Problem 39 - College Physics

Question

It has been suggested that we should use our power plants to generate energy in the off-hours (such as late at night) and store it for use during the day. One idea put forward is to store the energy in large flywheels. Suppose we want to build such a flywheel in the shape of a hollow cylinder of inner radius \(0.500 \mathrm{~m}\) and outer radius \(1.50 \mathrm{~m},\) using concrete of density \(2.20 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (a) If, for stability, such a heavy flywheel is limited to 1.75 seconds for each revolution and has negligible friction at its axle, what must be its length to store \(2.5 \mathrm{MJ}\) of energy in its rotational motion? (b) Suppose that by strengthening the frame you could safely double the flywheel's rate of spin. What length of flywheel would you need in that case? (Solve this part without reworking the entire problem!)

Step-by-Step Solution

Answer

Length is 3.95 m for part (a); for part (b), it is 0.9875 m.

Step 1: Understand the Problem

We need to find the length of a hollow cylindrical flywheel that stores 2.5 MJ of energy, given its rotational speed and density of concrete. The cylinder's energy storage is related to its moment of inertia and angular speed.

Step 2: Determine Moment of Inertia for the Hollow Cylinder

For a hollow cylindrical flywheel, the moment of inertia \( I \) is given by \( I = \frac{1}{2}m(R_1^2 + R_2^2) \), where \( m \) is the mass, \( R_1 \) is the inner radius and \( R_2 \) is the outer radius. Substitute \( R_1 = 0.500 \) m and \( R_2 = 1.50 \) m.

Step 3: Calculate Mass of the Flywheel

The mass \( m \) can be calculated using the volume and density. The volume \( V \) of the cylinder is \( V = \pi L (R_2^2 - R_1^2) \). Therefore, the mass \( m = \rho V = \pi \rho L (R_2^2 - R_1^2) \). Substitute \( \rho = 2.20 \times 10^3 \) kg/m\(^3\).

Step 4: Calculate Angular Speed

The angular speed \( \omega \) can be found from the given revolution time: \( \omega = \frac{2\pi}{T} \), where \( T = 1.75 \) seconds.

Step 5: Relate Rotational Energy to Moment of Inertia and Angular Speed

The rotational kinetic energy is given by \( KE = \frac{1}{2} I \omega^2 \). Set this equal to 2.5MJ \((2.5 \times 10^6\) Joules) and solve for \( L \) using \( I = \frac{1}{2}m(R_1^2 + R_2^2) \). Substitute \( I \) and \( m \) in the kinetic energy formula.

Step 6: Solve for Length in Part (a)

Calculate \( m \) using \( L \), then express \( I \) in terms of \( L \), and equate to 2.5MJ. Rearrange the equation to solve for \( L \).

Step 7: Adjust Angular Speed for Part (b)

If the rotational speed is doubled, then \( \omega' = 2\omega \). The kinetic energy increases by a factor of 4 (because energy \( KE = \frac{1}{2} I \omega^2 \)), meaning 4 times the rotational energy can be stored. Thus, solve the equation from Step 5 for \( L \) considering \( KE' = 4 \times 2.5 \times 10^6 \).

Key Concepts

Rotational Motion

Rotational motion is an intrinsic part of understanding how objects spin around an axis. Imagine a spinning top or the Earth rotating on its axis. This type of motion is characterized by a rotational or angular velocity, which describes how fast the object spins.

Just like linear motion, rotational motion can be described using different quantities such as angular displacement, angular velocity, and angular acceleration.

• Angular Displacement: It refers to the angle through which an object has rotated or moved in a circular path.
• Angular Velocity ( \(\omega\)): This is the rate of change of angular displacement, i.e., how fast an object is spinning or rotating.
• Angular Acceleration: It indicates how quickly the angular velocity is changing.

These concepts are useful when dealing with objects like flywheels, which can store mechanical energy through rotation.

Moment of Inertia

Moment of inertia is a crucial concept in rotational motion as it plays a role similar to mass in linear motion. It represents how difficult it is to change the rotational motion of an object.

In mathematical terms, moment of inertia for a rigid body is defined as: \[ I = \int r^2 \, dm \] where \(r\) is the distance from the axis of rotation to the mass element \(dm\).

• For a hollow cylindrical flywheel: The formula modifies to \( I = \frac{1}{2}m(R_1^2 + R_2^2) \), where \(R_1\) and \(R_2\) are the inner and outer radii respectively.
• It depends not only on the mass of the object but also on the shape and distribution of that mass around the rotational axis.

Understanding moment of inertia helps in predicting how a flywheel will behave under different physical conditions.

Angular Speed

Angular speed is a vital metric for describing how quickly an object is rotating. Much like speed in linear motion, angular speed is the scalar component of angular velocity.

It's computed using the relationship: \[ \omega = \frac{2\pi}{T} \] where \( T \) is the time taken for one complete revolution.

- In our flywheel example, knowing the rotational period (1.75 seconds per revolution) lets us calculate this speed. - This calculated speed informs us of how fast the energy stored in the flywheel is rotating.
Changing the angular speed, like doubling it, significantly alters the dynamics of energy storage, as the kinetic energy stored increases with the square of the angular speed.

Kinetic Energy

Kinetic energy in rotational motion is conceptually similar to that in linear motion. However, instead of relying on mass and velocity, it depends on the moment of inertia and angular speed.
The formula for rotational kinetic energy is given by: \[ KE = \frac{1}{2} I \omega^2 \] where:

• \(I\) is the moment of inertia,
• \(\omega\) is the angular speed.

- In our exercise, this equation helps us relate the flywheel's physical dimensions to the energy it can store.
Understanding this helps in practical applications such as energy storage, where optimizing the design for maximum energy storage with minimal size and material is desired.

Energy Storage

Energy storage in the context of rotational motion involves converting kinetic energy into a form that can be harnessed later. Flywheels are great examples of this concept. A flywheel stores energy by accelerating its rotation, holding on to kinetic energy until it is needed for use.

- This mechanism is efficient because it can absorb and release energy quickly with minimal loss.
- The stored energy is proportional to the moment of inertia and the square of the angular speed \((I\omega^2)\). - When the speed is doubled, theoretically, the energy capacity increases fourfold, demonstrating how effective enhancing rotation could be for energy storage solutions.These insights are pivotal when designing systems that use rotational motion to store energy, like the suggested flywheel in our exercise. By managing parameters like rotational speed and structural properties, energy storage can be optimized efficiently.