Calculating Inertia for a Rotating Disc with 16 Blades | Wk² = 120-188

[Buz]

I have been scratching my head for a few hours now. Problem is I don't believe my calculations one bit haha.

I have a circlular rotating disc mass (approx 80Lbs) radius of 13"

I have 16 Blades evenly spaced on this mass(approx weight of 3.2Lbs/each) Inside radius is 13" outside radius is 21"

I have came up with a number of 120Wk² and a number of 188Wk²

Someone save me from pulling my hair out lol

Thanks in advance

 

[SteamKing]

How did you come up with these numbers? What calculations did you do?

 

[Buz]

Thanks for response.

26" dia. cylinder
weight = 80Lbs
Radius = 13.3462

wk² = (Weight)(Gyro Radius)

Gyro Radius = 13.3462/12 = 1.11218/2 = 0.55609/2 = 0.8341 + 0.55609 = 0.4170

wk² = (80)(0.4170)
wk² = 33.365

Blades
weight = 3.2Lbs
In/out Radius = 13.3462/inner radius -- 21/outside radius

wk² = 3.2(1.1121+1.75)/2
wk² = 73.26

sum of inertia

Total wk² value = 106.625

---------------------------------

Other Calculation

Calculating this as a solid mass

Weight = 131.2Lbs
Radius = 21" / 12 = 1.75'

wk² = (0.5)(131.2)(1.75)
wk² = 114.8

Which one should apply?

 

[SteamKing]

Your calculations appear to be incorrect.

For instance, take the 80-lb. disk, radius 13" (round numbers for convenience).

The mass moment of inertia of a circular disk is (1/2)mR^2
For the disk, R = 13/12 = 1.0833 ft. and m = W/g = 80/32.2 = 2.48 slugs

MOI = 0.5 * 2.48 * 1.0833^2 = 1.458 slug-ft^2

The moment of inertia is also equal to the mass * gyradius^2, or MOI = mass * k^2, where k = gyradius

For the disk, k = SQRT (1.458 / 2.48) = 0.767 ft = 9.2 in.

To add the blades to the disk, you would need to estimate the MOI of a single blade and then use the parallel axis theorem to find the MOI of the disk and blades. The MOI of a blade is going to depend on the size and shape of the blade.