Unit 6 Notes: Rolling Motion and Orbital Motion (Energy + Angular Momentum Connections)

Rolling

What “rolling” means (and why it’s special)

Rolling motion is a combination of two motions happening at once:

1. Translation: the object’s center of mass moves forward (like a block sliding).
2. Rotation: the object spins around its center of mass.

A rolling wheel, cylinder, or sphere is never “just rotating” or “just translating” (unless it’s spinning in place or sliding without rotating). This matters because many AP Physics 1 problems ask you to connect linear quantities (like v and a of the center of mass) to rotational quantities (like \omega and \alpha), and to correctly account for energy that is split between translational and rotational kinetic energy.

Rolling without slipping: the key constraint

The most important case is rolling without slipping (also called pure rolling). It means the point of the object in contact with the surface is instantaneously at rest relative to the surface.

That condition creates a powerful geometric relationship between linear motion and rotation.

For an object of radius R rolling without slipping:

v_{cm} = \omega R

and (for tangential components):

a_{cm} = \alpha R

• v_{cm} is the speed of the center of mass.
• a_{cm} is the acceleration of the center of mass along the surface.
• \omega is angular speed.
• \alpha is angular acceleration.
• R is the radius.

Why this matters: in many problems you can’t solve using only linear dynamics or only rotational dynamics. The no-slip condition is the “bridge” that lets you combine them.

Common misconception (important): Students often think v_{cm} = \omega R is “always true for rotating objects.” It’s only true for rolling without slipping. A wheel spinning on ice might have large \omega but small v_{cm} if it slips.

The role of friction in rolling

Rolling without slipping usually requires static friction at the contact point. Static friction is what prevents relative motion between surfaces.

A subtle but crucial idea: in pure rolling on a stationary surface, static friction often does no mechanical work on the rolling object because the point of contact does not move relative to the ground. No displacement of the application point (in the ground frame) means no work at that contact.

That does not mean friction is unimportant—static friction can still:

• Provide a torque that changes the object’s rotation.
• Redistribute energy between translation and rotation.

Direction of friction depends on the situation, not “always opposite motion.” For example:

• Rolling down an incline without slipping: the object tends to slip downhill at the contact point; static friction typically points up the incline to prevent slipping.
• A driven wheel (car accelerating forward): the tire pushes backward on the ground; static friction from the ground on the tire points forward, accelerating the car.

If the required static friction exceeds \mu_s N, the object slips, friction becomes kinetic, and mechanical energy is typically dissipated as thermal energy.

Kinetic energy of a rolling object

A rolling object has two kinds of kinetic energy:

• Translational: movement of the center of mass
• Rotational: spinning about the center of mass

Total kinetic energy:

K_{total} = \frac{1}{2} m v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2

• m is mass.
• I_{cm} is the moment of inertia about the center of mass.

Using the rolling constraint v_{cm} = \omega R, you can rewrite rotational energy in terms of v_{cm}:

K_{total} = \frac{1}{2} m v_{cm}^2 + \frac{1}{2} I_{cm} \left(\frac{v_{cm}}{R}\right)^2

This is the core reason different shapes roll differently: I_{cm} depends on how mass is distributed.

Moment of inertia reference (common AP Physics 1 shapes)

Below are standard results about the symmetry axis through the center:

A helpful way to compare shapes is to write:

I_{cm} = \beta m R^2

where \beta is a shape factor (larger \beta means “harder to spin up”).

Rolling down an incline: why shape changes acceleration

Suppose an object rolls without slipping down an incline at angle \theta. Gravity pulls it down; static friction provides the torque needed for rotation.

You can solve this with energy or with Newton’s laws plus torque. A widely used result (for pure rolling, no energy losses) is:

a_{cm} = \frac{g\sin\theta}{1 + \frac{I_{cm}}{mR^2}}

Using I_{cm} = \beta mR^2:

a_{cm} = \frac{g\sin\theta}{1 + \beta}

Key interpretation:

• Larger I_{cm} (or larger \beta) means more energy must go into rotation, so translation speeds up more slowly.
• Therefore, on the same incline:
  • A solid sphere (small \beta) accelerates more than
  • A disk, which accelerates more than
  • A hoop (largest \beta among these).

This is a common conceptual question: “Which object reaches the bottom first?” The one with the smallest rotational inertia relative to mR^2 wins.

Worked example 1: acceleration comparison (hoop vs solid cylinder)

Situation: A hoop and a solid cylinder roll without slipping down the same incline of angle \theta.

For a hoop, I_{cm} = mR^2, so \beta = 1:

a_{hoop} = \frac{g\sin\theta}{1 + 1} = \frac{1}{2} g\sin\theta

For a solid cylinder, I_{cm} = \frac{1}{2} mR^2, so \beta = \frac{1}{2}:

a_{cyl} = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{2}{3} g\sin\theta

Since \frac{2}{3} g\sin\theta > \frac{1}{2} g\sin\theta, the solid cylinder accelerates faster and reaches the bottom first.

Worked example 2: speed after dropping a vertical height

Situation: A solid sphere rolls without slipping down from rest through a vertical drop h. Find its final center-of-mass speed v_{cm} (ignore energy losses).

Use conservation of mechanical energy:

• Initial energy: gravitational potential mgh
• Final energy: translational + rotational kinetic energy

mgh = \frac{1}{2} m v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2

For a solid sphere:

I_{cm} = \frac{2}{5} mR^2

Use rolling constraint \omega = \frac{v_{cm}}{R}:

mgh = \frac{1}{2} m v_{cm}^2 + \frac{1}{2} \left(\frac{2}{5} mR^2\right)\left(\frac{v_{cm}}{R}\right)^2

Simplify the rotational term:

mgh = \frac{1}{2} m v_{cm}^2 + \frac{1}{5} m v_{cm}^2

Combine coefficients:

mgh = \frac{7}{10} m v_{cm}^2

Solve:

v_{cm} = \sqrt{\frac{10}{7}gh}

What this teaches you: if the object were sliding without rotating, you’d get v = \sqrt{2gh}. Rolling gives a smaller translational speed because some energy becomes rotational kinetic energy.

Instantaneous axis idea (useful conceptual tool)

In pure rolling, the contact point is instantaneously at rest. You can think of the object as rotating about that contact point at that instant. This viewpoint can simplify some angular momentum reasoning, but be careful:

• The object is not physically pinned there.
• The “axis” changes continuously as it rolls.

This idea is most helpful when thinking about velocities of different points on the rolling object. For example, the top of a rolling wheel has speed (relative to ground):

v_{top} = v_{cm} + \omega R = 2v_{cm}

and the bottom point has instantaneous speed 0.

Exam Focus

• Typical question patterns
  • Compare accelerations/speeds of different rolling objects using I_{cm} (conceptual ranking or symbolic calculation).
  • Use energy to find final speed after a vertical drop, including both translational and rotational kinetic energy.
  • Determine whether static friction points up or down an incline and whether it does work.
• Common mistakes
  • Using v_{cm} = \omega R even when the problem states slipping occurs (or when kinetic friction is present).
  • Forgetting the rotational kinetic energy term and treating rolling like sliding.
  • Assuming friction always opposes the object’s overall motion rather than opposing impending slip at the contact point.

Motion of Orbiting Satellites

What an orbit is (physics definition, not just “going around”)

A satellite in orbit is an object whose motion is controlled primarily by gravity and whose path continuously “falls around” the central body rather than hitting it. In AP Physics 1, satellite motion is a powerful application of:

• Newton’s law of gravitation
• Uniform circular motion (for circular orbits)
• Conservation of energy and angular momentum (especially when comparing orbits)

A key mental model: an orbit is not “gravity disappears.” Gravity is very much present—it supplies the centripetal acceleration.

Circular orbit: gravity provides centripetal force

For a satellite of mass m in a circular orbit of radius r around a central body of mass M (like Earth), the gravitational force magnitude is:

F_g = \frac{GMm}{r^2}

This gravitational force points toward the center and acts as the centripetal force required for circular motion:

F_c = \frac{mv^2}{r}

Set them equal:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Cancel m (important result: orbital speed does not depend on satellite mass) and solve for orbital speed v:

v = \sqrt{\frac{GM}{r}}

• Larger r means smaller orbital speed.
• This can feel counterintuitive because “higher orbit” sounds like it should require “more speed.” The truth is: you need enough sideways speed to keep missing the planet, but the required speed decreases as gravity weakens with distance.

The centripetal acceleration is:

a_c = \frac{v^2}{r} = \frac{GM}{r^2}

This shows a deep connection: in circular orbit, the satellite’s inward acceleration equals the local gravitational field magnitude.

Orbital period: how long one revolution takes

The orbital period T for a circular orbit is distance per cycle divided by speed:

T = \frac{2\pi r}{v}

Substitute v = \sqrt{\frac{GM}{r}}:

T = 2\pi\sqrt{\frac{r^3}{GM}}

This is the circular-orbit version of the relationship behind Kepler’s third law: bigger orbits have much longer periods because T grows with r^{3/2}.

Energy in circular orbit: why “higher orbit” has higher energy

Gravitational potential energy (with zero at infinity) is:

U = -\frac{GMm}{r}

Kinetic energy in circular orbit uses v^2 = \frac{GM}{r}:

K = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM}{r} = \frac{GMm}{2r}

Total mechanical energy E = K + U:

E = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}

Important takeaways:

• Total energy in a circular orbit is negative: the satellite is gravitationally bound.
• If r increases, E becomes less negative, meaning the satellite has higher total energy.
• Yet its speed decreases with larger r. The “extra energy” shows up as a larger (less negative) potential energy.

Common misconception: “To go to a higher orbit you must speed up and stay faster.” In reality, you must add energy overall, but the final circular orbital speed is smaller at larger r.

Angular momentum and Kepler’s second law (conceptual link)

For a satellite under a central force (gravity toward the center), the torque about the center is zero because the force is along the radius vector. Zero torque means angular momentum is conserved.

For a particle of mass m moving with velocity perpendicular component v_\perp at distance r from the center, angular momentum magnitude is:

L = mrv_\perp

• In a circular orbit, v_\perp = v, so L = mrv.

Conservation of angular momentum is the physics behind Kepler’s second law (equal areas in equal times): when a satellite is closer to the planet in an elliptical orbit, it must move faster so that mrv_\perp stays constant.

AP Physics 1 typically treats elliptical orbits qualitatively: you should know speed is greatest at closest approach and smallest far away, and connect that to energy and angular momentum conservation.

Changing orbits: what a “burn” really does

When a rocket engine fires briefly (a burn), it provides an impulse that changes the satellite’s velocity. The direction of that velocity change matters:

• A tangential burn (speeding up along the direction of motion) changes the orbital energy strongly and is the usual way to raise the opposite side of the orbit.
• A radial burn (toward/away from the center) changes the shape differently but is less emphasized in introductory treatments.

A conceptual rule you can rely on in AP Physics 1: to move from one circular orbit to a higher circular orbit, you must add energy (do positive work). The satellite may temporarily speed up during the transfer, but the final circular speed at the higher radius is smaller.

Escape speed: the threshold to leave without coming back

Escape speed is the minimum speed needed at distance r from a central body to reach infinity with zero remaining speed, ignoring air resistance and assuming no further propulsion.

Use energy conservation with final total energy equal to zero:

\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0

Solve:

v_e = \sqrt{\frac{2GM}{r}}

Notice the relationship to circular orbit speed:

v_e = \sqrt{2}v

So escape speed is about 1.414 times the circular orbital speed at the same radius.

Worked example 1: orbital speed ratio at different radii

Situation: Two satellites orbit the same planet in circular orbits at radii r_1 and r_2. Find the ratio of their speeds.

From circular orbit speed:

v = \sqrt{\frac{GM}{r}}

So:

\frac{v_1}{v_2} = \sqrt{\frac{GM/r_1}{GM/r_2}} = \sqrt{\frac{r_2}{r_1}}

Interpretation: if r_2 is four times r_1, then v_1/v_2 = \sqrt{4} = 2, so the closer satellite moves twice as fast.

Worked example 2: energy required to move between circular orbits (symbolic)

Situation: A satellite of mass m moves from a circular orbit of radius r_1 to a circular orbit of radius r_2 around a planet of mass M. Find the change in total mechanical energy.

Total energy in a circular orbit:

E = -\frac{GMm}{2r}

So:

\Delta E = E_2 - E_1 = -\frac{GMm}{2r_2} + \frac{GMm}{2r_1}

If r_2 > r_1, then \frac{1}{r_2} < \frac{1}{r_1}, making \Delta E positive. You must add energy to raise the orbit.

This is a frequent reasoning target: you don’t need to memorize a transfer method to answer energy-comparison questions.

Common conceptual traps (and how to avoid them)

1. “There is no gravity in space.” Gravity decreases with distance but never suddenly turns off. Astronauts feel weightless because they are in free fall, not because gravity is zero.
2. Confusing altitude with orbital radius. The formula uses r measured from the planet’s center. If given altitude h above the surface, then:

r = R_{planet} + h

3. Thinking a stable orbit needs thrust. A circular orbit at constant radius requires no propulsion; gravity provides the centripetal acceleration.

Exam Focus

• Typical question patterns
  • Derive or apply v = \sqrt{\frac{GM}{r}} and T = 2\pi\sqrt{\frac{r^3}{GM}}, often through proportional reasoning (ratios) rather than heavy computation.
  • Use energy expressions U = -\frac{GMm}{r} and E = -\frac{GMm}{2r} to compare orbits or find energy changes.
  • Conceptual questions linking angular momentum conservation to changing speed in elliptical orbits (fastest near closest approach).
• Common mistakes
  • Setting GMm/r^2 equal to mv/r instead of mv^2/r (missing a power of v).
  • Forgetting that r is measured from the center of the planet, not from the surface.
  • Claiming that moving to a higher orbit means a higher final orbital speed, instead of recognizing the tradeoff: higher total energy but lower circular speed.