To determine the minimum rotational inertia of the flywheel in a single piston engine, we can use the principles of rotational dynamics. The key here is to relate the energy absorbed by the flywheel to its rotational inertia and the speed fluctuations. Let's break this down step by step.
Understanding the Problem
The flywheel is rotating at an average speed of 1500 RPM, and it absorbs 1000 J of energy during half a rotation. The maximum permissible speed fluctuation is ± 60 RPM. This means that the speed can vary from 1440 RPM to 1560 RPM. Our goal is to find the minimum rotational inertia (I) of the flywheel that allows it to handle this energy absorption without exceeding the speed limits.
Key Concepts
- Rotational Kinetic Energy: The kinetic energy (KE) of a rotating object is given by the formula: KE = (1/2) I ω², where I is the rotational inertia and ω is the angular velocity in radians per second.
- Angular Velocity: To convert RPM to radians per second, use the formula: ω = (RPM × 2π) / 60.
- Energy Absorption: The energy absorbed by the flywheel during the speed fluctuation must equal the change in kinetic energy.
Calculating Angular Velocities
First, we need to convert the average speed and the fluctuation limits from RPM to radians per second:
- Average speed (ω_avg) = (1500 × 2π) / 60 ≈ 157.08 rad/s
- Minimum speed (ω_min) = (1440 × 2π) / 60 ≈ 150.80 rad/s
- Maximum speed (ω_max) = (1560 × 2π) / 60 ≈ 163.36 rad/s
Change in Kinetic Energy
The change in kinetic energy (ΔKE) due to the speed fluctuation can be calculated as follows:
- ΔKE = KE_max - KE_min
- KE_max = (1/2) I ω_max²
- KE_min = (1/2) I ω_min²
Substituting these into the equation gives:
ΔKE = (1/2) I (ω_max² - ω_min²)
Setting Up the Equation
We know that the flywheel absorbs 1000 J of energy during this fluctuation, so we can set ΔKE equal to 1000 J:
(1/2) I (ω_max² - ω_min²) = 1000
Calculating the Difference in Angular Velocities
Now, let's calculate ω_max² and ω_min²:
- ω_max² ≈ (163.36)² ≈ 26670.43 rad²/s²
- ω_min² ≈ (150.80)² ≈ 22740.64 rad²/s²
Now, calculate the difference:
ω_max² - ω_min² ≈ 26670.43 - 22740.64 ≈ 3929.79 rad²/s²
Finding Rotational Inertia
Substituting this back into our energy equation:
(1/2) I (3929.79) = 1000
Now, solving for I:
I = (1000 × 2) / 3929.79 ≈ 0.508 kg·m²
Final Result
The minimum rotational inertia of the flywheel required to absorb 1000 J of energy during the specified speed fluctuations is approximately 0.508 kg·m². This value ensures that the flywheel can handle the energy changes without exceeding the maximum permissible speed fluctuations.
