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RPM
02-25-2006, 04:34 PM
I know there are some real physicists/mathematicians on the group, who will probably eat this problem for breakfast :-)
I am designing a machine which has a lightweight shaft that is turned by a small motor. I have to join the shaft to the motor with a coupling, and these come in different sizes, which are rated at different torque capacities, with units in inch-pounds.
My problem is that I have to select one size that will be able to handle the maximum torque put on it when the machine starts up. The two couplers I am looking at are rated at 3 inch-pounds, and 12 inch-pounds, and I would prefer the smaller one, with a shorter length, because of various aspects of the design.

The shaft, which is solid, weighs about 1 pound, is 1 inch in diameter, and the maximum torque comes from accelerating from 0 rpm to 400 rpm in approximately half a second.

My calculations so far, which may be completely wrong, are the angular velocity to be 83 radians per second, and my moment of inertia is 0.125 pound, but I can't see how to multiply these, and still get an answer in inch-pounds.
Any ideas?

Many thanks in advance

Richard in Los Angeles.

topct
02-25-2006, 04:58 PM
"The shaft, which is solid, weighs about 1 pound, is 1 inch in diameter, and the maximum torque comes from accelerating from 0 rpm to 400 rpm in approximately half a second."

Just the shaft? Surely it must be performing some kind of work.

[This message has been edited by topct (edited 02-25-2006).]

kf1002002
02-25-2006, 05:02 PM
The basic formula here is T=J*w where w is substituted for omega (rad/sec). consistent units for this are T in lbft, J in slugs where a slug is inertia in lbft^2 / g. Your inertia of .125 must be in some unit perhaps lb inch squared. Get this inertia right and it is a simple calculation. If you can clarify this I can whip it off quick and easy.
\
Ken

Leigh
02-25-2006, 05:54 PM
<font face="Verdana, Arial" size="2">Originally posted by kf1002002:
...where w is substituted for omega (rad/sec)</font>

Actually the character is ω which is a lower-case Omega. Not "substituted for". Minor clarification.

------------------
Leigh W3NLB

kf1002002
02-25-2006, 06:02 PM
Correction to my previous post-- I was going too fast.
The formula is T=J*alpha where alpha is angular acceleration in radians per second per second. In your case the speed runs up from 0 to 400 rpm in .5 sec or 0 to 41.9 radians per second in .5 second. this gives an acceleration of 83.8 radians per second per second which squares with your figure of 83 except that the "seconds" should be (seconds^2).

As for the inertia, if it is only the shaft of 1 lb and 1" dia, the formula is
J=M*(r^2)/2. In this formula M is in slugs, ie weight/g, r is shaft radius in feet.
So, to put numbers into your problem,
M=1/32.2 = .0311 slug
r=.5/12=.0417 ft
so: J=.027*10^-3 slugft^2(sure wish I could use mathcad here)

so T=.027*10^-3*83.8 ot T=.0023 lbft
or .0272 in lb

Check this arithmatic carefully as I have been known to make a mistake. Also, is it really a 1", 1lb shaft or is there something on the shaft? Thi seems like a small number.
Keep in mind the possibility of shock loads as this could be the load that breaks the coupling. If this is a significant risk we would need to know the motor inertia also plus make a few assumptions.

[This message has been edited by kf1002002 (edited 02-25-2006).]

kf1002002
02-25-2006, 06:08 PM
Leigh
sorry if I caused some confusion. I used w because I did not know how to type an omega in this editor, a problem which I see you've solved

Ken

mklotz
02-25-2006, 06:41 PM
Just one minor nit to pick.

The units of J, obvious from the formula used to compute it, are slug-ft^2,
not slugs as indicated in kf's post.

The units of the angular acceleration, alpha, are sec^-2.

Multiplying J*alpha produces a torque with units of slug-ft^2/sec^2 or
ft-(slug-ft/sec^2). One slug-ft/sec^2 is one pound-force (lbf.)

Aside: Damn to eternal hell the Imperial cretins who used the same unit, pound,
for both force and mass. Yet another reason to use metric.

So the J*alpha product has the final units of ft-lbf, the traditional unit for
torque. Multiplying by 12 (inches per foot) converts to in-lbf, as shown.

Regards, Marv

Home Shop Freeware - Tools for People Who Build Things
http://www.myvirtualnetwork.com/mklotz

kf1002002
02-25-2006, 08:02 PM
mklotz
I fully agree with you re imperial and metric.
In the work I did before I quit working for a living we seemed to use the word slugs for both w/g and for wk^2/g. I realized it was sort of loose use of the words but we seemed to understand each other. In this case I just carried it over without thinking it could cause a double take.

KF

barts
02-25-2006, 11:55 PM
As much as I like calculations (I'm a mech eng by training),
most coupling ratings are driven by some running torque value rather than peak load. Typicall, peak loads aren't known well enough to do much good, so the manufacturers tend to use "service factor" (fudge factor) for shock/vibration/etc.

Don't let this stop you, but you'll probably end up w/ a pretty
conservative design. Nothing wrong with that unless you're
building an airplane :-).

- Bart

Evan
02-26-2006, 01:36 AM
Everything you need to calculate it may be found here.

http://www.compumotor.com/catalog/catalogA/A60-A62.pdf

mklotz
02-26-2006, 09:30 AM
In the light of a new day, I've got another question about this computation.

The formula kf uses for the moment of inertia is for the moment about an axis
*transverse* to the shaft axis. The OP didn't specify so it's possible that
this is correct but one would normally expect that the shaft would rotate
*about* its axis, not perpendicular to it. If that's true, I think the formula
for the moment should be:

J = (M*d^2)/8

where:
M = 0.0311 slug
d = shaft diameter = 1 in = 1/12 ft

which would lead to an even smaller value of torque than kf computed.

Regardless, I have to agree with the other posters who suggested that this is
probably not the best way to establish the torque requirement for the coupler.
Most especially, what does the shaft drive and what torque load does the driven
element supply?

Regards, Marv

Home Shop Freeware - Tools for People Who Build Things
http://www.myvirtualnetwork.com/mklotz

kf1002002
02-26-2006, 11:06 AM
Marv
I spotted this too and I've reworked the calculation as you suggest. The new numbers are in the edited version of my posting. As everyone seems to suggest, the number is very small and we need to confirm that the shaft is not connected to anything.
The shaft weight is given as 1 lb but if it's steel it comes to more like 2.6 Lb. I think we need a better definition of the problem or RPM could use the 3 inlb coupling and have a conservative design which is more in line with what Bart says.
KF

RPM
02-26-2006, 03:09 PM
Thanks for all your help :-)
I knew i was in trouble when you started using 'slugs' in your formulas, something I hadn't heard of, nor could find on a long internet search.
The shaft is made from plastic, hence its light weight, and there is not a lot attached to it, hence the low figures. I anticipate that the running torque will be less than the start-up torque, so your calculations are spot on.
Thanks
Richard in Los Angeles

kf1002002
02-27-2006, 02:51 PM
RPM
A "slug" is more or less a colloquial term though it's evident from the postings that those in the business are familiar with it. It comes about because in the english and US system "pound" is used for both force and weight although these are different things, thus a new unit is needed for one or the other. The solution was to invent a "slug" as the unit of mass. A slug is simply the weight divided by "g" the acceleration due to gravity which is 32.2 ft/sec/sec. It must be used in problems based on "f=m*a" or its derivatives.
One way to look at it is if you go to the moon your weight is much smaller than on earth but you mass is still the same number of slugs as on earth. It is your earth weight divided by earth g or your moon weight divided by moon g which will come out the same. If there's a hole in my logic here I have no doubt we'll hear about it.
Ken

rklopp
02-27-2006, 04:45 PM
When selecting a coupling like this, typically you will want to select based on the larger of the max motor torque or the max inertial torque due to suddenly jamming the shaft while the motor is spinning at full speed. Max motor torque is typically the stall or breakdown torque, and should be available on datasheets for your motor. If it's possible to decelerate or jam the shaft when the motor is spinning at speed, then the coupling will have to bring the motor to a stop in the required time, or the coupling will fail. You may want that to happen, in which case the coupling is acting as a mechanical fuse to protect other parts of your system. If you don't want the coupling to fail, you will need to find out the intertia of the motor's rotor and everything else upstream of the coupling, accounting for any gearing or other speed ratios, and the compliance ("springiness") of the system. You can think of compliance as the airbag when your system crashes into a wall.

Of course there may be other components that will break at lower torques than your coupling, in which case, they act as the mechanical fuse, and your coupling need not take the full available torque.

You should apply a safety factor to whatever design torque you come up with. The safety factor should reflect the level of uncertainty in your design calculations and the risks associated with failure. I.e., if it'll kill somebody or you'll have to do a recall of 2 million units if the coupling breaks, you'd better have a big safety factor.