Torque needed to maintain a constant fall velocity

[jared1313]

BOTTOM LINE UP FRONT:
I need to roughly approximate the torque required from a rotating eddy current braking device in order to maintain a constant fall velocity of a suspended object on a pulley or cable drum.

LONGER DESCRIPTION:
I am working on a project where I have cable spooled around a pulley. At the free end of the cable is an object of some weight. With the pulley locked in place it will experience a torque of T=F*R where "F" is weight in pounds and "R" is radius of the pulley. When the pulley is unlocked, the weight will accelerate downward toward the ground. Assuming ideal conditions (no losses, friction, or resistances) the velocity at any given distance will be defined by v(d)=sqrt(2*a*d) where velocity (v) is a function of distance (d) and "a" is acceleration due to gravity. The RPM of the pulley can be defined as RPM=sqrt(2*a*d)/(2*pi*R) where "R" is the diameter of the pulley. From this point, how would I determine what opposing torque would be required to maintain some constant fall velocity?

I am not quite sure how to approach this and I have tried to google similar problems, but came up empty handed. Any suggestions are quite welcome.

Thank you in advance for any suggestion.

 

[spigor]

The torque to maintain a constant fall velocity will be the same as for a locked pulley. With the RPM increasing, the power required will also be increased. If an object should maintain a constant speed, no force must accelerate it. Since the weight is doing so, an opposite force must be used to cancel the weight out. I don't know if the force should be measured in pounds, as I'm not familiar with the imperial unit system. I'd say F=mg, where m is the mass in kg and g=9.81 m/s^2 (for Earth). I'm not saying that something is wrong, only that I cannot verify it on the spot.

 

[desertfox]

Hi

Unless the drag on the load equals and acts opposite the acceleration force your device will increase in velocity. Why can you not use a synchronous motor and use a brake for emergency conditions?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[electricpete]

When the pulley is unlocked, the weight will accelerate downward toward the ground. Assuming ideal conditions (no losses, friction, or resistances) the velocity at any given distance will be defined by v(d)=sqrt(2*a*d) where velocity (v) is a function of distance (d) and "a" is acceleration due to gravity. The RPM of the pulley can be defined as RPM=sqrt(2*a*d)/(2*pi*R) where "R" is the diameter of the pulley.

That would be free-fall at a constant acceleration.

From this point, how would I determine what opposing torque would be required to maintain some constant fall velocity?

What do you mean “from this point”? Are you transitioning from constant acceleration to constant velocity at some point during the fall? Can you explain more about that transition?

At any rate from Fnet=MA =0, the brake torque to maintain constant velocity means the force upward on the line F= Tbrake*R will be equal/opposite to the force from gravity F = m*g so trivially Tbrake = m*g / R

TLDR, the idealized conditions for constant velocity fall are obvious. The nature of your question is not (to me).


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(2B)+(2B)' ?

 

[TheTick]

Sounds like you need a damper. Typical damper behavior increases torque with speed.