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Inclined Plane Double Pulley Problem

  1. May 18, 2014 #1
    1. The problem statement, all variables and given/known data

    A double pulley consists of two separate pieces that are welded together; a hoop of radius 10.0 cm and a disk of radius 25.0 cm. The moment of inertia of the pulley is 0.160 kg-m^2. 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.

    Use force methods to answer the following questions.
    a) What direction are the blocks moving and why?
    b) How far will each block travel in their respective directions?
    c) Find the angular acceleration of the pulley?
    d) Find the acceleration of block m1.
    e) Find the acceleration of block m2.
    f) Find the tension in cable 1.
    g) Find the tension in cable 2.

    2vi4mmr.jpg


    2. Relevant equations



    3. The attempt at a solution

    So far I think I have found the acceleration of m2, which would indicate that it is moving downward and thus pulling m1 up the ramp. Using this formula, and also factoring in friction, I got 3.1 for the acceleration.

    a = ( m2*g - m1*g*sin(θ) - μk*m1*g*cos(θ) ) / ( m1 + m2 )
    a = ((24.0*9.81)-(15.0*9.81*sin(35°))-(0.25*15*9.81*cos(35°)))/39.0
    a= 3.1

    But does that actually mean m2 is pulling m1? Does the double pulley affect this at all?

    Overall I am pretty confused how the double pulley plays into this. The only problem I could find that was remotely similar to this was this one, but I still don't understand the difference between a regular pulley and double pulley here.

    As for b), I think the final position of m1 would be calculated with:

    xf = (2/k)[m2g - m1*g*sin(θ) - μk*m1*g*cos(θ)]

    So I could take xf - xi and find the distance traveled. But if I understand correctly that would be using energy, and I need to solve this with force methods.

    I also imagine the radii will affect the distance traveled, though I don't what equation to use.

    Guidance on double pulleys and determining distance-traveled greatly appreciated!
     
  2. jcsd
  3. May 18, 2014 #2

    tms

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    According to the description in the problem, the double pulley consists of two concentric pulleys that turn together. This implies that each weight has its own cable, and those cables are not connected to each other, but only interact through the pulley. And this implies that a single turn of the pulley will cause the weights to move different distances and at different speeds. Both parts of the pulley will have the same angular speed, however.
     
  4. May 18, 2014 #3

    ehild

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    You completely ignored the moment of inertia of the pulley. Neither you took into account that the pulley have an inner and an outer radius. The formula you used was invalid for this case.


    Draw the free-body diagram for both blocks and also on the pulley. Write the equations for the accelerations of the blocks and the angular acceleration for the pulley.

    Note that it is the tension of the string that pulls up m1 (not m2) .

    ehild
     
  5. May 18, 2014 #4
    Okay. So this will become relevant when I calculate the blocks' accelerations.

    Also, I am now factoring in the moment of inertia and the radii. For the acceleration of m1 I am now getting:

    a1 = ( (m2 * g * r / R) - (μ * m1 * g * cos(θ)) - (m1 * g * sin(θ)) ) / ( m1 + (m2 * (I/R)^2) + (I/R^2) )
    a1 = -1.35

    Where r and R are the little and big radii.

    a1 is negative. What does that mean?

    For distance-traveled I believe the distance of m2 is h, and m1 is h/sin(θ). Is this correct?
     
  6. May 18, 2014 #5

    ehild

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    How did you get that formula for a1? Explain!

    Start up from F=m1 a1. What forces act on m1?

    ehild
     
  7. May 18, 2014 #6
    Here's how I did it:

    Equations of motion:
    m1 * a1 = T1 - (μ * m1 * g * cos(θ)) - (m1 * g * sin(θ))
    m2 * a2 = m2 * g - T2
    I * α = T2 * r - T1 * R

    Deduced from equations:
    α = a1 / R = a2 / r

    Sub a2 with (a1 * r / R):
    m2 * a1 * r / R = m2 * g - T2

    Sub α with (a1 / R):
    I * a1 / R = T2 * r - T1 * R

    Rearrange equation to solve for T2:
    T2 = m2 * g - m2 * a1 * r / R

    Rearrange 3rd equation to put T1 on left:
    T1 = (T2 * r / R) - (I * a1 / R^2)

    Sub T2 with the equation from 2 steps above:
    T1 = ( (m2 * g) - (m2 * a1 * r / R) ) * (r / R) - ( I * a1 / R^2 )

    Multiply out:
    T1 = (m2 * g * r / R) - (m2 * a1 * (r / R)^2) - ( I * a1 / R^2 )

    Plug in T1:
    m1 * a1 = (m2 * g * r / R) - (m2 * a1 * (r / R)^2) - ( I * a1 / R^2 ) - (μ * m1 * g * cos(θ)) - (m1 * g * sin(θ))

    Do the algebra for a1:
    a1 * (m1 + m2 * (I / R)^2 + (I / R^2)) = (m2 * g * r / R) - (μ * m1 * g * cos(θ)) - (m1 * g * sin(θ))
    a1 = (m2 * g * r / R) - (μ * m1 * g * cos(θ)) - (m1 * g * sin(θ)) / (m1 + m2 * (I / R)^2 + (I / R^2))
     
    Last edited: May 18, 2014
  8. May 18, 2014 #7

    ehild

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    You made a mistake at the end (in red), It should be (r/R)2
    The derivation is much shorter, if you eliminate the tensions by multiplying the first equation by R, the second one by r and substitute for RT1 and rT2 into the second. Express the accelerations by the angular acceleration, and solve for it.



    ehild
     
    Last edited: May 18, 2014
  9. May 18, 2014 #8
    I tried doing this but I couldn't eliminate the tensions. In fact it ended up looking like a lot more work than the last equation I posted, so I guess I'm not doing it right:

    R * m1 * a1 = (R * T1) - (R * μ * m1 * g * cos(θ)) - (R * m1 * g * sin(θ))
    r * m2 * a2 = (r * m2 * g) - (r * T2)

    R * T1 = r * T2 - I * α
    r * T2 = R * T1 - I * α

    R * m1 * a1 = (r * T2) - (I * α) - (R * μ * m1 * g * cos(θ)) - (R * m1 * g * sin(θ))
    r * m2 * a2 = (r * m2 * g) - (R * T1) - (I * α)

    (I * α) = (r * T2) - (R * m1 * a1) - (R * μ * m1 * g * cos(θ)) - (R * m1 * g * sin(θ))
    (I * α) = (r * m2 * g) - (R * T1) - (r * m2 * a2)

    (r * m2 * g) - (R * T1) - (r * m2 * a2) = (r * T2) - (R * m1 * a1) - (R * μ * m1 * g * cos(θ)) - (R * m1 * g * sin(θ))

    Seems even more complex now because I need to eliminate the T's... somehow...

    If my attempt 2-posts-back is right (after fixing the error) then I'd like to move on with it. The answer I get is now a1 = -1.08. But I'm still not sure how to interpret the negative number that I got. What does being negative mean? Does that mean it's moving up or down the slope? How can I know?
     
    Last edited: May 18, 2014
  10. May 18, 2014 #9

    ehild

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    You assumed that m2 moves downward and you got a=-1.08 m/s^2. That means accelerating upward.
    If you tried to assume acceleration in the opposite direction, you would get some negative value again. That means, the blocks can not start moving from rest.
    In case they move they would decelerate. But the problem asks the direction of motion, which can not be determined from the given data. Either you made some mistake when copying the data of the problem writer gave wrong data.


    ehild
     
  11. May 18, 2014 #10
    It's more likely that I came up with the incorrect equation than that my professor gave a bad problem. I copied the problem exactly as-is.

    Maybe if I try solving it your way I will get a positive answer. (Which is apparently what I am supposed to get?) What am I doing wrong exactly 1-post-back? How can I eliminate T1 and T2? Is there some simpler method I am missing?
     
  12. May 18, 2014 #11

    ehild

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    You eliminated Iα instead of T1 and T2.

    [tex]m_1a_1=T_1-m_1g(\sinθ+μ\cosθ)[/tex]
    [tex]m_2a_2=m_2g-T_2[/tex]
    [tex]rT_2-RT_1=αI[/tex]

    a1=αR, a2=αR and isolate the tensions.

    [tex]T_1=m_1αR+m_1g(\sinθ+μ\cosθ)[/tex]
    [tex]T_2=m_2(g-αr)[/tex]

    Multiply T1 by R and T2 by r, substitute into the torque equation.
    [tex]rm_2(g-αr)-R(m_1αR+m_1g(\sinθ+μ\cosθ))=Iα[/tex]

    Expand, collect the terms with α.

    [tex]rm_2g-Rm_1g(\sinθ+μ\cosθ)=α(I+m_1R^2+m_2r^2)[/tex]
    [tex]α=\frac{g(rm_2-Rm_1(\sinθ+μ\cosθ))}{I+m_1R^2+m_2r^2}[/tex]


    Your numerical result was correct, it is negative. The problem is wrong somehow. Ask your professor.
    ehild
     
  13. May 18, 2014 #12
    A minor thing I realized: I was using centimeters rather than meters in the last equation. So the latest answer for a1 is a1 = -0.95.

    Sorry, I just realized this, but you said in your 2nd-to-last post that the blocks cannot start moving from rest. However the problem states "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." So they are not at rest, they are moving and are decelerating and will come to a stop. That changes things, right? Doesn't that mean we can find the directions of motion?

    After plugging in the numbers for your equation I got α = 3.33055. Which is quite similar to the crude answer of α = 3.0969 that I got by using this calculator: http://hyperphysics.phy-astr.gsu.edu/hbase/incpl2.html#c2, barring that that calculator doesn't have a double pulley. So I'm assuming α = 3.33055 is correct. And on the page I just linked to, they say "Taking downward as the positive direction for the hanging mass..." I guess because α is positive, that seems to imply that the hanging mass is moving downward...?

    So it seems as if m2 is moving down and m1 is being pulled up the ramp as a result? But how do I actually know that m2 is moving down? You said that I "assumed that m2 moves downward", but I was actually wondering whether it does move downward at all. I mean, m2 is heavier, so I think it does... :confused:
     
  14. May 18, 2014 #13

    ehild

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    I got other number, but it is also negative.

    We can not find out the direction of the motion. The blocks would not move without setting them moving . And you can not know if m2 moved initially downward or upward.

    How did you get positive value?

    No use to compare this problem with the much simpler one. There are two radii , two strings, two tensions, these change everything. In case of such problems, when you do not know the direction of motion, you assume some direction and calculate the acceleration. Note that the direction of friction is opposite to the velocity. If you get negative acceleration, you have to assume opposite motion.

    ehild
     
  15. May 18, 2014 #14
    Apparently I got the wrong number the first time. It's not α = 3.33055. It's α = 4.34596.

    [tex]α=\frac{9.81*(0.1*24-0.25*10*(\sin35°+0.25\cos35°))}{0.16+10*0.25^2+24*0.1^2}=4.34596[/tex]

    This is so confusing. It seems like such a simple question: "Which direction are they going?" I never would have imagined it was impossible to know... I thought the weights of the objects would determine which block pulled on which.
     
    Last edited: May 18, 2014
  16. May 18, 2014 #15

    haruspex

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    The question says "in the appropriate direction". What makes one direction more appropriate than the other? The interpretation so far seems to be "the direction they would move if released from rest". First, that's rather an assumption; secondly, it doesn't really help because we're not told the static coefficient; thirdly, as has been seen, if we assume static friction equals kinetic friction then the two torques on the pulley (ignoring friction) differ by less than the magnitude of the torque from maximum friction, so it would remain at rest.
    Another possible interpretation is "that direction for which one or other block reaches the ground". There is an implication that this happens, though it could be clearer.
     
  17. May 18, 2014 #16
    Alright, got an email from my professor. Apparently there was an error in the problem. The radius of the larger pulley is now 15cm instead of 25cm.

    [tex]α=\frac{9.81*(0.1*24-0.15*10*(\sin35°+0.25\cos35°))}{0.16+10*0.15^2+24*0.1^2}=19.3446[/tex]

    And using 0.15 instead of 0.25:

    [tex]a_1=\frac{m_2*g*r/R-μ*m_1*g*\cosθ-m_1*g*\sinθ}{m_1+m_2*(r/R)^2+I/R^2}[/tex]
    [tex]a_1=2.9016[/tex]

    Positive!

    ehild said "You assumed that m2 moves downward and you got a=-1.08 m/s^2. That means accelerating upward." So because my acceleration is now positive, that means accelerating downward. Which means m2 is moving downward. Which means m1 is moving up the ramp. Is this logic correct?
     
    Last edited: May 18, 2014
  18. May 18, 2014 #17

    haruspex

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    Yes.
     
  19. May 18, 2014 #18

    ehild

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    In the problem text, m1=15 kg. You calculated with 10 kg. Why?

    Remember, a1=Rα. You do not need to do all the calculation again.

    that is correct.

    Everybody can make mistakes, even your professor. But he noticed his error and emailed to the students, so all is good. You have learnt even form that mistake.

    ehild
     
  20. May 19, 2014 #19
    :eek: Close one!

    [tex]a_1=1.29[/tex]
    [tex]α=8.63[/tex]
     
  21. May 19, 2014 #20

    ehild

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    Much better :smile:. Go ahead.

    ehild
     
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