(adsbygoogle = window.adsbygoogle || []).push({}); 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]\tau=I\alpha[/tex]

[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]

[tex]F=ma_{t}[/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

Thanks,

vladimir

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# Homework Help: Find torque required to lift a mass

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