Calculate Acceleration of drum due to falling object 2

[CraneEng87]

Hello,

I'm trying to work out the acceleration of a drum as a load is dropped and need to account for mass moment of inertia.

I've attached a free body diagram of the system.

The load is instantaneously dropped causing the drum to un-spool wire rope as the load falls to the ground.

Assume the following:

Efficiency of sheave system = 0.942
Load = 34002 LBF
Drum Mass Moment of Inertia = 24532.57 LBM-FT^2
Radius of drum to rope = 2.946 ft

Find Angular Acceleration of Drum:

I'm not sure how to relate these components together and working in the imperial system for a problem like this is killing me. I can convert to SI if needed.

I'd appreciate any help.

 

[BUGGAR]

I was going to give this a shot until I noticed this problem is in almost every physics book I have.

 

[drawoh]

CraneEng87,

What is the inertia of the sheave?

Is efficiency a valid concept for a sheave? Can we compare power in to power out? I think what you need is the friction torque.

How did all your numbers get to be so accurate?

Handy tip: Pounds force and pounds mass are evil. A pound is a unit of force, convertible to Newtons. In the English system of units, mass is m=w/g. Do the substitution, and you can use metric equations with English units. Otherwise, you have to figure out, equation by equation, where g goes.

--
JHG

 

[CraneEng87]

BUGGAR,

Thanks for your productive input I always appreciate posters such as yourself..... If you would like to reference a specific book I would be glad to look it up, however this problem is not in any of my current references.

Drawoh,

You can ignore the sheave inertia and the efficiency is a "reeving system efficiency" calculated using bearing friction coefficients, reeving design and number of sheaves. I can work in the efficiency where it's needed. The numbers are the result of other calculations such as dynamic rope load and standard mass moment of inertia formulas assuming the drum is a thick walled cylinder.

I have attempted to use (T/I = a) however as you called out the use of LBF and LBM is a pain and I must be working the Gc into the equation improperly. The overall end state is to determine drum speed after a certain amount of time. So I have an initial angular velocity and a time for the formula (w = wo + at) I need to work out angular acceleration.

 

[drawoh]

CraneEng87,

Draw a free body diagram. Work out the force your sheave exerts on your cable. Efficiency has no meaning in this problem. Work out the force on your pulley, then the torque.

This is simple. BUGGAR has a point.

--
JHG

 

[rb1957]

drawoh had an excellent point for people not used to working with the imperial system. The basic unit of mass is slug = 32 lbm.

i think the sheave efficiency plays a part, in that there's some friction on the sleave.

you know the dynamics of the block, it's simple.

the dynamics of the sheave and the drum are a bit of work to figure out. eg, you have the rotational inertia of the drum so you know that's going to be spinning, but the sheave will also be rotating (no inertia, but i think that's where the efficiency factor fits in).

this does look a lot like homework ? but it might be a "simple" work problem that someone has asked you to solve (to see how much you know). it could be a problem you've seen in passing and are trying to solve it yourself for your own interest and learning ?

you should be able to get the dynamics of the drum. in a perfect world, the speed of the cable at the drum would be the same as the block, but we have an imperfect sheave. in a perfect world, the potential energy lost by the block is equal to the rotational energy of the drum; but we have an imperfect sheave.

and, btw, it's better not to post snarky replies.

another day in paradise, or is paradise one day closer ?

 

[moon161]

Atwood machine:

https://en.wikipedia.org/wiki/Atwood_machine

 

[rb1957]

not quite the same ... that device transfers kinetic and potential energy between the two masses

another day in paradise, or is paradise one day closer ?

 

[CraneEng87]

rb1957,

Thanks for the reply, I will clear the air and attempt to explain myself. I tried to simplify the problem to avoid a long drawn out explanation, however I may have simplified it too much.

The problem is a work related problem where we are fitting a disc brake to a hoist drum. The manufacturer of the disc brake in Germany has provided calculations, we as the primary contractor are required to review these calculations prior to sending them on to the customer. Our senior mechanical engineer has attempted to make sense of the German calculations and has been unable to replicate it. So he has gone ahead and done the calcs his own way in imperial units to attempt to achieve the same results. His results are similar but not exactly the same so he has passed them on to me to verify. I am attempting to verify his calcs and figure out the German calcs as well.

I have attached the calcs from Germany if anyone can make sense of what they did that would help. The formula I am working on is the "overspeed, lowering with load".

The constants I think I have worked out:

187.35 = (9.81 * 60) / pi
39.24 = 2 x 2 x 9.81

I need to break their formula down so that it is clear how gravity, inertia, rope load, and torque have been accounted for with an answer coming out in RPM.

 

[drawoh]

drawoh had an excellent point for people not used to working with the imperial system. The basic unit of mass is slug = 32 lbm.

m = w/g = 1lb/(32.2ft/sec[sup]2[/sup]) = 3.1[×]10[sup]-3[/sup]lb.sec[sup]2[/sup]/ft = 3.1[×]10[sup]-3[/sup] slugs per pound weight.

1 slug = 32.2 ft/lb.sec[sup]2[/sup]

Unit balances keep me out of trouble.

--
JHG

 

[moon161]

"not quite the same ... that device transfers kinetic and potential energy between the two masses "

zero out M2, add a pulley, there you are.

 

[kjoiner]

I just glanced at this quickly, but wouldn't the weight vary as well since more cable is pulled out, travels over the sheave and ends up vertical? Depending on the cable diameter and weight per foot it may be small, but if you're dropping it a long distance, it starts to add up. Just a thought.

Kyle

 

[CraneEng87]

kjoiner,

You are right the rope weight is 6.56 LB/FT and does start to add up however the time of drop for this application is very short and the added rope weight is negligible.

 

[rb1957]

personally, love periods for thousands and commas for decimals !

would the rope tension be less than the weight (less by the efficiency of the sheave) or more ?

so St*dt/2*nu is the torque the rope is applying to the drum (there's one of the "2*")
the 60/p you have is to convert the result to rpm, well it's 60/2pi ... and there's your other "2*"
they using dt in the denominator, but i think they really want radius ...
see attachment


another day in paradise, or is paradise one day closer ?

 

[chicopee]

"...His results are similar but not exactly the same so he has passed them on to me to verify..." I would be very surprised if your results were the same. However, since your results seem to be in the same "dugout; not ball park" I would make sure that your braking system is overly sized to cope with your results as you must not forget that heat will be generated during the brake application. If you should attempt to have an instantaneous stoppage of the load, there will be a theoretical minimum load intensification of a minimum of twice the weight of the load.

 

[3DDave]

I basically agree with Buggar on this. It's an entry level dynamics question as originally posed with the only complication being that tension in the rope is not equal on each side of the sheave.

The easy method is to look at the energy of the components. The rate of lowering the weight is tied to the angular speed of the drum times the radius of the drum so that the difference in (m*g*delta-h) will equal the difference in energy of the weight 1/2*m*v^2 plus the difference in rotational energy of the drum 1/2*I*omega^2. Since there's a relation between omega and v, they can be combined into a single equation.

The only tricky thing is that as the rope pays out it decreases the I value of the drum and increases the m value of the material in motion, so a more accurate equation includes alpha, the angle the drum has turned through.

Funny thing is, the energy to be absorbed by any brake is just mg delta-h. I think it doesn't matter how massive or how much inertia there is as long as the brake can absorb the energy in the time required.

 

[rb1957]

i'll correct my rearrangement of terms to ...

N = time/(1/omega_double_dot + r/g) ...

not really sure what that means, physically !?

another day in paradise, or is paradise one day closer ?

 

[hydtools]

I can get close to the idea with torque impulse = change in rotary momentum. But I don't end with the same equation or result as posted.

For the drum: torque * time = J * omega-final - J * omega-initial = J( omega-final - omega-initial)
omega-final = omega-initial + torque*time/J in rad/sec

torque = SsubT * dsubT/2

omega = n*2pi/60 = n/9.549, n in rpm, omega in rad/sec

nsubT-prime = nsubT + 9.549*(SsubT * dsubT * tsubo)/(2*JsubR) in rpm

Ted

 

[hydtools]

I erred and left out initial and final rotational momentum of the falling weight.

Vector equation: initial system momentum plus torque impulse equals final system momentum. Results in the equation posted in OP's previous attachment.

Ted

 

[rb1957]

where does the r/g term come in ?

another day in paradise, or is paradise one day closer ?