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Blocks and pulley with mass - rotational dynamics

  1. Jun 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Two blocks of masses m1 = 8.00 kg and m2 = 12.00 kg are connected by a string of insignificant mass. The string lies on a cylinder of mass 0.50 kg and a radius of 0.15 m. The coefficient of kinetic friction between m1 and the surface of an inclined plane is 0.10. The angle of that plane is 30°. (See figure here http://imgur.com/WtpJoye ) If the blocks were stationery at the beginning and they move 4 m, calculate:

    a) The initial mechanical energy of the system.
    b) The final mechanical energy of the system after travelling 4 m.
    c) The velocity of the blocks after travelling 4 m.

    2. Relevant equations

    3. The attempt at a solution

    For part c), I have tried the following
    (W being the weight of m1)
    [tex]
    EI = PEf + KEf + Wfr\\
    m2·g·h = m1·g·h + \frac{1}{2}·m1·v^2 + \frac{1}{2}·m2·v^2 + \frac{1}{2}·w^2 + W·sen(30°)·4·0.10
    [/tex]

    Which after more manipulation becomes

    [tex]
    v = \sqrt{\frac{m2·g·h-m1·g·h-9.8·m1·sen(30°)·4·0.10}{\frac{1}{2}·m1 + \frac{1}{2}·m2 + \frac{1}{4}·mp}}
    [/tex]
    Where mp is the mass of the pulley or cylinder.
    The result I get after changing the values is v = 3.73 m/s

    Is this correct? Also, how can I solve the parts a) and b) ? I am not really sure about what to do next or if what I'm doing is right.

    I would really appreciate any help you could give me to solve this correctly. Thanks.
     
  2. jcsd
  3. Jun 10, 2014 #2
    Note that the KE of rotation of the cylinder is given by

    [itex]\frac{1}{2}[/itex]Iω[itex]^{2}[/itex] where I is the moment of inertia of the cylinder.

    Also it is not clear where you have taken the reference level from which the PE[itex]_{grav}[/itex] is to be measured.
     
  4. Jun 10, 2014 #3

    Nathanael

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    He has appropriately involved the rotational inertia and rotational KE (not in his original equation, but in his final equation it appears correct.)
    I don't think it matters? (Only the change in GPE is relevant)
     
  5. Jun 10, 2014 #4

    Nathanael

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    I only see two possible problems:

    First, why would it be sin(30°) and not cos(30°)? (I'm talking about this term: [itex]9.8·m_1·sin(30°) · 4 ·0.10[/itex])

    Next, you seem to have neglected the fact that the change in height would not be the same for [itex]m_1[/itex] and [itex]m_2[/itex] (you used "h" for both of them)

    By "h" you mean "4" right ???

    for [itex]m_2[/itex] it would go down by 4 (and thus have [itex]|ΔE_{grav}|=4m_2·g[/itex])

    but for [itex]m_1[/itex] it would only have [itex]|ΔE_{grav}|=4m_1·g·sin(30°)=2·m_1·g[/itex]
    (the reason is that the distance of 4 meters that it travels is the distance of the "hypotenuse" whereas the Δh travelled is only 2 meters)
     
    Last edited: Jun 10, 2014
  6. Jun 10, 2014 #5
    My post refers to the first step of the original post.
     
  7. Jun 10, 2014 #6

    Nathanael

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    Yes I figured, that's why I was letting you know that he did it correctly (as far as I can tell) in his final step

    (so his first step was possibly a typo or something)
     
  8. Jun 10, 2014 #7
    v = [itex]\sqrt{(m_{2}g4 - 0.1m_{1}g4cos30 - m_{1}g4sin30)/(m_{1}/2 + m_{2}/2 + m_{p}/4)}[/itex]

    as Nathanael said.
     
  9. Jun 10, 2014 #8
    Hi, thanks for your help. I just noticed that I didnt consider the Y-axis movement for m1 to be different from the one of m2. I didn't define PEgrav because I'm just looking for the change in energy, the problem isn't very specific on that one.

    However, now I'm getting 5.3 m/s as the result of the above given equation. Can you confirm? Thank you again.
     
  10. Jun 10, 2014 #9

    Nathanael

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    I get 5.4 m/s

    Did you take care of the sin\cos problem in the frictional term?

    Or perhaps you simply forgot to round up to 5.4? (It was 5.38)
     
  11. Jun 11, 2014 #10
    I got 5.34 actually, I'll check out the calculations. Thanks for the help!
     
  12. Jun 13, 2014 #11
    post later, forgot the friction
     
    Last edited: Jun 13, 2014
  13. Jun 14, 2014 #12
    a)
    Initial energy = 0
    c + b)
    A = incline angle = 30 °
    g = 9.81 (m/s)/s
    r = pulley radius = 0.15 metres
    M = pulley mass = 0.5 kg
    M1 = 8 kg
    M2 = 12 kg
    µ = M1 kinetic friction co-efficient = 0.1

    Pulley wheel moment of inertia (i) = ½ * m * r ² = 0.005625 kg – m ²

    Calculate the net force on the system :
    f1 = gravitational force created by M2 = M2 * g = 117.72 N
    f2 = gravitational force created by M1 = M1 * g * sine A = 39.24 N
    f3 = friction force of block M1 = M1 * g * cosine A * µ = 6.7966 N

    The (net) force driving the system = f1 – ( f2 + f3 ) = 71.6834 N

    Equivalent mass of the pulley = 0.25 kg

    Total equivalent system mass = 8 + 12 + 0.25 = 20.25 kg

    System acceleration (a) :
    a = net force / total equivalent mass
    a = 71.6834 / 20.25
    a = 3.8756 ( m / s ) / s

    Block velocity (v) after 4 m :
    v = square root ( u ² + ( 2 * a * s ) )
    v = square root ( 2 * 3.8756 * 4 )
    v = 5.3216 m/s (ANSWER c)

    Final system KE :
    KE (M1) = ½ * M1 * v ²
    = 113.2775 Joules
    KE (m2) = ½ * M2 * v ²
    = 169.9163 Joules
    KE pulley wheel = ½ * i * ( ( v / r ) ² )
    = 3.5399 Joules

    Total system final energy :
    = 113.2775 + 169.9163 + 3.5399
    = 286.7337 Joules (ANSWER b)

    You can check answer b by subtracting (the PE gained by M2 + The work done by the friction force) from the PE translated by M2 :
    = ( 12 * 9.81 * 4 ) - ( ( 8 * 9.81 * Sine A * 2 ) + ( f3 * 4 ) )
    = 286.7337 Joules

    Note : calcs done on excel, answers rounded on this reply.
     
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