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Find torque required to lift a mass

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data
    A crane contains a hollow drum of mass 150kg and radius 0.8m that is driven by an engine to wind up a cable. The cable passes over a solid cylindrical 30kg pulley 0.3m in radius to lift a 2000N weight. How much torque must the engine apply to the drum to lift the weight with an acceleration of 1ms^(-2)
    2. Relevant equations
    As I am only currently dealing with 1D rotationl motion the equations are
    [tex]\tau=FR\sin\theta[/tex] where i suppose theta is 90 degrees
    [tex]a_{t}=\alpha R[/tex]
    [tex]I_{pulley} = \frac{1}{2}M_{p}R_{p}^2[/tex]
    [tex]I_{drum} = M_{d}R_{d}^2[/tex]
    3. The attempt at a solution
    The difficulty I have with this question is the pulley and drum. I'm not sure how to factor both of these in and not sure where the mass comes into the overall scheme of things either, though I'm sure its got something to do with it.
    Now I know
    [tex]I_{pulley} = \frac{1}{2} \times 30 \times 0.3^2 = 1.35 kg m^2[/tex]
    [tex]I_{drum} = 150 \times 0.8^2 = 96 kg m^2[/tex]
    [tex] m = \frac{2000}{9.8} = 204.1kg[/tex]
    The tension in the rope between the pulley and the mass is given by
    [tex] T-mg=ma [/tex] where [tex] a = 1ms^-2[/tex] so
    [tex] T = mg+ma = 2204.1N [/tex]
    This is where things become unclear, after this I am just guessing
    [tex]\tau_{pulley} = T R_{pulley} = 2204.1 \times 0.3 = 661.23Nm[/tex]
    assuming tension in rope is the same between drum and pulley then
    [tex]\tau_{drum} = T R_{drum} = 2204.1 \times 0.8 = 1763.28Nm[/tex]
    or perhaps
    [tex]\tau_{pulley} = I_{pulley} \alpha = I_{pulley} \frac{a_{t}}{R_p}=1.35 \times \frac{1}{0.3}=4.5Nm[/tex]
    I am out of ideas after this point.
    answer in back of book = 1900Nm

  2. jcsd
  3. Nov 21, 2008 #2


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    More than one torque is acting on the drum. So think in terms of net torque and angular acceleration, just like you do with force and linear acceleration.

    Actually, it isn't. The rope exerts a torque on the pulley due to friction. This is the torque that causes the pulley rotation to accelerate. Likewise (Newton's 3rd Law), the pulley exerts a force on the rope, changing the tension value before & after the pulley.
  4. Nov 21, 2008 #3
    still not sure how to go about it
    The only guess I could make is that (assuming that the entire rope is accelerating at the same rate of [tex]1ms^{-2}[/tex]
    \tau_{net} = \tau_{drum} + \tau_{pulley}
    which leads to about 3 possibilities I can think of
    we can do
    \tau_{net} = M_{drum} R_{drum}^2 \alpha_{drum} + \frac{1}{2}M_{pulley}R_{pulley}^2 \alpha_{pulley} = 124.5Nm
    where [tex]\alpha_{drum}=\frac{1}{R_{drum}}[/tex] and [tex]\alpha_{pulley}=\frac{1}{R_{pulley}}[/tex]
    \tau_{net} = M_{drum} R_{drum}^2 \alpha_{drum} + TR_{pulley} = 781Nm
    or even
    \tau_{net} = T R_{drum} + \frac{1}{2}M_{pulley}R_{pulley}^2 \alpha_{pulley} = 1767Nm

    None of the above arrives at the right answer but they are only guesses. My first guess makes the most sense to me, however it is wrong.

    By the way, does anyone know how to get rid of that white ... out of the tex equations?
  5. Nov 22, 2008 #4


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    I'm not sure which object this taunet refers to? It almost sounds like it's for the rope, but that doesn't make any sense since the rope is not rotating.

    We are trying to set up a
    . . torquenet = I * alpha
    equation for the drum, to answer the question.

    To do that, we need the tension in the rope at the drum. Since the rope provides a net torque on the pulley, we'll need to set up
    . . torquenet = I * alpha
    for the pulley.

    So, draw a free body diagram for the pulley. Hint: think of the rope as exerting two forces, in different directions, on the pulley. One force is from the weight-to-pulley section of rope, the 2nd force is from the pulley-to-drum section of rope. They have different tensions; solve the equation for the pulley-to-drum tension.

    It's a bug, and unfortunately there's nothing members like us can do to get rid of it. The forum administrators are aware of the problem, but I guess it's not an easy one to fix.
    Last edited: Nov 22, 2008
  6. Nov 23, 2008 #5
    ok here goes another try
    let [tex]T_{1}[/tex] be the tension in the rope between the drum and the pulley
    let [tex]T_{2}[/tex] be the tension in the rope between the pulley and the mass
    so using Mr Newton on the 2000N mass
    [tex]T_{2} - 2000 = ma = 204.1 \times 1[/tex]
    [tex]T_{2} = 2204.1[/tex]
    now for the net torque on the pulley
    [tex]\tau_{net} = (T_{1} - T_{2}) R_{p} = I_{p} \alpha_{p}[/tex]
    [tex]T_{1} = \frac{I_{p}\alpha_{p}}{R_{p}} + T_{2}= \frac{\frac{1}{2}M_{p} R_{p}^2\alpha_{p}}{R_{p}} + T_{2}=\frac{1}{2}M_{p}R_{p}\alpha_{p}+T_{2}[/tex]
    now I shall take [tex]a_{t} = 1ms^{-2}[/tex] so [tex]\alpha_{p} = \frac{1}{R_{p}}[/tex]
    which gives
    [tex]T_{1} = \frac{1}{2}M_{p} + T_{2}=15+2204.1=2219.1N[/tex]
    now the net torque on the drum is
    [tex]\tau_{net} = T_{1} R_{d} = 2219.1N \times 0.8m = 1775Nm[/tex]
    which is wrong.
    Not sure where the [tex]I\alpha[/tex] thing comes into it for the net torque of the drum. I suppose it must pop in somewhere because my answer doesn't depend on the mass of the drum
  7. Nov 23, 2008 #6


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    You're very close!

    The question being asked is

    In other words, there's an engine and it contributes to the net torque on the drum.

    Set up a torquenet = I * alpha equation for the drum, similar to what you did for the pulley.
  8. Nov 24, 2008 #7
    ok i think i have finally got it
    heres a quick sketch of what i did
    [tex]T_{2} - mg = ma[/tex]

    net torque of pulley:
    \tau_{net} = (T_{1} - T_{2}) R_{p} = \frac{1}{2}M_{p}R_{p}^2\alpha_{p} [/tex]
    [tex]T_{1} = \frac{1}{2}M_{p}+T_{2}[/tex]

    net torque of drum:
    \tau_{net} = (F_{engine} - T_{1}) R_{d} = M_{d}R_{d}^2\alpha_{d} [/tex]
    [tex]F_{engine} = M_{d}+T_{1}[/tex]

    torque required by engine
    [tex]\tau_{engine} = F_{engine} R_{drum}
    which gave me about 1895Nm

    thanks for your help Redbelly98
    greatly appreciated, would never have solved it without your help
  9. Nov 24, 2008 #8


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    Looks pretty good. There are a couple of "R" terms missing, don't know if you just forgot about them in your post or they are missing from your written solution as well:

    ½ Mp → ½ Mp Rp

    Md → Md Rd

    Usually, checking the units (and explicitly including the 1/s2 angular acceleration) will catch little errors like that.

    You're welcome!
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