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Blocks on Pullys and Torque

  1. Dec 31, 2005 #1

    G01

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    A 4.0 kg block is suspended 1.0 m off the ground and is attached via a pulley and massless rope to a 2.0 kg mass held to the ground. The pulley is .12m in diameter and and has a mass of 2.0 kg. Friction between the axel and pulley exerts a torque or .5 Nm. How long does it take the 4.0 kg block to hit the ground.
    Okay, first Happy New Year!!!!!! Why am I posting this on New Years Day? Because if I don't it will never stop bugging me!!:confused:
    First I found the net Torque.
    [tex] \sum \tau = \tau_{w of 4.0 kg} - \tau_{2.0 kg} - \tau_{friction}[/tex]
    [tex] \sum \tau = 2.4 - 1.176 - .5 [/tex]
    [tex] \sum \tau = .724 N*m [/tex]

    I don't see a problem so far.

    [tex] I = .5Mr^2 = .0036 kg m^2 [/tex]

    [tex] \alpha = \frac{\sum \tau}{I} [/tex]

    [tex] \alpha = 201.11 \frac{rad}{s^2} [/tex]

    Ok now, the block falls 1.0m which is equal to 16.53 radians here.

    [tex] \frac{.38 m}{2\pi rad} = \frac{1.0m}{x rad} [/tex]

    [tex] x = 16.53 rad [/tex]

    Then:

    [tex] \theta_f = \theta_0 + \omega_0 t + .5\alpha t^2 [/tex]

    If I solve this for t I get .41s, but the answer is 1.11s. Where did I go wrong? Thanks for the help.
     
  2. jcsd
  3. Jan 1, 2006 #2

    Fermat

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    I don't know how you got these particular values, but I would say that this is where you went wrong.
    Let T1 be the tension in the rope attached to the mass m1 = 2.0 kg.
    Let T2 be the tension in the rope attached to the mass m2 = 4.0 kg.
    Then the net torque is given by,
    [tex] \sum \tau = T_2*r - T_1*r - 0.5 [/tex]
    [tex] \sum \tau = (T_2 - T_1)r - 0.5 [/tex]
    Now you have to construct an eqn of (linear) motion for the two blocks using the unknown tensions T1 and T2 using newton's 2nd law.
     
    Last edited: Jan 1, 2006
  4. Jan 1, 2006 #3

    G01

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    Thank you and happy new year
     
  5. Jan 1, 2006 #4

    G01

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    I'm sorry I'm still kinda lost here. I have no problem with the rotational motion part of the problem but I'm still lost as to how to find an equation for T2- T1

    I tried using Newton's Third Law:

    [tex] \Sigma F_4 = w_4 -w_2 - T_{tot} = 4a [/tex]

    [tex] \Sigma F_2 = w_4 - w_2 + T_{tot} = 2a [/tex]

    [tex] \Sigma F_{system} = w_4- w_2 +T_{tot}- T_{tot} = 6a [/tex]

    Here I can solve for the acceleration as if ther was no friction and then plugin in a in the first equaton to find T. I think I can do this because the tension is the same whether there is friction at the axel or not. Then i can fill in T in the torque equation to find the net torque including friction and from there use newton II to find [tex] \alpha [/tex] and then use rotational kinematics to solve for time. I still don't get the right answer what did I do wrong?
     
    Last edited: Jan 1, 2006
  6. Jan 1, 2006 #5

    Fermat

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    I still don't clearly follow your working - I'm not sure what Ttot is.

    Anyway ...

    Try it like this,

    Particle 1

    Force up = T1
    Force down = m1g

    Using Newton's 2nd Law: F=ma,

    T1 - m1g = m1a ---------------(1)

    Particle 2

    Force up = T2
    Force down = m2g

    Using Newton's 2nd Law: F=ma,

    m2g - T2 = m2a ---------------(2)

    adding (1) and (2),

    (m2 - m1)g - (T2 - 1) = (m1 + m2)a

    Now you already have an expresion for (T2 - T1), back in Post #2
    Substitute for that into the previuous eqn and solve for a.
     
  7. Jan 1, 2006 #6

    G01

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    Ok what you said is exactly what I did but I can't solve for a if I substitute the first equation because i'll have 2 variables [tex] \Sigma \tau [/tex] and a. Also, shouldn't T2 - T1 = 0 because they are an action reaction pair?
     
    Last edited: Jan 1, 2006
  8. Jan 1, 2006 #7

    Fermat

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    [tex]\sum \tau = I\alpha[/tex]

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

    No, T2 and T1 are not equal.
    They are both used for acelerating different sized masses at the same accln, so will be different sized themselves.

    Moderated by the friction, yes, but still different.
     
  9. Jan 1, 2006 #8

    G01

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    Sigh ok heres what I got now.

    [tex] \sum \tau = I\alpha = (T_2 - T_1)r - 0.5 [/tex]

    [tex] (T_1 - T_2) = (m_2 - m_1)g - (m_2 + m_1)a [/tex]

    So:

    [tex] \sum\tau = I\alpha = (m_2 - m_1)gr - (m_2 +m_1)ar -0.5 [/tex]

    [tex] Ia = 2gr^2 - 6ar^2 -0.5r [/tex]

    the mass of this pulley is 2kg and its [tex] I = .5mr^2 [/tex] so [tex] I = r^2 [/tex] So we can cancel out the I with the r^2:

    [tex] a = 2g - 6a -\frac{0.5}{r} [/tex]

    [tex] 7a = 2g -\frac{0.5}{r} [/tex]

    so: [tex] a = 11.27 m/s [/tex]

    Here [tex] x_f = 1/2at^2 [/tex]

    Solving for t I get t= .42s Where did I go wrong this time. The correct answer is 1.11s. Sorry Im sick today and obviously not thinking straight.......
     
    Last edited: Jan 1, 2006
  10. Jan 1, 2006 #9

    G01

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    NEVER MIND!!!!!!!!!! forgot to divide by 7 lol. I got the right answer. Thanks alot Fermat!
     
  11. Jan 1, 2006 #10

    Fermat

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    Great!

    Now it is a Happy New Year.

    And if it makes you feel any better, I used 0.05 as the torque for quite a while before I found my error!!
     
  12. Jan 1, 2006 #11

    G01

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    Yes it is a happy new year now. still though i could have swore T2 - T1 were an action reaction pair and had to be equal in magnitude? Of course I could just be rusty on Newton's third law....
     
  13. Jan 1, 2006 #12

    Doc Al

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    3rd law pairs

    3rd law pairs are forces that two objects exert on each other: The pulley and mass 1 exert a force T1 on each other (opposite directions, of course); similarly, the pulley and mass 2 exert a force T2 on each other. (Note that the rope is massless, so you can treat it as just transmitting the force between the two bodies.)

    If the pulley were massless and frictionless, then T1 would equal T2.
     
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