A flywheel of mass 500 kg and diameter 1 m makes 500 rpm. Assuming the mass to be concentrated along the rim, calculate (i) angular velocity (ii) moment of inertia and (ii) rotational K.E. of the flywheel.

Answer

To solve the problem step by step, we will calculate the angular velocity, moment of inertia, and rotational kinetic energy of the flywheel.

### Step 1: Calculate Angular Velocity (ω)

1. **Given Data**:
- Rotations per minute (RPM) = 500
- To convert RPM to radians per second, we use the formula:
\[ \omega = \text{RPM} \times \frac{2\pi \text{ radians}}{1 \text{ rotation}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} \]

2. **Calculation**:
\[ \omega = 500 \times \frac{2\pi}{60} = \frac{1000\pi}{60} = \frac{50\pi}{3} \text{ rad/s} \]
Approximating π as 3.14, we get:
\[ \omega \approx 52.36 \text{ rad/s} \]

### Step 2: Calculate Moment of Inertia (I)

1. **Given Data**:
- Mass (m) = 500 kg
- Diameter = 1 m, thus radius (r) = 0.5 m

2. **Formula for Moment of Inertia for a ring**:
\[ I = m r^2 \]

3. **Calculation**:
\[ I = 500 \times (0.5)^2 = 500 \times 0.25 = 125 \text{ kg m}^2 \]

### Step 3: Calculate Rotational Kinetic Energy (K.E.)

1. **Formula for Rotational Kinetic Energy**:
\[ K.E. = \frac{1}{2} I \omega^2 \]

2. **Calculation**:
\[ K.E. = \frac{1}{2} \times 125 \times \left(\frac{50\pi}{3}\right)^2 \]
First, calculate \(\left(\frac{50\pi}{3}\right)^2\):
\[ \left(\frac{50\pi}{3}\right)^2 = \frac{2500\pi^2}{9} \]
Now substituting this into the K.E. formula:
\[ K.E. = \frac{1}{2} \times 125 \times \frac{2500\pi^2}{9} = \frac{125 \times 2500\pi^2}{18} \]
Approximating \(\pi^2 \approx 9.87\):
\[ K.E. \approx \frac{125 \times 2500 \times 9.87}{18} \approx 1.715 \times 10^5 \text{ Joules} \]

### Final Answers:
1. Angular Velocity (ω) ≈ 52.36 rad/s
2. Moment of Inertia (I) = 125 kg m²
3. Rotational K.E. ≈ 1.715 × 10^5 Joules