Question

In experiments at the Princeton Plasma Physics Laboratory, a plasma of hydrogen atoms is heated to over 500 million degrees Celsius (about 25 times hotter than the center of the Sun) and confined for tens of milliseconds by powerful magnetic fields \((100,000\) times greater than the Earth's magnetic field). For each experimental run, a huge amount of energy is required over a fraction of a second, which translates into a power requirement that would cause a blackout if electricity from the normal grid were to be used to power the experiment. Instead, kinetic energy is stored in a colossal flywheel, which is a spinning solid cylinder with a radius of \(3.00 \mathrm{~m}\) and mass of \(1.18 \cdot 10^{6} \mathrm{~kg}\). Electrical energy from the power grid starts the flywheel spinning, and it takes 10.0 min to reach an angular speed of \(1.95 \mathrm{rad} / \mathrm{s}\). Once the flywheel reaches this angular speed, all of its energy can be drawn off very quickly to power an experimental run. What is the mechanical energy stored in the flywheel when it spins at \(1.95 \mathrm{rad} / \mathrm{s}\) ? What is the average torque required to accelerate the flywheel from rest to \(1.95 \mathrm{rad} / \mathrm{s}\) in \(10.0 \mathrm{~min} ?\)

Short answer

Solution: Step 1: We first find the moment of inertia of the flywheel using the formula \(I = \frac{1}{2}MR^2\). Given mass \(M = 1.18 \cdot 10^6 \mathrm{~kg}\), and radius \(R = 3.00 \mathrm{~m}\), we calculate the moment of inertia: \(I = \frac{1}{2}(1.18 \cdot 10^6 \mathrm{~kg})(3.00 \mathrm{~m})^2\). Step 2: The mechanical energy stored in the flywheel is the rotational kinetic energy, which we calculate using the formula \(K = \frac{1}{2}I\omega^2\). Given final angular speed \(\omega = 1.95 \mathrm{rad/s}\), we calculate the mechanical energy stored: \(K = \frac{1}{2}I(1.95 \mathrm{rad/s})^2\). Step 3: To find the average torque, we first calculate the angular acceleration \(\alpha\) using the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\). Given change in angular speed \(\Delta \omega = 1.95 \mathrm{rad/s}\) (from rest) and time \(\Delta t = 10 \mathrm{~min} = 600 \mathrm{s}\), we calculate the angular acceleration: \(\alpha = \frac{1.95 \mathrm{rad/s}}{600 \mathrm{s}}\). Step 4: We can now find the average torque required to accelerate the flywheel using the formula \(\tau = I\alpha\), where \(\tau\) is the torque. Using the values found in the previous steps, we calculate the average torque: \(\tau = I\alpha\).

Step-by-step solution

Find the moment of inertia of the flywheel

Since the flywheel is a solid cylinder, its moment of inertia can be calculated using the formula \(I = \frac{1}{2}MR^2\), where \(M\) is the mass of the cylinder and \(R\) is its radius. In this case, \(M = 1.18 \cdot 10^6 \mathrm{~kg}\) and \(R = 3.00 \mathrm{~m}\). Calculate the moment of inertia as follows: \(I = \frac{1}{2}(1.18 \cdot 10^6 \mathrm{~kg})(3.00 \mathrm{~m})^2\).

Calculate the mechanical energy stored in the flywheel

The mechanical energy stored in the flywheel is the rotational kinetic energy, which can be found using the formula \(K = \frac{1}{2}I\omega^2\), where \(I\) is the moment of inertia calculated in step 1 and \(\omega\) is the final angular speed. The given angular speed is \(1.95 \mathrm{rad/s}\). Calculate the mechanical energy stored as follows: \(K = \frac{1}{2}I(1.95 \mathrm{rad/s})^2\).

Calculate the angular acceleration

To find the average torque, we first need to find the angular acceleration \(\alpha\). This can be determined using the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular speed and \(\Delta t\) is the time taken to accelerate. In this case, \(\Delta \omega = 1.95 \mathrm{rad/s}\) (from rest) and the given time \(\Delta t = 10 \mathrm{~min} = 600 \mathrm{s}\). Calculate the angular acceleration as follows: \(\alpha = \frac{1.95 \mathrm{rad/s}}{600 \mathrm{s}}\).

Calculate the average torque

Now we can find the average torque required to accelerate the flywheel using the formula \(\tau = I\alpha\), where \(\tau\) is the torque, \(I\) is the moment of inertia calculated in step 1, and \(\alpha\) is the angular acceleration calculated in step 3. Calculate the average torque as follows: \(\tau = I\alpha\) (use the values found in the previous steps).

Key concepts

Moment of Inertia

The moment of inertia is a fundamental concept in rotational dynamics, playing a role similar to that of mass in linear motion. It represents the resistance that a body displays to changes in its rotation.

For solid objects, the moment of inertia depends on the mass distribution relative to the axis of rotation. Different shapes have distinct moments of inertia formulas. In our example, a flywheel is essentially a solid cylinder, so its moment of inertia is computed as \(I = \frac{1}{2}MR^2\), with \(M\) being the mass and \(R\) the radius of the cylinder.

As the flywheel's mass or radius increases, its moment of inertia grows larger, meaning more torque would be needed to achieve the same angular acceleration. This fundamental property plays a crucial role in ensuring that the flywheel can store substantial amounts of rotational kinetic energy.

Angular Acceleration

Angular acceleration describes how quickly a rotational speed changes with time. It's analogous to linear acceleration but applies to rotational motion. The symbol \(\alpha\) represents angular acceleration, typically measured in radians per second squared.

In the case of the flywheel, we need to determine the rate at which it accelerates from rest to its final angular speed. The angular acceleration is found using the relationship \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time over which this change occurs.

In practice, angular acceleration is crucial because it indicates the rate at which a rotating object speeds up or slows down, which is directly related to the forces or torques applied to it.

Torque

Torque is the measure of how much a force acting on an object causes that object to rotate. Think of it as the rotational equivalent of force. The formula for average torque is given by \(\tau = I\alpha\), connecting it directly to moment of inertia \(I\) and angular acceleration \(\alpha\).

When calculating the average torque required for the flywheel's acceleration, we apply the values of its moment of inertia and the angular acceleration we've determined. The interplay between torque, moment of inertia, and angular acceleration encapsulates the dynamic aspects of rotational motion. For our spinning flywheel, the required torque dictates the energy input needed to reach its operating angular speed, reflecting the inherent connection between the concepts of torque, energy, and rotational motion.