley's angular acceleration

    [tex]\alpha = a/ r[/tex]

    using the acceleration of the blocks as the a

    [tex]\alpha = 0.06m/s^2 / .05m = 1.2s^-2 [/tex]

    direction=> is in the page? since I'm thinking right hand rule and the pulley turns clockwise.

    e) rotational inertia?

    have the [tex] \alpha = 1.2 s^-2[/tex]

    [tex](T_1-T_2)r=I \alpha[/tex]

    [tex] T_1 = 4.875N [/tex]

    [tex] T_2 = 4.5402N [/tex]

    [tex]\frac{4.875N- 4.5402N} {1.2s^-2} = 0.279N/s^-2 [/tex]
    Not sure about the units...

    f) rotational kinetic E after 5.00s

    [tex] K_r = \frac {1} {2} I \omega^2 [/tex]

    [tex] K_r = \frac {1} {2} I \theta/t [/tex]

    basically I'm not sure how to get this.

  13. Nov 25, 2007 #12

    Doc Al

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    All they want to know is whether its clockwise or counterclockwise. (I wouldn't worry about vectors here.)

    Redo this. In calculating the torque, you left off the radius.

    Since the basic formula for I is mr^2, the simplest units for I would be kg-m^2.


    Not good. Use kinematics to find the angular velocity of the pulley after 5.00 s. You already found the angular acceleration, so use it.
  14. Nov 25, 2007 #13


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    Thank You Doc Al for all your help :smile:
  15. Dec 4, 2007 #14
    I'm a little confused here (of course I can't get past the simple stuff) but I keep getting 0.03m/s^2 instead of 0.06. What did I do wrong here:

    Assuming the acceleration is constant



    So put d/t in the a equation, and i end up with

    =75cm/5^2 s^2
    Last edited by a moderator: Dec 4, 2007
  16. Dec 4, 2007 #15

    Doc Al

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    In this equation, v stands for the final velocity (since it starts from rest).

    This equation gives you the average velocity, not the final velocity. For constant acceleration starting from rest, the average velocity equals the final velocity divided by 2. That's why your answer is off by a factor of 2.

    The proper equation to use would be:
    d = 1/2 a t^2, or:
    a = 2d/t^2
  17. Feb 29, 2008 #16
    Why is acceleration towards the pulley considered positive? Wouldn't that make gravity negative?
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