A rotating flywheel can be used to store energy. If it is required to store 1.00 × 106 J of energy when rotating at 400 rad/s, what is the moment of inertial of the wheel in kg m2?单项选择题
A
6.25
B
12.5
C
25.0
D
50.0

Question

A rotating flywheel can be used to store energy. If it is required to store 1.00 × 106 J of energy when rotating at 400 rad/s, what is the moment of inertial of the wheel in kg m2?

BlackTom AI · Accepted Answer

Rotational kinetic energy of a rigid body is given by K = (1/2) I ω^2, where I is the moment of inertia and ω is the angular speed.

1) Relevant concepts
- Rotational kinetic energy formula K = (1/2) I ω^2.
- Solve this expression for I when K and ω are known.

2) Step-by-step reasoning
We are given K = 1.00 × 10^6 J and ω = 400 rad/s. Rearranging the formula to solve for I:
I = 2K / ω^2.

3) Calculations with intermediate steps
- Compute ω^2: ω^2 = (400 rad/s)^2 = 400 × 400 = 160000 = 1.6 × 10^5 s^-2.
- Compute numerator 2K: 2K = 2 × (1.00 × 10^6 J) = 2.00 × 10^6 J.
- Divide: I = (2.00 × 10^6) / (1.6 × 10^5) = (2.00 / 1.6) × 10^(6−5)
  = 1.25 × 10^1 = 12.5 kg·m^2.

4) Why this answer is correct
Substituting I = 12.5 kg·m^2 back into K = (1/2) I ω^2 gives K = 0.5 × 12.5 × 400^2 = 6.25 × 160000 = 1.00 × 10^6 J, which matches the required stored energy. Thus the computed I yields the specified energy at the given angular speed.

5) Why the other options are incorrect
- 6.25 kg·m^2: half the correct moment; using this would give only 0.5 × 6.25 × 400^2 = 5.00 × 10^5 J (half the required energy).
- 25.0 kg·m^2: twice the correct moment; would store 2.00 × 10^6 J (double the required energy).
- 50.0 kg·m^2: four times the correct moment; would store 4.00 × 10^6 J (four times the required energy).

6) Conclusion
The correct moment of inertia is 12.5 kg·m^2, which matches the provided answer.

A rotating flywheel can be used to store energy. If it is required to store 1.00 × 106 J of energy when rotating at 400 rad/s, what is the moment of inertial of the wheel in kg m2?单项选择题

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