Question

A flywheel is designed for engine speed variation from $190\text{ rps}$ to $210\text{ rps}$. Calculate the inertia of the wheel if the kinetic energy stored in the flywheel is $400\text{ N}\cdot\text{m}$.

• A. $0.02\text{ kg}\cdot\text{m}^2$
• B. $0.01\text{ kg}\cdot\text{m}^2$
• C. $0.2\text{ kg}\cdot\text{m}^2$
• D. $0.1\text{ kg}\cdot\text{m}^2$

Hint:

To simplify calculations involving flywheel energy changes, always use the difference of squares identity: $\omega_{\text{max}}^2 - \omega_{\text{min}}^2 = (\omega_{\text{max}} - \omega_{\text{min}})(\omega_{\text{max}} + \omega_{\text{min}})$. This turns long squaring steps into a quick multiplication.

Answer 1

The Correct Option is A

Concept: A flywheel acts as an energy reservoir that smooths out fluctuations in engine speed by storing kinetic energy during periods of excess power and releasing it when power demand exceeds production. The fluctuation of energy $\Delta E$ during a cycle is related to the flywheel's mass moment of inertia $I$ and its operating speed range by the kinetic energy equation: \[ \Delta E = \frac{1}{2} I \left( \omega_{\text{max}}^2 - \omega_{\text{min}}^2 \right) \] where $\omega_{\text{max}}$ and $\omega_{\text{min}}$ represent the maximum and minimum angular velocities expressed in radians per second (rad/s).

Step 1: Converting rotational speed units from rps to rad/s.
The problem gives the engine speeds in revolutions per second (rps). We convert these to angular velocities using the relation $\omega = 2\pi N$:

• Maximum speed, $N_{\text{max}} = 210\text{ rps} \quad \Rightarrow \quad \omega_{\text{max}} = 2\pi \times 210 = 420\pi\text{ rad/s}$

• Minimum speed, $N_{\text{min}} = 190\text{ rps} \quad \Rightarrow \quad \omega_{\text{min}} = 2\pi \times 190 = 380\pi\text{ rad/s}$

Step 2: Setting up the energy balance equation.
The fluctuation of kinetic energy stored in the flywheel is given as: \[ \Delta E = 400\text{ N}\cdot\text{m} \] Substitute these values into our energy fluctuation formula: \[ 400 = \frac{1}{2} \cdot I \cdot \left[ (420\pi)^2 - (380\pi)^2 \right] \] Multiply both sides by 2 to simplify the fraction: \[ 800 = I \cdot \pi^2 \cdot \left( 420^2 - 380^2 \right) \]

Step 3: Factoring and simplifying the squared terms.
Using the difference of squares identity ($a^2 - b^2 = (a-b)(a+b)$) to simplify the calculation: \[ 420^2 - 380^2 = (420 - 380)(420 + 380) = (40)(800) = 32,000 \] Substitute this back into the equation: \[ 800 = I \cdot \pi^2 \cdot 32,000 \]

Step 4: Solving for the mass moment of inertia $I$.
Isolate the inertia variable $I$: \[ I = \frac{800}{32,000 \cdot \pi^2} = \frac{1}{40\pi^2} \] Using the approximation $\pi^2 \approx 9.8696$: \[ I = \frac{1}{40 \times 9.8696} = \frac{1}{394.78} \approx 0.00253\text{ kg}\cdot\text{m}^2 \]