Rotational kinetic energy Revision Notes for AQA A-Level Physics

Rotational kinetic energy

Introduction to rotational energy

When an object rotates, it possesses kinetic energy due to its rotational motion. This rotational kinetic energy is a form of energy that depends on how fast the object is rotating and how its mass is distributed around the axis of rotation.

To understand rotational kinetic energy, we begin by examining the simplest case: a single point mass moving in a circular path. Consider a point mass $m_1$ located at a distance $r_1$ from a fixed point O, moving in a circle with linear speed $v_1$ . This particle has kinetic energy that can be calculated using the standard formula for linear kinetic energy.

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Understanding rotational kinetic energy starts with analyzing the motion of a single particle in circular motion, which we can then extend to entire rotating bodies.

Deriving the rotational kinetic energy formula

For our point mass moving in circular motion, the kinetic energy is initially expressed as:

$E_k = \frac{1}{2}mv_1^2$

However, we know that the linear velocity of a particle moving in a circle is related to the angular velocity through the equation $v = r\omega$ , where $\omega$ (omega) is the angular velocity in radians per second. The linear velocity $v_1$ of our point mass can therefore be written as $v_1 = r_1\omega$ .

Substituting this relationship into the kinetic energy equation gives:

$E_k = \frac{1}{2}m_1(r_1\omega)^2$

This can be rearranged by expanding the squared term:

$E_k = \frac{1}{2}m_1r_1^2\omega^2$

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This expression reveals something important. When we compare it to the standard kinetic energy formula for linear motion, we notice that angular velocity $\omega$ has replaced linear velocity $v$ . Additionally, the term $m_1r_1^2$ has effectively replaced the mass $m$ . This quantity $m_1r_1^2$ is actually the moment of inertia of the point mass about the axis of rotation, a concept that appears throughout rotational dynamics.

Rotational kinetic energy for extended bodies

Real objects are not point masses but extended bodies comprising many particles. Consider an extended object rotating about an axis through point O. This object can be thought of as consisting of many point masses $m_1$ , $m_2$ , $m_3$ , and so on, each at perpendicular distances $r_1$ , $r_2$ , $r_3$ , etc., from the axis of rotation.

Each of these point masses contributes to the total rotational kinetic energy. The particle at distance $r_1$ with mass $m_1$ contributes $\frac{1}{2}m_1r_1^2\omega^2$ , the particle at distance $r_2$ with mass $m_2$ contributes $\frac{1}{2}m_2r_2^2\omega^2$ , and so forth. The total rotational kinetic energy is therefore the sum of all these individual contributions:

$E_k = \frac{1}{2}m_1r_1^2\omega^2 + \frac{1}{2}m_2r_2^2\omega^2 + \frac{1}{2}m_3r_3^2\omega^2 + ...$

Since all parts of a rigid body rotate with the same angular velocity $\omega$ , we can factor this out:

$E_k = \frac{1}{2}\omega^2(m_1r_1^2 + m_2r_2^2 + m_3r_3^2 + ...)$

The term in brackets $(m_1r_1^2 + m_2r_2^2 + m_3r_3^2 + ...)$ is the moment of inertia $I$ of the entire body about the axis of rotation. This allows us to express the rotational kinetic energy in its most useful form:

$E_k = \frac{1}{2}I\omega^2$

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This formula is the rotational equivalent of the linear kinetic energy formula $E_k = \frac{1}{2}mv^2$ , with moment of inertia $I$ playing the role of mass, and angular velocity $\omega$ playing the role of linear velocity.

Flywheels and energy storage

A flywheel is a mechanical device designed to store rotational kinetic energy. By examining the formula $E_k = \frac{1}{2}I\omega^2$ , we can identify how to maximize the energy storage capacity of a flywheel. The rotational kinetic energy depends on two key factors:

Angular velocity ( $\omega$ ): The rotational kinetic energy increases with the square of the angular velocity. Doubling the angular velocity quadruples the stored energy.
Moment of inertia ( $I$ ): A larger moment of inertia means more energy can be stored at a given angular velocity.

To maximize energy storage, both $I$ and $\omega$ must be as large as possible. A large moment of inertia is achieved by positioning as much mass as possible far from the axis of rotation.

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There are practical limitations to flywheel design. As the flywheel spins faster, the structural integrity of the device becomes critical. The material must be strong enough to withstand the forces generated at high rotational speeds without breaking apart.

The relationship between rotational kinetic energy and angular velocity squared has important implications. Just as linear kinetic energy depends on $v^2$ , the energy required to decelerate and stop a rotating object increases dramatically with angular velocity. This has consequences for braking distances and the forces needed to stop rotating machinery.

Combined linear and rotational motion

Many objects in real-world situations undergo both translational (linear) and rotational motion simultaneously. A typical example is a ball rolling along a surface. When an object is rolling rather than sliding, it possesses two forms of kinetic energy:

Linear kinetic energy: due to the motion of the object's center of mass ( $E_k = \frac{1}{2}mv^2$ )
Rotational kinetic energy: due to the rotation about the center of mass ( $E_k = \frac{1}{2}I\omega^2$ )

The total kinetic energy of a rolling object is the sum of these two components. This is an important concept because it affects how energy is distributed in rolling motion and influences phenomena such as which object will roll down a slope faster.

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For objects that roll without slipping, both forms of kinetic energy contribute to the total energy, and neither can be neglected when calculating the object's total energy.

Understanding rolling motion

When an object rolls without slipping, there is a specific relationship between its linear and angular motion. The linear velocity $v$ of the center of mass is related to the angular velocity $\omega$ by $v = r\omega$ , where $r$ is the radius of the rolling object.

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At the point of contact with the surface (the bottom of a rolling disc, for instance), the tangential velocity due to rotation must be equal and opposite to the linear velocity. If this condition were not met, the object would be slipping rather than rolling. This means that the tangential velocity at the contact point is zero relative to the surface, while the top of the object moves at twice the linear velocity of the center of mass.

Worked example: disc and hoop comparison

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Worked Example: Comparing Energy Storage in Different Shapes

Consider a disc that is spun to 100 rpm initially and then later to 300 rpm. The disc has a radius of 50 cm and a mass of 10 kg.

Step 1: Calculate the moment of inertia of the disc

For a disc rotating about its center, $I = \frac{1}{2}mr^2$ . Substituting the values:

$I = \frac{1}{2} \times 10 \text{ kg} \times (0.5 \text{ m})^2 = 1.25 \text{ kg} \cdot \text{m}^2$

Step 2: Convert angular velocities to rad/s

At 100 rpm: $\omega = 100 \times \frac{2\pi}{60} = 10.5 \text{ rad} \cdot \text{s}^{-1}$
At 300 rpm: $\omega = 300 \times \frac{2\pi}{60} = 31.4 \text{ rad} \cdot \text{s}^{-1}$

Step 3: Calculate rotational kinetic energy

Using $E_k = \frac{1}{2}I\omega^2$ :

At 100 rpm: $E_k = \frac{1}{2} \times 1.25 \times (10.5)^2 = 68.9 \text{ J}$
At 300 rpm: $E_k = \frac{1}{2} \times 1.25 \times (31.4)^2 = 616 \text{ J}$

This demonstrates the dramatic effect of angular velocity on kinetic energy. Tripling the angular velocity increases the kinetic energy by a factor of nine.

Step 4: Compare with a hoop of the same mass and radius

For a hoop, $I = mr^2$ , so:

$I = 10 \text{ kg} \times (0.5 \text{ m})^2 = 2.5 \text{ kg} \cdot \text{m}^2$

The rotational kinetic energies become:

At 100 rpm: $E_k = \frac{1}{2} \times 2.5 \times (10.5)^2 = 138 \text{ J}$
At 300 rpm: $E_k = \frac{1}{2} \times 2.5 \times (31.4)^2 = 1230 \text{ J}$

The hoop stores more energy than the disc at the same angular velocity because its moment of inertia is larger. This confirms that maximum energy storage requires both the largest possible angular velocity and the largest possible moment of inertia.

Worked example: rolling disc with combined energy

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Worked Example: Total Kinetic Energy of a Rolling Disc

A uniform disc has a mass of 500 g and a diameter of 30.0 cm. It rolls on its edge along a horizontal table, traveling 45.0 cm in 1 second. We need to find the total kinetic energy.

Since the disc is both rotating and translating, its total energy includes both forms of kinetic energy.

Step 1: Calculate linear kinetic energy

The linear velocity is:

$v = \frac{\text{distance}}{\text{time}} = \frac{0.450 \text{ m}}{1 \text{ s}} = 0.450 \text{ m} \cdot \text{s}^{-1}$

Linear kinetic energy:

$E_k(\text{linear}) = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \text{ kg} \times (0.450 \text{ m} \cdot \text{s}^{-1})^2 = 0.0506 \text{ J}$

Step 2: Calculate angular velocity

For a rolling disc without slipping, the relationship between linear and angular velocity is $\omega = \frac{v}{r}$ . The radius of the disc is 0.150 m (half the diameter), so:

$\omega = \frac{0.450 \text{ m} \cdot \text{s}^{-1}}{0.150 \text{ m}} = 3.00 \text{ rad} \cdot \text{s}^{-1}$

Step 3: Calculate moment of inertia

For a disc rotating about its center:

$I = \frac{1}{2}mr^2 = \frac{1}{2} \times 0.500 \text{ kg} \times (0.150 \text{ m})^2 = 0.0056 \text{ kg} \cdot \text{m}^2$

Step 4: Calculate rotational kinetic energy

$E_k(\text{rotational}) = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.0056 \text{ kg} \cdot \text{m}^2 \times (3.00 \text{ rad} \cdot \text{s}^{-1})^2 = 0.0252 \text{ J}$

Step 5: Calculate total energy

\text{Total energy} &=& E_k(\text{linear}) + E_k(\text{rotational}) \\ &=& 0.0506 \text{ J} + 0.0252 \text{ J} \\ &=& 0.0758 \text{ J} \end{array}$$ :success[This example illustrates that for a rolling disc, the total kinetic energy is distributed between translational and rotational motion.] The disc has approximately twice as much linear kinetic energy as rotational kinetic energy, which is a characteristic of a disc rolling without slipping.

Summary

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Key Points to Remember:

Rotational kinetic energy is given by $E_k = \frac{1}{2}I\omega^2$ , where $I$ is the moment of inertia and $\omega$ is the angular velocity.
The rotational kinetic energy formula is analogous to linear kinetic energy ( $\frac{1}{2}mv^2$ ), with moment of inertia replacing mass and angular velocity replacing linear velocity.
Flywheels store rotational kinetic energy by having a large moment of inertia (mass distributed far from the axis) and spinning at high angular velocities.
Objects that roll have both linear kinetic energy ( $\frac{1}{2}mv^2$ ) and rotational kinetic energy ( $\frac{1}{2}I\omega^2$ ), and the total energy is the sum of both.
Rotational kinetic energy increases with the square of angular velocity, meaning doubling the rotation speed quadruples the kinetic energy.

Rotational kinetic energy (AQA A-Level Physics): Revision Notes