- #1
riseofphoenix
- 294
- 2
KErotational = (1/2)Iω2
420 = (1/2)Iω2
420 = (1/2)I(402)
420/(402) = (1/2)I
0.2625 = (1/2)I
2(0.2625) = I
0.525 = I
0.53 = I
Last edited:
What is the moment of inertia of this rotating object?
- Thread starter riseofphoenix
- Start date
AI Thread Summary
The discussion focuses on calculating the moment of inertia (I) of a rotating object using the formula for rotational kinetic energy (KErotational = (1/2)Iω²). The calculation shows that with a given kinetic energy of 420 and an angular velocity of 40 rad/s, the moment of inertia is determined to be approximately 0.53. Participants clarify that kinetic energy cannot be equated to angular momentum (L) due to differing units. The expression for rotational kinetic energy is confirmed as correct, reinforcing the relationship between I and ω. Overall, the calculations and concepts regarding rotational dynamics are validated.
KErotational = (1/2)Iω2
420 = (1/2)Iω2
420 = (1/2)I(402)
420/(402) = (1/2)I
0.2625 = (1/2)I
2(0.2625) = I
0.525 = I
0.53 = I
- #2
Doc Al
Mentor
- 45,578
- 2,426
You cannot set KE = L. The units don't even match.
What's the expression for rotational KE in terms of ω?
What's the expression for rotational KE in terms of ω?
- #3
riseofphoenix
- 294
- 2
Doc Al said:
(1/2)I(v/r)?
I'm not sure if there is another equation for rotational KE
- #4
riseofphoenix
- 294
- 2
KErotational = (1/2)Iω2
420 = (1/2)Iω2
420 = (1/2)I(402)
420/(402) = (1/2)I
0.2625 = (1/2)I
2(0.2625) = I
0.525 = I
Is that right?
420 = (1/2)Iω2
420 = (1/2)I(402)
420/(402) = (1/2)I
0.2625 = (1/2)I
2(0.2625) = I
0.525 = I
Is that right?
- #5
Doc Al
Mentor
- 45,578
- 2,426
Looks good.
- #6
riseofphoenix
- 294
- 2
Thanks!
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