Rolling, Torque, and Angular Momentum
Wheels that are translation and rotation at once — and a third great conservation law for the spinning world
- How smooth rolling is the sum of pure translation and pure rotation, with \(v_{\text{com}} = \omega R\).
- The kinetic energy of rolling (\(K = \tfrac{1}{2}I_{\text{com}}\omega ^{2} + \tfrac{1}{2}Mv_{\text{com}}^{2}\)) and the role of friction in rolling down a ramp.
- Torque as a vector, defined by the cross product \(\tau = r \times F\).
- Angular momentum of a particle (\(\ell = r \times p\)) and of a rigid body (\(L = I\omega\)), and Newton's second law in angular form \(\tau _{\text{net}} = dL/dt\).
- The conservation of angular momentum and how a spinning gyroscope precesses.
What Is Physics?
The most consequential application of rotation is the rolling of wheels — the technology behind everything from the logs that moved the Easter Island statues to today's cars, bikes, and skateboards. Rolling looks settled, yet it still yields surprises (in-line skates, street luge, the self-righting Segway). Our way into the physics is a clean decomposition: treat rolling as the combination of two motions we already understand — translation of the center of mass and rotation about it.
Rolling as Translation and Rotation Combined
For an object that rolls smoothly (no slipping or bouncing), the center \(O\) moves a distance \(s\) while the body turns through angle \(\theta\), with \(s = \theta R\). Differentiating links the center's speed to the spin:
Pure rotation (every rim point at speed \(\omega R = v_{\text{com}}\)) plus pure translation (every point at \(v_{\text{com}}\)) adds up to the real rolling motion.
At the contact point \(P\) the two motions cancel: it is instantaneously at rest. At the top \(T\) they add: it moves at \(2v_{\text{com}}\) — the blur you see on a rolling wheel's upper spokes.
The Kinetic Energy of Rolling
Viewing rolling as pure rotation about \(P\) gives \(K = \tfrac{1}{2}I_{P}\omega ^{2}\). The parallel-axis theorem (\(I_{P} = I_{\text{com}} + MR^{2}\)) and \(v_{\text{com}} = \omega R\) split this into two clean pieces:
The Forces of Rolling
A wheel rolling at constant speed has no tendency to slide, so no friction acts. But the moment a net force speeds it up or slows it, the contact point tends to slip, and static friction appears to prevent that. For smooth rolling, differentiating \(v_{\text{com}} = \omega R\) links the accelerations:
For a round body rolling down a ramp of angle \(\theta\), applying both \(F_{\text{net}} = Ma\) (along the ramp) and \(\tau _{\text{net}} = I\alpha\) (about the center, where only friction has a moment arm) and combining gives a single result:
The factor \(I_{\text{com}}/MR^{2}\) is the great equalizer: it is independent of mass and radius and depends only on shape. A hoop (large \(I/MR^{2} = 1\)) accelerates slowest; a solid sphere (\(\tfrac{2}{5}\)) fastest. Mass and radius cancel — any sphere beats any hoop down the same ramp.
The Yo-Yo
A yo-yo is a ramp problem stood on end. As it rolls down its string it trades gravitational potential energy for both translational and rotational kinetic energy. The analysis is identical to the ramp, with three changes: the "ramp" is vertical (\(\theta = 90^{\circ}\)), the rolling radius is the axle radius \(R_{0}\), and the slowing force is the string tension rather than friction.
Torque Revisited
In Chapter 10 torque was a signed scalar for rotation about a fixed axis. Now we generalize it to a particle moving along any path relative to a fixed point \(O\), making torque a true vector:
Angular Momentum
Angular momentum is to linear momentum what torque is to force. For a particle of momentum \(p = mv\) at position \(r\) from \(O\):
Newton's Second Law in Angular Form
Guided by \(F_{\text{net}} = dp/dt\), the angular analogue is exactly what you would guess — and differentiating \(\ell = m(r \times v)\) proves it (the \(v \times v\) term vanishes, leaving \(r \times ma\)):
The net torque on a particle equals the time rate of change of its angular momentum. Both must be measured about the same point — usually the origin.
The Angular Momentum of a System of Particles
For a system, the total angular momentum is the vector sum \(L = \sum \ell _{i}\). Internal torques cancel in third-law pairs, so only external torques can change \(L\):
The Angular Momentum of a Rigid Body Rotating About a Fixed Axis
Summing the axial components of \(\ell _{i}\) over all mass elements of a body spinning about a fixed axis (using \(v_{i} = \omega r_{i}\)) factors out \(\omega\) and recovers the rotational inertia:
Conservation of Angular Momentum
If no net external torque acts, \(dL/dt = 0\), so the total angular momentum is constant. This is the third great conservation law, alongside energy and linear momentum.
For a body that redistributes its mass, \(I_{i}\omega _{i} = I_{f}\omega _{f}\). Pull mass inward and \(I\) drops, so \(\omega\) must rise. No exception to this law has ever been found — it holds in relativity and quantum physics too.
Precession of a Gyroscope
A non-spinning gyroscope simply topples. A rapidly spinning one, released with its shaft horizontal, instead sweeps slowly around a vertical axis — it precesses. The gravitational torque cannot grow the fixed-magnitude \(L\); it can only swing its direction. Since \(L\) points along the shaft, the shaft itself rotates.
Putting It to Work
Problem. A uniform ball (\(M = 6.00 \mathrm{kg}\), \(I_{\text{com}} = \tfrac{2}{5}MR^{2}\)) rolls smoothly from rest down a 30.0° ramp, descending a vertical height \(h = 1.20 m\). Find (a) its speed at the bottom and (b) the friction force.
Solution. Mechanical energy is conserved (friction does no thermal work in smooth rolling). With \(K_{f} = \tfrac{1}{2}I_{\text{com}}\omega ^{2} + \tfrac{1}{2}Mv^{2}\) and \(\omega = v/R\):
The speed is independent of \(M\) and \(R\) — any uniform ball arrives equally fast. The friction force needs \(M\) but not \(R\).
Problem. A force of magnitude 2.0 N acts on a particle at position \(r\) (\(r = 3.0 m\)) in the xz plane. Find the torque magnitude about O when the angle between \(r\) and the force is (a) 150°, (b) 120°, (c) 90°.
Directions come from the right-hand rule: sweep \(r\) into \(F\) and read the thumb (into the page, out of the page, etc.). Torque is largest when the force is perpendicular to \(r\).
Problem. Particle 1 (\(p_{1} = 5.0 \mathrm{kg}\cdot m/s\)) passes 2.0 m from O, swinging counterclockwise; particle 2 (\(p_{2} = 2.0 \mathrm{kg}\cdot m/s\)) passes 4.0 m from O, swinging clockwise. Find the net angular momentum about O.
Solution. Use \(\ell = r_{\perp}p\) with signs from the rotation sense (CCW +, CW −).
The positive sum means the system's net angular momentum points out of the page.
Problem. A student on a frictionless stool holds a spinning wheel (\(I_{\text{wh}} = 1.2 \mathrm{kg}\cdot m^{2}\)) with its angular momentum pointing up, \(\omega _{\text{wh}} = 3.9 \mathrm{rev}/s\). He flips the wheel over. The student-plus-stool system has \(I_{b} = 6.8 \mathrm{kg}\cdot m^{2}\). How fast does he end up rotating?
Solution. No external vertical torque acts, so the total vertical angular momentum is conserved. Flipping reverses the wheel's contribution; with the student initially at rest:
The positive result means the student spins counterclockwise (seen from above). Flipping the wheel back would stop him.
Problem. A cockroach of mass \(m\) rides a disk of mass \(6.00m\) and radius \(R\) spinning at \(\omega _{i} = 1.50 \mathrm{rad}/s\). Starting at \(r = 0.800R\), it crawls to the rim. Find the final angular speed.
Solution. No external torque, so \(L\) is conserved. Disk: \(I_{d} = \tfrac{1}{2}(6.00m)R^{2} = 3.00mR^{2}\). Roach (a particle): \(mr^{2}\), giving \(0.64mR^{2}\) initially, \(mR^{2}\) finally.
Moving mass outward raises \(I\), so the spin slows — and the unknowns \(m\) and \(R\) cancel entirely.
Chapter Summary
\(v_{\text{com}} = \omega R\) and \(a_{\text{com}} = \alpha R\) for smooth rolling. Contact point is momentarily at rest; the top moves at \(2v_{\text{com}}\).
\(K = \tfrac{1}{2}I_{\text{com}}\omega ^{2} + \tfrac{1}{2}Mv_{\text{com}}^{2}\) — rotation about the com plus translation of the com.
\(a_{\text{com}} = -g \sin \theta /(1 + I_{\text{com}}/MR^{2})\). Shape alone sets the ranking; mass and radius cancel.
\(\tau = r \times F\), magnitude \(rF \sin \varphi\), direction by the right-hand rule, defined about a point.
Particle: \(\ell = r \times p\). Rigid body: \(L = I\omega\). Dynamics: \(\tau _{\text{net}} = dL/dt\).
Zero net torque ⟹ \(L_{i} = L_{f}\) (so \(I_{i}\omega _{i} = I_{f}\omega _{f}\)). A spinning gyroscope precesses at \(\Omega = Mgr/I\omega\).
Problems
Rolling problems usually yield to energy (\(K = \tfrac{1}{2}I_{\text{com}}\omega ^{2} + \tfrac{1}{2}Mv^{2}\)) or to the paired force/torque laws. Angular-momentum problems split cleanly: use \(\tau _{\text{net}} = dL/dt\) when a torque acts, and conservation (\(I_{i}\omega _{i} = I_{f}\omega _{f}\)) when none does. Take \(g = 9.8 m/s^{2}\).
- A 140 kg hoop rolls along a floor with its center moving at 0.150 m/s. How much work must be done to stop it?
- A uniform solid sphere rolls down an incline. (a) What incline angle gives the center an acceleration of 0.10g? (b) Would a frictionless block on that same angle accelerate more, less, or equally? Why?
- A 1000 kg car has four 10 kg wheels (each a uniform disk). What fraction of the car's total kinetic energy is rotational? Why is the wheel radius irrelevant?
- A solid cylinder (radius 10 cm, mass 12 kg) starts from rest and rolls 6.0 m down a roof inclined at 30°. (a) Find its angular speed as it leaves the roof. (b) If the roof edge is 5.0 m high, how far horizontally does it land?
- A yo-yo has rotational inertia 950 g·cm², mass 120 g, and axle radius 3.2 mm, on a 120 cm string. Find (a) its linear acceleration and (b) the time to roll to the string's end from rest.
- In unit-vector notation, find the net torque about the origin on a flea at (0, 4.0 m, 5.0 m) when forces \(F_{1} = (3.0 N)\hat{k}\) and \(F_{2} = (-2.0 N)\hat{\jmath}\) act on it.
- A 0.25 kg object has position \(r = (2.0\hat{\imath} - 2.0\hat{k}) m\) and velocity \(v = (-5.0\hat{\imath} + 5.0\hat{k}) m/s\), with force \(F = (4.0 N)\hat{\jmath}\). Find (a) its angular momentum and (b) the torque on it, both about the origin.
- A man on a frictionless rotating platform spins at 1.2 rev/s with arms out; the system inertia is 6.0 kg·m². He pulls the bricks in to make it 2.0 kg·m². Find (a) the new angular speed and (b) the ratio of new to old kinetic energy. (c) Where did the extra energy come from?
- Two disks on a common axle couple together. Disk 1 (3.30 kg·m²) spins CCW at 450 rev/min; disk 2 (6.60 kg·m²) spins CCW at 900 rev/min. Find their common angular speed after coupling.
- A gyroscope is a uniform disk of radius 50 cm at the center of an 11 cm massless axle supported at one end. If it spins at 1000 rev/min, what is its precession rate?