Torque and Power (looks pretty simple)

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SUMMARY

The discussion focuses on calculating the torque and lifting capacity of an industrial motor with an output of 225 kW operating at an angular speed of 4260 RPM. The torque developed by the motor is determined to be 504 Nm using the formula τ = P/ω. The calculations reveal that the motor can lift a weight of approximately 77.9 kg at a linear velocity of 147 m/s, although this scenario is deemed unrealistic due to the power requirements and the assumptions made regarding the mass of the drum and cables.

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Homework Statement


(a) Compute the torque developed by an industrial motor whose output is 225kW at an angular speed of 4260 RPM. (b) A drum with negligible mass, 0.660 m in diameter, is attached to the motor shaft and the power output of the motor is used to raise the weight hanging from the drum. How heavy can the motor lift at a constant speed? (c) At what constant speed will the weight rise?

The Attempt at a Solution


\omega = 4260 rev/min * \frac{2\pi}{60} = 446 rad/s
(a)P = \tau\omega
\tau = \frac{P}{\omega} = \frac{225000}{446} = 504 Nm
(b) This is where I got confused. I'm not sure how to figure this out. The only thing I could think of was v = r\omega = (0.660 m)(446 rad/s) = 294 m/s. But that doesn't answer the question. Any help would be greatly appreciated.

*EDIT* I thought of something:
\tau = Fl
F = \frac{\tau}{l} = \frac{504 Nm}{0.660 m} = 764N
m = 764 / 9.81 = 77.9 kg

77.9 kg moving at 294 m/s.. that sounds incredibly wrong.

I think I'm going about this wrong. Putting a max weight on motor would use up a lot of the power, so it would go really slow, if not just be able to keep the weight from falling. I really don't understand this one at all.
 
Last edited:
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odie5533 said:
(a)P = \tau\omega
\tau = \frac{P}{\omega} = \frac{225000}{446} = 504 Nm

This seems to be fine.

\tau = Fl
F = \frac{\tau}{l} = \frac{504 Nm}{0.660 m} = 764N
m = 764 / 9.81 = 77.9 kg

Be careful: if l is the moment arm for the torque, that will be the radius of the drum. The mass will be twice that much.

The linear velocity of the edge of the drum would be (0.33 m)·(446 rad/sec) = 147 m/sec.

Well, let's see if this is remotely credible as an answer. Raising 156 kg at 147 m/sec would require a power of

P = mgv = (156 kg)·(9.81 m/(sec^2))·(147 m/sec) = 225,000 W.

So that does check. It sounds excessive, but recall that 225 kW = 302 horsepower. Consider how fast a '60s "muscle car", which easily massed over 1000 kg, could be pushed up to...

What is a bit unrealistic about this problem is that a motor that powerful used in a winch would not be doing lifting and hauling with massless cables and take-up drums. (I think the drum would also be a bit bigger than 2 ft. across...)
 
Last edited:
Thanks for the help on the problem, and the explanation of why it's possible, which really helped me understand this problem better. Thanks again!
 

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