Unit 6.4 | Rolling and Orbiting Satellites
- 18 minutes
- 3 formulas
- 24 questions
- 3 videos
A rolling ball moves and spins, and an orbiting planet keeps both its energy and its angular momentum. This lesson ends Unit 6 with both.
Unit 6 Breakdown
In this lesson:
- Rolling without slipping: [katex] v = R\omega [/katex], and kinetic energy that is part translational, part rotational
- Why a solid sphere beats a hoop down a ramp, and why a sliding block beats them both
- Orbits: a satellite keeps both its angular momentum and its mechanical energy; circular orbit speed and escape speed
- The 4 problem types: the rolling energy split, rolling down a ramp, circular orbits, and elliptical orbits and escape
Introduction
This lesson covers rolling and orbiting satellites, the last two topics of Unit 6. Both use everything from the unit at once: rotational kinetic energy, angular momentum and their conservation laws.
A rolling ball moves and spins, so its kinetic energy has two parts (Unit 6.1). A planet circling a star has angular momentum about the star, and nothing exerts a torque about the star to change it (Unit 6.3).
Rolling Without Slipping
When a wheel or ball rolls without slipping, the point touching the ground is momentarily at rest. The center moves forward one circumference for each full turn, so the center’s speed and the spin are locked together (Unit 5.1):
[katex] v_{cm} = R\omega [/katex]
The kinetic energy is the translational energy of the center of mass plus the rotational energy about it:
[katex] K = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 [/katex]
Write the rotational inertia as [katex] I = \beta MR^2 [/katex], where [katex] \beta [/katex] is the number in front for each shape. With [katex] \omega = \frac{v}{R} [/katex] the energy becomes [katex] K = \tfrac{1}{2}(1 + \beta)Mv^2 [/katex]. The radius cancels, and the shape decides the split:
| Rolling object | [katex] \beta [/katex] | Rotational share of K | Speed after a drop h |
| Block sliding on ice (no rolling) | 0 | 0 | [katex] \sqrt{2gh} [/katex] |
| Solid sphere | [katex] \tfrac{2}{5} [/katex] | [katex] \tfrac{2}{7} [/katex] (29%) | [katex] \sqrt{\tfrac{10}{7}gh} [/katex] |
| Solid cylinder or disk | [katex] \tfrac{1}{2} [/katex] | [katex] \tfrac{1}{3} [/katex] (33%) | [katex] \sqrt{\tfrac{4}{3}gh} [/katex] |
| Hoop or thin-walled pipe | 1 | [katex] \tfrac{1}{2} [/katex] (50%) | [katex] \sqrt{gh} [/katex] |
Static friction is what makes a ball roll instead of slide, but it does no work on a ball that rolls without slipping: the contact point does not move. So mechanical energy is conserved for rolling without slipping.
Every rolling and orbit question on the AP exam is one of the four types below.
| Problem type | The clue | The setup |
| 1. The rolling energy split | “rolls without slipping”, “ratio of rotational to total” | [katex] K = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 [/katex] with [katex] v = R\omega [/katex] |
| 2. Rolling down a ramp | released from rest at height [katex] h [/katex], “which reaches the bottom first” | [katex] Mgh = \tfrac{1}{2}(1 + \beta)Mv^2 [/katex] |
| 3. Circular orbits | a satellite in a circular orbit of radius [katex] r [/katex] | [katex] \frac{GMm}{r^2} = \frac{mv^2}{r} [/katex], so [katex] v = \sqrt{\frac{GM}{r}} [/katex] |
| 4. Elliptical orbits and escape | “closest” and “farthest” points, “minimum speed to escape” | [katex] mv_pr_p = mv_ar_a [/katex] and [katex] \tfrac{1}{2}mv^2 – \frac{GMm}{r} [/katex] constant |
Problem Type 1: The rolling energy split
How to spot it: An object rolls without slipping, and the question compares its translational and rotational kinetic energy, or asks for its total kinetic energy.
Setup: Use [katex] \omega = \frac{v}{R} [/katex] to write both parts in terms of [katex] v [/katex]: [katex] K_{trans} = \tfrac{1}{2}Mv^2 [/katex], [katex] K_{rot} = \tfrac{1}{2}\beta Mv^2 [/katex]. The ratio [katex] \frac{K_{rot}}{K_{trans}} = \beta [/katex] depends only on the shape.
A [katex] 5.0\,\text{kg} [/katex] bowling ball (a solid sphere) rolls without slipping at [katex] 4.0\,\text{m/s} [/katex]. Find its translational, rotational and total kinetic energy.
[katex] K_{trans} = \tfrac{1}{2}(5.0)(4.0)^2 = 40 \text{ J} [/katex]. [katex] K_{rot} = \tfrac{1}{2}\left(\tfrac{2}{5}MR^2\right)\frac{v^2}{R^2} = \tfrac{1}{5}(5.0)(4.0)^2 = 16 \text{ J} [/katex]. Total: [katex] 56\,\text{J} [/katex]. The radius was never needed.
AP trap: Writing [katex] K = \tfrac{1}{2}Mv^2 [/katex] for a rolling object. That is only the translational part. Rolling objects always carry extra kinetic energy in the spin.
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Problem Type 2: Rolling down a ramp
How to spot it: Objects released from rest roll down a ramp (or roll up one): “speed at the bottom”, “how high does it go”, “which reaches the bottom first”.
Setup: Energy: [katex] Mgh = \tfrac{1}{2}(1 + \beta)Mv^2 [/katex], so [katex] v = \sqrt{\frac{2gh}{1 + \beta}} [/katex]. Mass and radius cancel. The smaller [katex] \beta [/katex] (mass closer to the center), the faster it goes and the sooner it arrives.
A solid sphere, a solid cylinder and a hoop roll without slipping down a [katex] 1.5\,\text{m} [/katex] high ramp, and a block slides down an identical frictionless ramp. Find each speed at the bottom.
- Block: [katex] v = \sqrt{2(9.8)(1.5)} \approx 5.4 \text{ m/s} [/katex]. All of the energy goes into moving forward.
- Solid sphere: [katex] v = \sqrt{\frac{2(9.8)(1.5)}{1.4}} \approx 4.6 \text{ m/s} [/katex].
- Solid cylinder: [katex] v = \sqrt{\frac{2(9.8)(1.5)}{1.5}} \approx 4.4 \text{ m/s} [/katex].
- Hoop: [katex] v = \sqrt{\frac{2(9.8)(1.5)}{2}} \approx 3.8 \text{ m/s} [/katex].
The order of finishing is the same: block, sphere, cylinder, hoop. Each has the same [katex] Mgh [/katex] to spend. The more of it that goes into spinning, the less is left for moving down the ramp.
A ball rolling up a ramp is the same problem backwards: it climbs to [katex] h = \frac{(1 + \beta)v^2}{2g} [/katex], higher than a sliding block with the same speed, because its spin energy also turns into height. But if the ramp is frictionless, nothing can slow the spin: the ball keeps its rotational energy and rises only as high as a sliding block would, [katex] \frac{v^2}{2g} [/katex].
▶ Watch: a worked rolling problem, and the friction that makes an object roll (starts at 18:45).
AP trap: Thinking a heavier or bigger ball wins. For rolling without slipping, mass and radius cancel. Only the shape, [katex] \beta [/katex], matters. Two solid spheres of any size tie.
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Orbiting Satellites
A satellite (or moon, or planet) orbiting a much more massive body feels only gravity, and gravity always points at the center of the big body. A force aimed at a point exerts no torque about that point. So for the satellite:
- Angular momentum about the central body is conserved: [katex] L = mvr_\perp [/katex] stays constant all the way around.
- Mechanical energy is conserved, with the gravitational potential energy [katex] U_G = -\frac{GMm}{r} [/katex] (zero far away, negative closer in): [katex] \tfrac{1}{2}mv^2 – \frac{GMm}{r} [/katex] stays constant.
In a circular orbit, [katex] r [/katex] and [katex] v [/katex] never change, so both laws hold trivially. In an elliptical orbit both quantities trade back and forth, and the two conservation laws tell you how.
Problem Type 3: Circular orbits
How to spot it: A satellite or moon in a circular orbit: its speed, period, kinetic energy or angular momentum, often compared between two orbits.
Setup: Gravity provides the centripetal force (Unit 2): [katex] \frac{GMm}{r^2} = \frac{mv^2}{r} [/katex], so [katex] v = \sqrt{\frac{GM}{r}} [/katex]. Then [katex] L = mvr [/katex], [katex] K = \frac{GMm}{2r} [/katex] and [katex] E = K + U_G = -\frac{GMm}{2r} [/katex].
A satellite orbits Earth ([katex] M = 5.97 \times 10^{24} \text{ kg} [/katex]) in a circle of radius [katex] 7.0 \times 10^{6} \text{ m} [/katex]. Find its speed and period. How does its speed change if it moves to an orbit 4 times as wide?
[katex] v = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.0 \times 10^{6}}} \approx 7.5 \times 10^{3} \text{ m/s} [/katex]. Period: [katex] T = \frac{2\pi r}{v} \approx 5.8 \times 10^{3} \text{ s} [/katex], about 97 minutes. Four times the radius gives [katex] \frac{1}{\sqrt{4}} [/katex], half the speed.
The satellite’s own mass cancels: every object in that orbit moves at the same speed. A higher orbit is slower, but its total energy [katex] -\frac{GMm}{2r} [/katex] is less negative, so moving up takes energy.
AP trap: Thinking a heavier satellite needs to go faster to stay up. The mass of the satellite cancels from [katex] v = \sqrt{\frac{GM}{r}} [/katex]. Only the central mass and the radius matter.
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Problem Type 4: Elliptical orbits and escape speed
How to spot it: An orbit with a closest and a farthest point, a comet swinging past the Sun, or “the minimum launch speed to escape”.
Setup: Angular momentum between any two points: [katex] mv_1r_{\perp 1} = mv_2r_{\perp 2} [/katex] (at the closest and farthest points, [katex] r_\perp [/katex] is just [katex] r [/katex]). Energy between any two points: [katex] \tfrac{1}{2}mv_1^2 – \frac{GMm}{r_1} = \tfrac{1}{2}mv_2^2 – \frac{GMm}{r_2} [/katex]. To escape, the object must reach [katex] r \to \infty [/katex] with [katex] v \ge 0 [/katex].
A satellite moves at [katex] 8.0\,\text{km/s} [/katex] at its closest point to a planet, [katex] 8.0 \times 10^{6} \text{ m} [/katex] from the planet’s center. Its farthest point is [katex] 1.6 \times 10^{7} \text{ m} [/katex] from the center. How fast is it moving there?
At both points the velocity is perpendicular to the radius, so [katex] (8.0)(8.0 \times 10^{6}) = v_a(1.6 \times 10^{7}) [/katex] and [katex] v_a = 4.0 \text{ km/s} [/katex]. Twice as far, half as fast. Energy agrees: as it climbs away it trades kinetic energy for potential energy.
Escape speed
An object escapes if it can reach infinite distance, where [katex] U_G = 0 [/katex], without running out of kinetic energy. The minimum launch speed from the surface (radius [katex] R [/katex]) makes the total energy exactly zero:
[katex] \tfrac{1}{2}mv_{esc}^2 – \frac{GMm}{R} = 0 \quad\Rightarrow\quad v_{esc} = \sqrt{\frac{2GM}{R}} [/katex]
For Earth, [katex] v_{esc} = \sqrt{\frac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^{6}}} \approx 11.2 \text{ km/s} [/katex]. Like orbit speed, it does not depend on the mass of the object being launched.
AP trap: Using [katex] mgh [/katex] for orbits. [katex] mgh [/katex] only works near the surface, where [katex] g [/katex] is constant. Far from a planet use [katex] U_G = -\frac{GMm}{r} [/katex], which goes to zero far away and is negative everywhere else.
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Unit 6 Wrap-Up
- Spinning objects carry kinetic energy, [katex] \tfrac{1}{2}I\omega^2 [/katex], and a torque does work [katex] \tau\Delta\theta [/katex] (6.1).
- Angular momentum is [katex] I\omega [/katex] for a spinning object and [katex] mvr_\perp [/katex] for a moving one; an angular impulse [katex] \tau\Delta t [/katex] changes it (6.2).
- With no net external torque, angular momentum is conserved, even when kinetic energy is not (6.3).
- Rolling splits kinetic energy between moving and spinning, and orbits conserve both angular momentum and energy (6.4).
Review the whole unit in the rotational energy and angular momentum speed review, or watch the rotation section of my speed review:
Ready for the whole unit? Take the unit review:
Unit 7 Preview
Next up is Unit 7: Oscillations. Springs and pendulums repeat the same motion over and over, and the energy ideas from Units 3 and 6 explain how.
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This article was written by Jason