91Ó°ÊÓ

A flywheel is a mechanical device specifically designed to efficiently store rotational energy. By the very nature of their operation, flywheels can accumulate energy during periods when there is an excess supply, which allows them to release this energy quickly during times of high demand. This is particularly useful in industrial settings.

Consider the flywheel in our problem: it has a diameter of 1.5 meters and a mass of 250 kg. Large mass and dimensions enable it to store a significant amount of energy. It's like a massive "battery" that operates through the principles of rotational dynamics. The larger and heavier the wheel, the more energy it can store. The process of storing and releasing energy involves spinning up the flywheel to a certain angular velocity (speed), which brings us to our next topic.

Angular velocity is how fast an object rotates or spins, and it's a core concept in understanding flywheels. Measured in radians per second (\(rad/s\), it tells us how quickly we can achieve desired operations. Here, we start with an angular velocity given in revolutions per minute, or RPM, which is common for stating motor speeds.

To convert RPM to rad/s, we use the formula: \[\omega = RPM \times \frac{2\pi}{60}\]where \(\omega\) is the angular velocity in rad/s. In our exercise with a maximum speed of 1200 RPM, the angular velocity is 125.66 rad/s. This conversion is crucial for calculating other parameters like moment of inertia and rotational kinetic energy.

The moment of inertia is an object's resistance to changes in its rotation. Think of it as rotational mass. The bigger the moment of inertia, the harder it is to change the object's rotational speed. For a flywheel, which is typically a disk, the moment of inertia \(I\) is given by:\[I = 0.5 \times m \times r^2\]where \(m\) is mass and \(r\) is the radius of the flywheel.

In our example, with a mass of 250 kg and radius of 0.75 m (half of the diameter), the moment of inertia is 70.31 kg·m². This tells us how much torque is needed for the wheel to spin up to the top speed, which is a crucial factor in our calculations to understand the time it takes for the flywheel to reach its maximum angular speed and how much energy it can store.

Power and energy are definitive aspects of flywheel operations, encapsulating both stored energy and energy transfer rates. Energy stored in a flywheel is computed as rotational kinetic energy using:\[KE = 0.5 \times I \times \omega^2\]where \(KE\) is the kinetic energy.

In our scenario, when calculated, this results in a substantial 556,362 Joules of stored energy. When the flywheel delivers energy to a machine, half of this energy is transferred in 2 seconds, taking the energy transfer rate—known as power—to be:\[P = \frac{Energy}{Time} = 139,090.5 \, W\]Understanding the interplay between torque, power, and angular velocity, as captured in \(P = \tau \times \omega\), is essential. This relationship allows us to solve for torque exerted by the flywheel—1107 Nm in this problem—which demonstrates how efficiently energy can be transmitted to machines.