Energy and Momentum of Rotating Systems
Unit 6 of AP Physics 1, worth 5–8% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Rotational kinetic energy, torque and work, angular momentum and impulse, conservation of angular momentum, rolling, orbits and satellites.
How this unit is tested
Start every rotation problem by deciding whether it is an energy question, a momentum question, or both. Energy questions ask for a speed or height and are solved with $W=\Delta K$ or conservation of mechanical energy, where rotating objects carry rotational KE $\frac12 I\omega^2$ in addition to (or instead of) translational KE. Momentum questions ask what happens over a time interval, or what stays fixed when something rearranges its mass distribution (a skater, a collapsing star, a collision on a turntable) — these are solved with the angular impulse-momentum theorem or conservation of angular momentum.
For rolling problems, always write down the rolling constraint $v_{cm}=r\omega$ first: it is what lets you replace $\omega$ with $v_{cm}/r$ and turn two unknowns into one. Remember that total kinetic energy of a rolling object is translational plus rotational, and that the moment of inertia (how the mass is distributed relative to the axis) determines what fraction of the energy goes into each — this is why objects with different shapes race down a ramp at different speeds even though they all conserve energy.
For orbit problems, recognize that gravity is the centripetal force: setting $\frac{GMm}{r^2}=\frac{mv^2}{r}$ lets you solve for orbital speed, and combining that with $v=\frac{2\pi r}{T}$ gives Kepler's third law, $T^2\propto r^3$. Because gravity is always directed along the radius, it exerts zero torque about the central body, so angular momentum is conserved throughout an orbit — this is the physical reason behind Kepler's second law (equal areas in equal times) and explains why an object speeds up near closest approach.
Whenever a problem changes shape, radius, or mass distribution partway through, ask which quantity is actually conserved (usually angular momentum, not angular velocity and not kinetic energy) before writing any equation.
What you have to know
Rotational kinetic energy
$K_{rot}=\frac12 I\omega^2$, where $I$ is the moment of inertia about the rotation axis and $\omega$ is angular speed.
Work-energy theorem for rotation
$W_{net}=\tau\,\Delta\theta=\Delta K_{rot}$ for a constant net torque acting through angular displacement $\Delta\theta$.
Angular momentum
For a rigid body about a fixed axis, $L=I\omega$. For a particle, $L=mvr\sin\theta$, where $\theta$ is the angle between the velocity and the position vector from the reference point.
Angular impulse-momentum theorem
$\tau_{net}\,\Delta t=\Delta L$. When the net external torque on a system is zero, angular momentum is conserved: $L_i=L_f$.
Rolling without slipping
The constraint $v_{cm}=r\omega$ links translational and rotational motion; total kinetic energy is $K=\frac12 m v_{cm}^2+\frac12 I\omega^2$.
Circular orbit condition and Kepler's third law
Gravity supplies the centripetal force: $\frac{GMm}{r^2}=\frac{mv^2}{r}$, giving orbital speed $v=\sqrt{GM/r}$. This leads to $T^2\propto r^3$ for objects orbiting the same central mass.
14 practice questions
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A rotating wheel has moment of inertia $I=4\,kg\cdot m^2$ and angular speed $\omega=3\,rad/s$. What is its rotational kinetic energy?
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Answer. 18 J
$K_{rot}=\frac12I\omega^2=\frac12(4)(3)^2=\frac12(4)(9)=18\,J$. -
A constant torque of 5 N·m rotates a wheel through one full revolution (2π rad) starting from rest. How much work is done on the wheel by this torque?
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Answer. About 31.4 J
$W=\tau\Delta\theta=5(2\pi)\approx31.4\,J$. By the work-energy theorem, this equals the wheel's gain in rotational kinetic energy. -
A 2 kg particle moves in a circle of radius 0.5 m at a constant speed of 3 m/s. What is the magnitude of its angular momentum about the center of the circle?
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Answer. 3 kg·m²/s
In circular motion the velocity is always perpendicular to the radius, so $\theta=90°$ and $L=mvr\sin\theta=mvr=(2)(3)(0.5)=3\,kg\cdot m^2/s$. -
A net torque of 6 N·m acts for 4 s on a wheel initially at rest, with moment of inertia 8 kg·m². What is the wheel's angular speed at the end of 4 s?
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Answer. 3 rad/s
Angular impulse: $\Delta L=\tau\Delta t=(6)(4)=24\,kg\cdot m^2/s$. Since $\Delta L=I\Delta\omega$ and $\omega_i=0$, $\omega_f=24/8=3\,rad/s$. -
A figure skater spinning at 2 rad/s with arms extended (I = 5 kg·m²) pulls her arms in, reducing her moment of inertia to 2 kg·m². Assuming no external torque, what is her new angular speed?
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Answer. 5 rad/s
Angular momentum is conserved: $I_i\omega_i=I_f\omega_f\Rightarrow(5)(2)=(2)\omega_f\Rightarrow\omega_f=5\,rad/s$. -
In the previous scenario, what happens to the skater's rotational kinetic energy as she pulls her arms in?
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Answer. It increases, because she does positive work pulling her arms inward against the outward tendency of her arms.
Since $K_{rot}=L^2/(2I)$ and $L$ is fixed while $I$ decreases, $K_{rot}$ must increase. The extra energy comes from the muscular work the skater does, not from an external torque. -
A uniform hoop (I = MR²) is released from rest at the top of a ramp of height 2.0 m and rolls without slipping to the bottom. Find its speed at the bottom.
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Answer. About 4.4 m/s
Energy conservation: $Mgh=\frac12Mv^2+\frac12(MR^2)(v/R)^2=\frac12Mv^2+\frac12Mv^2=Mv^2$. So $v=\sqrt{gh}=\sqrt{(9.8)(2.0)}\approx4.43\,m/s$. -
Two identical-looking spheres of the same mass and radius are released simultaneously from rest at the top of an incline: one slides down a frictionless track, the other rolls without slipping down an identical incline. Which reaches the bottom with the greater speed?
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Answer. The sliding sphere
For the slider, all gravitational PE converts to translational KE: $v=\sqrt{2gh}$. For the roller, some PE goes into rotational KE, leaving less for translational motion, so its speed is smaller: $v=\sqrt{2gh/(1+I/MR^2)}$. -
Explain why a planet's angular momentum about the Sun is conserved throughout its elliptical orbit, and state what this implies about its orbital speed.
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Answer. Gravity is a central force always directed along the line to the Sun, so it exerts zero torque about the Sun; angular momentum is therefore conserved, meaning the planet moves faster when close to the Sun and slower when far away (Kepler's second law).
Torque is $\tau=rF\sin\theta$; since the gravitational force is parallel (or antiparallel) to the radius vector, $\theta=0°$ and $\tau=0$. With no net torque, $L=mvr\sin\theta$ stays constant, forcing $v$ to increase as $r$ decreases near perihelion. -
A satellite orbits Earth in a circular path of radius $r=7.0\times10^6\,m$. Using $GM_{Earth}=3.99\times10^{14}\,m^3/s^2$, find its orbital speed.
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Answer. About 7.5×10³ m/s
Gravity provides the centripetal force: $\frac{GMm}{r^2}=\frac{mv^2}{r}\Rightarrow v=\sqrt{GM/r}=\sqrt{3.99\times10^{14}/7.0\times10^6}\approx7550\,m/s$. -
Satellite A orbits Earth at radius r, and Satellite B orbits at radius 4r. What is the ratio of their orbital periods, T_B/T_A?
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Answer. 8
Kepler's third law gives $T^2\propto r^3$, so $T_B/T_A=(r_B/r_A)^{3/2}=4^{3/2}=8$. -
A solid disk (I = ½MR²) and a thin ring (I = MR²) of equal mass and radius are released from rest simultaneously at the top of the same ramp and roll without slipping. Which reaches the bottom first?
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Answer. The disk
For rolling objects, $v_{bottom}=\sqrt{2gh/(1+I/MR^2)}$. The disk's smaller ratio $I/MR^2=0.5$ (versus the ring's 1) means less energy is diverted into rotation, giving it a larger translational speed and shorter travel time. -
A yo-yo of mass M and moment of inertia I about its center is released from rest, with its string wound around an axle of radius r. Using energy conservation, find its center-of-mass speed after falling a height h (ignore string mass).
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Answer. $v=\sqrt{\dfrac{2Mgh}{M+I/r^2}}$
Energy conservation gives $Mgh=\frac12Mv^2+\frac12I\omega^2$, and the string constraint gives $\omega=v/r$. Substituting and solving: $Mgh=\frac{v^2}{2}(M+I/r^2)$, so $v=\sqrt{2Mgh/(M+I/r^2)}$. -
A spinning disk (I = 0.40 kg·m², ω = 10 rad/s) is dropped onto an identical stationary disk sharing the same vertical axis; they stick together. Find the common final angular speed.
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Answer. 5 rad/s
No external torque acts about the shared axis during the collision, so angular momentum is conserved: $L_i=I\omega=(0.40)(10)=4\,kg\cdot m^2/s$. With combined $I_{total}=0.80\,kg\cdot m^2$, $\omega_f=4/0.80=5\,rad/s$; kinetic energy is not conserved in this perfectly inelastic rotational collision.
What people get wrong
- Computing only $\frac12mv^2$ for a rolling object's kinetic energy. Instead, always add the rotational term $\frac12I\omega^2$ using the rolling constraint $\omega=v/r$.
- Assuming angular speed stays constant when moment of inertia changes (e.g., a skater pulling in arms). Instead, recognize that angular momentum $L=I\omega$ is what's conserved, so $\omega$ must increase as $I$ decreases.
- Using $L=mvr$ even when the velocity isn't perpendicular to the position vector. Instead, use $L=mvr\sin\theta$, or identify the perpendicular distance (lever arm) directly.
- Treating the static friction force in rolling-without-slipping problems as if it dissipates energy. Instead, remember static friction does zero work (the contact point has zero velocity), so mechanical energy is still conserved.
- Using $g=9.8\,m/s^2$ for orbital problems far from Earth's surface. Instead, compute gravitational acceleration at the actual orbital radius using $\frac{GM}{r^2}$, since $g$ decreases with altitude.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Physics 1. Free, and no account needed to start.