Unit 6: Energy and Momentum of Rotating Systems

This unit explores the extension of energy and momentum principles to rotational systems. Students learn how torque, angular momentum, and rotational kinetic energy govern the behavior of rotating objects. The unit also applies these ideas to real-world systems such as rolling motion and orbiting satellites.

Rotational Kinetic Energy

Rotational kinetic energy represents the energy an object has due to its spinning motion. It is given by the formula \(K_{rot} = \tfrac{1}{2}I\omega^2\), where \(I\) is the moment of inertia and \(\omega\) is angular velocity. The moment of inertia depends not just on the object’s mass but on how that mass is distributed relative to the axis of rotation.
Objects that are rolling have both translational kinetic energy (\(\tfrac{1}{2}mv^2\)) and rotational kinetic energy. This means their total kinetic energy is the sum of the two forms. As a result, objects with larger moments of inertia roll more slowly down ramps because more of their energy is stored in rotation rather than translation.
This concept explains why a solid sphere, a hollow sphere, and a hoop roll at different speeds when released from the same height. Even though they have the same mass and radius, their different mass distributions produce different rotational kinetic energies, affecting how fast they accelerate.

Torque and Work

Torque (\(\tau\)) is the rotational equivalent of force and measures an object’s tendency to rotate. The work done by torque is given by \(W = \tau \theta\), where \(\theta\) is the angular displacement. This formula mirrors linear work, \(W = Fd\), showing that torque plays the same role in rotational motion that force does in linear motion.
Power in rotational systems can be expressed as \(P = \tau \omega\), which shows how quickly torque is doing work to change an object’s angular speed. This relationship is important in real systems such as engines, where torque and angular velocity together determine mechanical output.
The direction and effectiveness of torque depend on both the magnitude of the force and the perpendicular distance (lever arm) from the axis of rotation. This explains why using a longer wrench makes it easier to loosen a tight bolt — the same force applied at a longer lever arm produces more torque.
Torque can also be expressed using vector notation as \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{r} \) is the position vector from the axis of rotation to the point of force application. The cross product means torque depends on both the magnitude of the force and the angle between the force and the lever arm. Maximum torque occurs when the force is applied perpendicular to the lever arm.
In real-world applications, torque often works against resistive forces like friction, meaning not all the work done is converted into rotational kinetic energy. For example, in a car engine, part of the torque produced is lost as heat due to friction between moving parts. This distinction is important for understanding efficiency in mechanical systems.
When analyzing work done by torque, it’s also useful to think about angular displacement in radians. Because one radian corresponds to an arc length equal to the radius, using radians ensures that \(W = \tau \theta\) is dimensionally consistent with the linear work equation \(W = Fd\). This highlights why radians are the standard unit in rotational motion.

Angular Momentum and Angular Impulse

Angular momentum (\(L\)) is the rotational counterpart of linear momentum. It is defined as \(L = I\omega\) for rigid bodies or more generally as \(L = r \times p\). The quantity depends on both the distribution of mass and the angular velocity of the object.
The angular impulse-momentum theorem states that \(\tau \Delta t = \Delta L\). This means that the product of torque and the time it acts produces a change in angular momentum, just as linear impulse (\(F \Delta t\)) changes linear momentum. The longer or stronger a torque acts, the greater the effect on rotational motion.
Angular impulse and angular momentum are critical in activities like hitting a baseball or applying brakes on a rotating wheel. A short, powerful torque can cause a large change in angular momentum, while a small torque acting over a long time can achieve the same effect.

Conservation of Angular Momentum

The law of conservation of angular momentum states that if no external torque acts on a system, the system’s angular momentum remains constant. This is similar to the conservation of linear momentum in the absence of external forces. It is a fundamental principle that applies to many physical systems.
One common example is an ice skater pulling in her arms to spin faster. By reducing her moment of inertia, she increases her angular velocity so that \(I\omega\) remains constant. This demonstrates how changes in mass distribution affect angular velocity when angular momentum is conserved.
Another example is the collapse of a star into a neutron star. As the radius decreases dramatically, the moment of inertia becomes much smaller, causing the rotational speed to increase drastically. This principle explains why many astronomical objects spin so rapidly after collapsing.

Rolling Motion

Rolling without slipping occurs when an object’s linear velocity and angular velocity are linked by the condition \(v = r\omega\). This means the object’s point of contact with the surface is momentarily at rest, allowing smooth rolling rather than sliding. This condition is crucial for analyzing rolling motion problems.
The total kinetic energy of a rolling object is the sum of translational and rotational components. For example, a rolling solid sphere has more of its energy in translational motion compared to a hoop of the same mass and radius, so it will reach the bottom of a ramp faster.
Mass distribution plays a key role: objects with more mass concentrated farther from the axis of rotation (larger \(I\)) roll more slowly. This explains why a hoop accelerates more slowly down an incline compared to a disk or sphere, even if their sizes are identical.
Friction plays a critical role in rolling without slipping. Contrary to common belief, static friction (not kinetic friction) is what enables rolling, since the point of contact does not slide. Without static friction, an object would slide instead of roll, losing the energy distribution that defines rolling motion.
Real rolling systems rarely roll without some energy loss. Energy is often lost to rolling resistance, which comes from deformation of the object or the surface. For example, a basketball eventually stops rolling on the floor because some energy is dissipated as heat in the ball’s rubber and the floor surface.
The acceleration of rolling objects down a ramp depends not only on gravitational force but also on their moment of inertia. This is why a solid sphere will beat a hollow sphere in a race down the same incline — the solid sphere has a smaller moment of inertia relative to its mass, leaving more energy available for translation.
In rolling without slipping, the condition \( v = r\omega \) connects the object’s linear velocity to its angular velocity. This means the speed of the object’s center of mass is directly tied to how fast it rotates. If the wheel spins faster, its center of mass must also move faster to maintain the no-slip condition.
This relationship can be visualized by looking at a single point on the rim of the wheel. At the top of the wheel, the point moves at \( v + r\omega \), while at the bottom, the point is instantaneously at rest relative to the surface. This explains why the bottom of a rolling wheel does not slide even though the wheel itself is moving forward.
When analyzing energy, the \( v = r\omega \) condition ensures that the distribution between translational and rotational kinetic energy is consistent. For example, a rolling sphere’s total kinetic energy is split between its forward velocity (\(\frac{1}{2}mv^2\)) and its rotational speed (\(\frac{1}{2}I\omega^2\)). Without linking velocity to angular velocity, solving rolling problems correctly would not be possible.

Motion of Orbiting Satellites

Satellites in orbit provide a real-world example of angular momentum conservation. Since there is negligible external torque acting on them, their angular momentum remains constant. This principle allows satellites to maintain stable orbits around Earth.
Orbital motion is governed by the balance between gravitational force and centripetal force. The required orbital velocity is derived from setting \(F_g = F_c\): \(\frac{GMm}{r^2} = \frac{mv^2}{r}\), which simplifies to \(v = \sqrt{\frac{GM}{r}}\). This velocity ensures the satellite continuously “falls around” Earth without colliding into it.
Kepler’s laws also connect to this concept: as satellites move closer to Earth, they orbit faster due to conservation of angular momentum. This explains why the International Space Station moves quickly at low Earth orbit while distant satellites orbit more slowly.
Satellites have both kinetic and gravitational potential energy. The total mechanical energy in a stable circular orbit is negative, given by \( E = -\frac{GMm}{2r} \). This negative energy indicates the satellite is bound to the planet. Understanding this balance explains why satellites cannot escape orbit without additional energy.
The speed of a satellite in orbit depends on its distance from the center of the Earth. Using \( v = \sqrt{\frac{GM}{r}} \), we see that closer satellites must move faster to balance the stronger gravitational pull. This explains why the International Space Station orbits Earth in about 90 minutes while geostationary satellites take 24 hours.
Kepler’s second law — that a line from the planet to the sun sweeps out equal areas in equal times — is a direct consequence of angular momentum conservation. Even though AP Physics 1 does not require a full derivation, understanding this connection helps explain why planets move faster when closer to the sun in their elliptical orbits.

Work and Rotational Kinetic Energy

Work in rotational systems is given by \(W = \tau\theta\), where torque plays the same role as force in linear systems. This formula shows that when torque acts through an angular displacement, energy is transferred to the system.
Rotational kinetic energy is expressed as \(K_{rot} = \tfrac{1}{2}I\omega^2\). Just like translational kinetic energy depends on mass and velocity, rotational kinetic energy depends on moment of inertia and angular velocity.
Rolling objects often possess both translational and rotational kinetic energy, so their total kinetic energy is the sum of both. Recognizing this is key when applying energy conservation in rolling motion problems.

Power in Rotational Systems

Power in rotational systems is given by \(P = \tau\omega\), showing how quickly torque is doing work. This is analogous to \(P = Fv\) in linear systems, linking torque with angular velocity instead of force with linear velocity.
Rotational power is crucial in systems such as engines and turbines, where both torque and angular speed determine performance. A high-torque, low-speed machine may produce the same power as a low-torque, high-speed one.
This relationship helps engineers design machines for specific purposes, ensuring they provide enough torque and speed to accomplish a task efficiently.

Angular Momentum

Angular momentum is defined as \(L = I\omega\) for rigid bodies and as \(L = r \times p\) in general. It measures how much rotational motion an object has, combining both its mass distribution and angular velocity.
Angular momentum plays the same role in rotational systems that linear momentum plays in linear systems. A larger moment of inertia or faster angular velocity increases angular momentum, making it harder to change an object’s spin.
Real-world applications include ice skaters pulling in their arms to spin faster or planets orbiting the sun with nearly constant angular momentum due to negligible external torque.

Impulse and Angular Impulse

The angular impulse-momentum theorem states that \(\tau\Delta t = \Delta L\). This means the product of torque and the time it acts changes an object’s angular momentum, just like linear impulse changes linear momentum.
This principle explains why applying a small torque for a long time can produce the same change as applying a large torque briefly. It helps in analyzing systems like brakes, sports swings, or rotating machinery.
For example, in baseball, the force of the bat acting over a short contact time creates a large angular impulse, changing the ball’s angular momentum dramatically.

Rolling Motion and Energy Conservation

Rolling without slipping requires the condition \(v = r\omega\), ensuring the point of contact with the ground is momentarily at rest. This prevents energy loss from sliding friction and allows energy conservation to be applied cleanly.
The total kinetic energy of a rolling object includes both translational and rotational components. Objects with smaller moments of inertia (like a solid sphere) accelerate faster down ramps than objects with larger ones (like a hoop).
Understanding how mass distribution affects rolling speed helps explain everyday phenomena such as why a bicycle wheel maintains speed or why different sports balls roll differently.

Collisions Involving Rotation

In collisions involving rotating bodies, angular momentum is conserved if there is no net external torque. This principle applies whether the collisions are elastic or inelastic.
Elastic rotational collisions conserve both angular momentum and rotational kinetic energy, while inelastic ones conserve only angular momentum. This distinction is crucial for analyzing outcomes.
Examples include billiard balls spinning after impact or figure skaters colliding during a spin, both of which require careful application of conservation laws.

Torque and Center of Mass Review

Torque depends on both the force applied and the lever arm (distance from the axis). The greater the lever arm, the more torque the same force produces, making it easier to rotate objects.
Center of mass plays a critical role in rotational stability. An object remains balanced when its center of mass is directly above its base of support, but it topples if displaced beyond it.
This principle explains why tightrope walkers use poles to lower their center of mass and why vehicles are more stable with a wider wheelbase.

Example Problem 1: Rotational Dynamics of a Rolling Cylinder

Context: A solid cylinder of mass \( M = 3.0\, \text{kg} \) and radius \( R = 0.40\, \text{m} \) is released from rest at the top of a \( 2.5\, \text{m} \)-long ramp inclined at \( 30^\circ \). The cylinder rolls down the ramp without slipping.

(a) Calculate the moment of inertia of the cylinder about its central axis.
Use the formula for a solid cylinder: \( I = \frac{1}{2}MR^2 \)
\( I = \frac{1}{2}(3.0)(0.40)^2 = \frac{1}{2}(3.0)(0.16) = 0.24\, \text{kg} \cdot \text{m}^2 \)
(b) Determine the final speed of the cylinder at the bottom of the ramp.
Use conservation of energy: \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \), and \( \omega = \frac{v}{R} \)
Substituting: \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)\frac{v^2}{R^2} = \frac{3}{4}mv^2 \)
Solve for \( v \): \( v = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4(9.8)(2.5 \sin 30^\circ)}{3}} = \sqrt{\frac{49}{3}} ≈ 4.04\, \text{m/s} \)
(c) What is the angular velocity of the cylinder at the bottom?
\( \omega = \frac{v}{R} = \frac{4.04}{0.40} = 10.1\, \text{rad/s} \)

Example Problem 2: Angular Momentum and Conservation

Context: A child of mass \( m = 40\, \text{kg} \) stands on the edge of a frictionless merry-go-round of radius \( R = 2.0\, \text{m} \), initially at rest. The merry-go-round has a moment of inertia \( I = 200\, \text{kg} \cdot \text{m}^2 \). The child begins to walk tangentially at \( v = 1.5\, \text{m/s} \) relative to the ground.

(a) What is the angular velocity of the merry-go-round after the child begins walking?
Apply conservation of angular momentum: \( L_i = 0 \Rightarrow L_f = 0 \)
Child: \( L = mRv \), MGR: \( L = -I\omega \)
\( mRv = I\omega \Rightarrow \omega = \frac{mRv}{I} = \frac{(40)(2.0)(1.5)}{200} = 0.60\, \text{rad/s} \)
(b) What is the total kinetic energy of the system?
Child’s KE: \( \frac{1}{2}mv^2 = \frac{1}{2}(40)(1.5)^2 = 45\, \text{J} \)
Merry-go-round’s KE: \( \frac{1}{2}I\omega^2 = \frac{1}{2}(200)(0.60)^2 = 36\, \text{J} \)
Total: \( 45 + 36 = 81\, \text{J} \)

Example Problem 3: Rotational Collision

Context: A disk of mass \( M = 2.0\, \text{kg} \) and radius \( R = 0.30\, \text{m} \) is spinning freely at \( \omega_i = 10\, \text{rad/s} \) on a frictionless axle. A small block of mass \( m = 0.50\, \text{kg} \) is gently dropped onto the edge of the disk and sticks to it.

(a) What is the initial angular momentum of the system?
Disk’s moment of inertia: \( I = \frac{1}{2}MR^2 = 0.09\, \text{kg} \cdot \text{m}^2 \)
Initial angular momentum: \( L_i = I\omega = (0.09)(10) = 0.9\, \text{kg} \cdot \text{m}^2/\text{s} \)
(b) What is the final angular velocity of the system?
Final I: \( I_f = 0.09 + mR^2 = 0.09 + (0.5)(0.09) = 0.135\, \text{kg} \cdot \text{m}^2 \)
\( \omega_f = \frac{L_i}{I_f} = \frac{0.9}{0.135} = 6.67\, \text{rad/s} \)
(c) How much rotational kinetic energy is lost in the collision?
Initial KE: \( \frac{1}{2}(0.09)(10)^2 = 4.5\, \text{J} \)
Final KE: \( \frac{1}{2}(0.135)(6.67)^2 ≈ 3.0\, \text{J} \)
Loss: \( 4.5 - 3.0 = 1.5\, \text{J} \), due to inelastic sticking

Common Misconceptions in Energy and Momentum of Rotating Systems

Many students believe that torque and force are the same thing. While both are related, torque depends not only on the magnitude of the force but also on the lever arm (the perpendicular distance from the axis of rotation). A small force applied far from the axis can create the same torque as a large force applied close to the axis.
It is a common mistake to think that heavier rolling objects always roll faster down a ramp. In reality, the speed depends on the object’s moment of inertia, which is determined by how its mass is distributed. A lighter object with a smaller moment of inertia can sometimes roll faster than a heavier one.
Students often forget that rolling objects possess both translational and rotational kinetic energy. Leaving out one of these energy components when applying conservation of energy leads to incorrect results. For example, treating a rolling ball as if it only has translational energy will underestimate its total energy.
Another misconception is that angular momentum only depends on angular velocity. In fact, it also depends on the object’s moment of inertia, meaning that redistributing mass (such as an ice skater pulling in their arms) can change angular velocity while keeping angular momentum constant.
In collision problems, many students assume kinetic energy is always conserved. However, in rotational collisions, angular momentum is conserved, but kinetic energy may not be, especially if the collision is inelastic. Forgetting this distinction often leads to errors in solving rotational collision problems.