When the speed of a flywheel is increased from 240 r.p.m. to 360 r.p.m., the energy spent is 1936 J.
What is te moment of inertia of the flywheel?

Options

A. `4.9kg-m^(2)`
B. `9.8kg-m^(2)`
C. `2kg-m^(2)`
D. `15kg-m^(2)`

Explanation

To solve the problem, we need to find the moment of inertia (I) of the flywheel given that the speed is increased from 240 r.p.m. to 360 r.p.m. and the energy spent is 1936 J. We will use the formula for kinetic energy and the relationship between angular velocity and moment of inertia.

### Step-by-Step Solution:

1. **Convert the angular velocities from r.p.m. to rad/s:**
- The formula to convert revolutions per minute (r.p.m.) to radians per second (rad/s) is:
\[ \omega = \text{r.p.m.} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \]
- For the initial speed (240 r.p.m.):
\[ \omega_i = 240 \times \frac{2\pi}{60} = 8\pi \text{ rad/s} \]
- For the final speed (360 r.p.m.):
\[ \omega_f = 360 \times \frac{2\pi}{60} = 12\pi \text{ rad/s} \]

2. **Calculate the change in kinetic energy:**
- The kinetic energy (K.E.) is given by:
\[ K.E. = \frac{1}{2} I \omega^2 \]
- The initial kinetic energy (K.E._i) is:
\[ K.E._i = \frac{1}{2} I \omega_i^2 = \frac{1}{2} I (8\pi)^2 = \frac{1}{2} I (64\pi^2) \]
- The final kinetic energy (K.E._f) is:
\[ K.E._f = \frac{1}{2} I \omega_f^2 = \frac{1}{2} I (12\pi)^2 = \frac{1}{2} I (144\pi^2) \]

3. **Determine the energy spent:**
- The energy spent (ΔK.E.) is the difference between the final and initial kinetic energy:
\[ \Delta K.E. = K.E._f - K.E._i = \frac{1}{2} I (144\pi^2) - \frac{1}{2} I (64\pi^2) \]
- Simplifying this gives:
\[ \Delta K.E. = \frac{1}{2} I (144\pi^2 - 64\pi^2) = \frac{1}{2} I (80\pi^2) \]

4. **Set the energy spent equal to the given energy:**
- We know that the energy spent is 1936 J:
\[ \frac{1}{2} I (80\pi^2) = 1936 \]

5. **Solve for the moment of inertia (I):**
- Rearranging gives:
\[ I = \frac{1936 \times 2}{80\pi^2} \]
- Calculating this:
\[ I = \frac{3872}{80\pi^2} \]
- Using \(\pi \approx 3.14\):
\[ I \approx \frac{3872}{80 \times (3.14)^2} \approx \frac{3872}{80 \times 9.8596} \approx \frac{3872}{788.768} \approx 4.9 \text{ kg m}^2 \]

### Final Answer:
The moment of inertia of the flywheel is approximately \(4.9 \text{ kg m}^2\).

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