Question

For the specified flywheel design, the variation of the engine speed is from 210 rad/s to 190 rad/s. During the cycle process, the change in kinetic energy is found to be 400 N-m, then the moment of inertia of the flywheel in kg-m² is:

• A. 0.40
• B. 0.30
• C. 0.20
• D. 0.10

Hint:

When solving for the moment of inertia from kinetic energy changes, remember that the change in kinetic energy is proportional to the difference in the square of the angular velocities.

Answer 1

The Correct Option is D

Step 1: Kinetic Energy Formula
The change in kinetic energy (\( \Delta KE \)) of a rotating body is given by the formula: \[ \Delta KE = \frac{1}{2} I (\omega_2^2 - \omega_1^2) \] where: \( I \) is the moment of inertia,
\( \omega_1 \) and \( \omega_2 \) are the initial and final angular velocities, respectively.
Step 2: Substituting the Known Values
From the problem:
Initial angular velocity \( \omega_1 = 210 \, \text{rad/s} \),
Final angular velocity \( \omega_2 = 190 \, \text{rad/s} \),
Change in kinetic energy \( \Delta KE = 400 \, \text{N-m} \).
Substituting into the kinetic energy formula: \[ 400 = \frac{1}{2} I \left( (190)^2 - (210)^2 \right) \]
Step 3: Simplifying the Expression
First, calculate the difference of the squares: \[ (190)^2 - (210)^2 = (190 - 210)(190 + 210) = (-20)(400) = -8000 \] Thus, the equation becomes: \[ 400 = \frac{1}{2} I (-8000) \] \[ 400 = -4000 I \] \[ I = \frac{400}{4000} = 0.10 \, \text{kg-m}^2 \]
Step 4: Conclusion
So, the moment of inertia of the flywheel is \( \boxed{0.10} \, \text{kg-m}^2 \).