To reduce its rotational inertia, an engineer proposes drilling three holes in a large wheel of radius R. The holes are to be centered midway between the center and edge of the wheel. What should be their radii (in terms of R) if drilling the holes is to reduce the rotational inertia by 19%?

Answer 1

A flywheel for energy storage in a hybrid gasoline-electric car is a ring-shaped rotor of mass 50 kg, designed to rotate at 30000 rpm. At that speed it stores 10 MJ of energy (equivalent to about a cup of gasoline). What is the ring's diameter?

Answer 2

My mistake, wrong button sorry John!

Answer 3

No problem. For my question, I know the goal is to basically subtract the inertia's of the 3 holes from the whole disk's inertia, 0.5*MR^2, but I think what I'm stuck on is how to find that value, or relate it to what we know. The final inertia needs to be 0.81(0.5*MR^2). I'm not sure how to apply the "center of masses of each hole" being R/2 away from the big/main center of mass because I don't know where the unknown radii lengths of those circles fit into the relation.

Answer 4

Hmm this is probably wrong but I was thinking of something like this:
M1 = the mass that was in one of the holes
R2 = the radius of one of the holes
M = mass of the whole original disk
R = radius of whole original disk

I think the following relation makes sense:
M1(R/2)^2 = 0.19M(R)^2 / 2 / 3

Basically what I'm thinking is the inertia of one of "hole's matter" should be a third of the total inertia we're looking for. Then for M1, since the whole disk is uniform density, I think M1 and M are proportional to the differing areas, meaning:
M1 / M = pi(R2)^2 / pi(R)^2

If that's right, you can solve for M1, plug it into the first relation, and then you're left with R2 and R terms, and can simply solve for R2 in terms of a constant times R. When I did so I got 0.3559...

...but I think this is incorrect b/c I ran this through my "answer checker" and it said 0.3559 was incorrect, so I'm really stuck ;\, please help if anyone can.

Answer 5

m(hole)= m, M(plate)=M
We must treat the holes as the objects of negative mass.
The moment of inertia is
I=I(plate) – 3I(hole).
The plate is disk => I(plate)=M•R²/2.
The moment of inertia of the hole using the parallel axis theorem is
I(hole)= m•r²/2 + m• (R/2)² =
=(m/2){r² +R² /2},
The moment of inertia of the plate with the holes is
I= M•R²/2- 3• (m/2){ r² +R² /2}.
Since the plate is uniform, its mass is proportional to area, so
m/M =πr²/πR²=r²/R²,
m=M(r²/R²).
I= M•R²/2 - 3• (Mr²/2R²){ r² +R² /2}
Initial moment of inertia (without the holes) was I₀= M•R²/2.
Then (I-I₀)/I₀=0.19
{M•R²/2 - 3• (Mr²/2R²){ r² +R² /2}- M•R²/2}/ (M•R²/2)=0.19
It gives
3r²(2r² +R²)/2R⁴ = 0.19
If r² = x
6x²+3R²-0.38R⁴=0
x=0.605 R²
r=0.78R
I’m not sure in my calculation, check it.