To solve this problem, we need to address both parts (a) and (b) step by step.

Given Data:

• Kinetic Energy (E): 2000 ft-lb
• Mean Diameter (D): 36 inches
• Rated Speed (N): 66 RPM
• Coefficient of Fluctuation (C): 0.2
• Rim Width to Thickness Ratio: 4:1
• Specific Weight of Cast Iron: 0.26 lb/in³

Conversions:

1. Convert kinetic energy from ft-lb to in-lb (since 1 ft-lb = 12 in-lb): $E = 2000 \times 12 = 24000 \text{ in-lb}$
2. Mean Radius (r): $r = \frac{D}{2} = \frac{36}{2} = 18 \text{ inches}$

Part (a): Determine the weight of the rim

The kinetic energy stored in the flywheel can be expressed by: $E = \frac{1}{2} I \omega^2 C$ Where:

• $I$ is the moment of inertia
• $\omega$ is the angular velocity in radians per second

1. Convert RPM to radians per second: $\omega = \frac{2\pi N}{60} = \frac{2 \times 3.1416 \times 66}{60} \approx 6.91 \text{ rad/s}$

2. Moment of inertia for a rim-shaped flywheel: $I = m r^2$ Where $m$ is the mass of the rim.

3. Using the equation for kinetic energy: $E = \frac{1}{2} m r^2 \omega^2 C$ Solving for $m$: $m = \frac{2E}{r^2 \omega^2 C} = \frac{2 \times 24000}{18^2 \times 6.91^2 \times 0.2} \text{ in-lb}$ $m = \frac{48000}{324 \times 47.75 \times 0.2} \approx \frac{48000}{3090.3} \approx 15.53 \text{ lb}$

4. Weight of the rim (W): $W = m \times g$ Here, $g$ is the acceleration due to gravity. In lb, the weight is equivalent to mass since $g$ cancels out.

Thus, the weight of the rim is approximately 15.53 lb.

Part (b): Determine the rim width and thickness

1. Let the rim thickness be $t$, then the rim width $b = 4t$.

2. Volume of the rim $V$: $V = 2\pi r b t \approx 2 \times 3.1416 \times 18 \times 4t \times t = 144 \pi t^2 \text{ in}^3$

3. Using the specific weight of cast iron: $W = \text{Specific Weight} \times V$ $15.53 = 0.26 \times 144 \pi t^2$ Solving for $t$: $t^2 = \frac{15.53}{0.26 \times 144 \times 3.1416} \approx \frac{15.53}{117.81} \approx 0.1318 \text{ in}^2$ $t \approx \sqrt{0.1318} \approx 0.363 \text{ inches}$

4. Rim width $b = 4t$: $b = 4 \times 0.363 \approx 1.452 \text{ inches}$

Thus, the rim thickness is approximately 0.363 inches, and the rim width is approximately 1.452 inches.

Summary:

• (a) The weight of the rim is 15.53 lb.
• (b) The rim thickness is 0.363 inches, and the rim width is 1.452 inches.

Would you like further details or have any questions?

Here are 5 related questions:

1. How would the required weight of the rim change if the coefficient of fluctuation was different?
2. What would happen to the thickness and width if the rim's diameter increased?
3. How does the specific weight of a material influence the design of a flywheel?
4. What is the impact of the rated speed on the flywheel's kinetic energy?
5. Why is it important to neglect the effect of arms and hub in this calculation?

Tip: When solving rotational dynamics problems, always ensure consistent units, especially when working with energy and inertia.