how to calculate required hp for machine

OldEngr63

I have estimated the total MMOI as 16.5 kg-m^2, and on that basis, Bob_S value of 41 watts appears to be a reasonable estimate for the total power required to accelerate the roll (I calculated slightly less). The total energy of the spinning roll at 100 rpm is 903 J.

As to realism, now, these numbers are so small that I rather think that they are close to meaningless. The friction losses in the gear train, the bearings, the windage, etc. will consume more than what we have computed for the change in the energy state of the rotating mass. Thus I think that considerably more power will be required to overcome the system losses than to power the roll. Therefore, I think that these calculations are not realistic for evaluating the power required.

Bob S

First, you have to do the math.
If the stored rotational energy after 60 seconds is
[tex] E=\frac{1}{2}I\omega^2=\frac{1}{2}mr^{2}\omega^2 = 1234 \text{ Joules}[/tex] then you have to add energy at the rate of 1234/60 joules/sec = 21 watts for 60 seconds.
This does not include motor inefficiency, friction losses, not using the right gear ratios, etc. Please check my math.

OldEngr63

@ Bob_S: I did do the math, in considerably more detail than it appears you have done. It appears that you are simply saying that
I = MR² = 17.956 kg-m²
Note that the radius value you used is to the outer diameter of the rubber layer, well past the last of the steel structure. This gives an excessively large estimate for R.
By a more detailed reckoning, based on the drawing, I came up with the figure that I gave previously
I = 16.5 kg-m²
I think that you have somewhat over estimated the MMOI by putting it all at the rubber layer, well outside the steel, considering that part of it is internal structure.

N = 100 rpm → ω = 10.47 rad/s
ω² = 109.66 1/s²

Because of the difference in the estimates for I, we differ substantially in the total energy content of the rotating roll. This is where our differences lie.

My comments about these values being minor compared to the losses in the system and therefore not useful for determining drive power requirements still stand.

Bob S

You are correct. I decided to put the maximum value in [itex] I=mr^2 [/itex], which is exactly twice the moment of inertia for a solid cylinder. It adds a few % to the result, as you found out.