Torque and Power (looks pretty simple)

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The discussion centers on calculating torque and lifting capacity for an industrial motor with an output of 225 kW at 4260 RPM. The torque is computed as 504 Nm, and the force exerted by the motor is determined to be 764 N, allowing for a maximum lift of approximately 77.9 kg. However, there is confusion regarding the speed at which the weight can be raised, with initial calculations suggesting an unrealistic velocity of 294 m/s. Clarifications indicate that the linear velocity at the drum's edge is actually 147 m/s, which aligns with the power output when lifting a heavier weight. The conversation highlights the impracticality of the scenario, considering the motor's specifications and the nature of lifting applications.
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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.
 
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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...)
 
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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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