Rotational or Moment of Inertia Help 1

[morbark]

To all,
I have a 10,000 lb mass that is about 36 inches in diameter that is going to have to be accelerated from rest to about 300 rpms in roughly 10 seconds.
My question is...I guess I can figure it's "I" value (ie 1/2*m*r^2), but where does the acceleration of this body come into play? Isn't that quite important?
We're trying to determine the greatest amount of torque that will be needed to get this beast moving!
Thanks so much,
Bill

 

[prex]

The equation of interest, by similarity with Newton's law, is
T=Ia_{[ignore]&theta[/ignore];}
where T is the torque, I the moment of inertia and a_{[ignore]&theta[/ignore];} the angular acceleration (radians/sec²).
Using a constant acceleration over the 10 seconds you'll get a first approximation: the peak acceleration will be higher, especially if you don't want to brake the mass and can't accept an overspeed.

prex

http://www.xcalcs.com

Online tools for structural design

 

[morbark]

Prex...could you please expand on:

"the peak acceleration will be higher, especially if you don't want to brake the mass and can't accept an overspeed".

I don't need to brake it....and overspeed is not an issue.
Why is this important?

Thanks,
Bill

 

[morbark]

Also...in calculating this, I get 1.62million in lbs for "I", and consequently 305million in lbs for torque at an acceleration of 188.5 rad/s/s.
Geez this seems huge enough to not be real!
Bill

 

[pjhype]

morbark,

The angular acceleration = 300rev/min * 1min/60sec * 2*pi rad/rev divided by 10 sec = 3 rad/s^2 (you forgot to convert rev/min to rev/sec)

The moment of inertia = 1/2 * m * r^2 = 1/2 * 10,000 lb * 1 sl/386.4 lb * 36^2 = 16,770 sl-in^2 (you forgot to convert lb to mass elements, sl)

Torque = 16,670 sl-in^2 * 3 rad/sec^2 = 50,000 in-lb.

Note that this is valid if you have a 10,000 lb concentrated weight at a radius of 36 in. The required torque will change according to the geometry of the weight (e.g. if it is a disk).

pj

 

[Neilmo]

Hi morbark,

I calculate the torque required like this -

I = M*r^2/2
I = 10000*1.5^2/2 = 11250 lb.ft^2

w = 2*Pi*revs/60
w = 2*Pi*300/60 = 31.416rads/sec

a = w/t
a = 31.416/10 = 3.1416rads/sec^2

T = I*a/g
T = 11250*3.1416/32.2 = 1097.6lb.ft

And the power required to accelerate -
P = T*revs/5250
P = 1097.6*300/5250 = 62.7hp

Regards,
Neilmo

 

[morbark]

pj...you're right!
only thing is that 36 is the diameter, not the radius. So if you take 18 for "r"...I get 12577 in lbs.
Only one more ? though...how do you go from multiplying (sl-in) X (rad/s^2) to get (in-lbs) for the resulting units?
Bill

 

[desertfox]

Hi morbark

You can calculate the torque required from the change in angular momentum using the following formula:-


T*t = I*(W2-W1)


T= torque

I= moment of inertia

W1,W2 are the initial and final angular velocities of
the rotating mass

t= time taken to reach final angular velocity

using your figures and assuming a solid disc as the rotating part:-

I=mk^2 where k =radius of gyration

k=r/2^0,5 assuming r=radius of 18"=0.4572m

k = 0.4572m/sq root of 2 = 0.3233m

so mass = 10000lb/2.2 to give mass in kg

m= 4545.4545 kg

I = 4545.4545 x 0.3233^2 = 475.1 kgm^2


so T*10 = 475.1(W2-W1)

now at an radius of 0.4572m (18&quot and a final rpm of 300

the angular velocity = (300/60)*2*3.142= 31.42 rad/sec


therefore T= 475.1*(31.42-0)/(10)

T = 1492.76 Nm or (1098 lbft)


hope this helps

regards

desertfox

 

[pjhype]

morbark,

Thanks for correcting the radius. This result is consistent with desertfox and Neilmo. When you multiply inertia in sl-in^2 by angular acceleration in rad/s^2, you will get (sl-in/s^2)*in. (sl-in/s^2) can be stated in lbs, and your units become in-lbs.

pj

 

[morbark]

To all....looks like, and I calculate (w/ everyone's help) we're looking at about 13k in-lbs. That's great, and will work well w/ our components! Many thanks to all.....you're all very smart.....& I'm VERY grateful!
Regards,
Bill

 

[manam]

If you read the first post carefully, it says the sphere is of diameter 36 inches.. Now the moment of inertia of a solid sphere is... (2/5)*m*r^2 and not (1/2)..well the factor in the front is 0.4 instead of 0.5. So maybe you will need even lesser torque.

Also, do you really want to ignore the Static Friction altogether in this case..

 

[Neilmo]

manam,

I see no mention of `sphere` in the first post. Morbark has given I = m*r^2/2 so I think we have all assumed that it is a `disk`.

Regards,
Neilmo