Mistake Master
Mistake Master
A skater pulls her arms in and spins faster. A blob drops onto a spinning disk and the whole thing reorganizes. A wheel slows when its axle drags. One rule covers all three: pick a system, check whether anything outside applies a torque, and if nothing does, the system's total angular momentum stays the same. The traps are which torques count as external, and what does and does not change when $I$ shifts.
In Topic 6.3 we wrote the rotational version of impulse and momentum:
$$J_{ext} = \Delta L_{sys}.$$The angular impulse delivered by external torques over an interval equals the change in the system's angular momentum across that interval. Topic 6.4 is the special case where the external torques add to zero, so $J_{ext} = 0$:
$$\tau_{net,ext} = 0 \;\;\Longrightarrow\;\; L_{f,sys} = L_{i,sys}.$$If no external torque acts on the chosen system across an interval, the system's total angular momentum at the end equals its total at the start. This is the rotational version of linear momentum conservation from Topic 4.3, where a zero net external force conserved total linear momentum. The same idea, in rotational variables.
Two questions decide every problem in this topic. What is the system? The system boundary is what makes a torque count as internal or external. Are the external torques zero? If yes, write $L_f = L_i$. If no, use $J = \Delta L$ from 6.3 instead. Almost every miss comes from picking the wrong system or misclassifying a torque.
For a system made of $N$ parts, the system's total angular momentum about a chosen axis is the signed sum of the parts' angular momenta about that same axis:
$$L_{tot} = \sum_{i=1}^{N} L_i.$$Each $L_i$ uses the same sign convention (we use CCW positive throughout). A rigid body contributes $L = I\omega$. A point particle moving in a straight line contributes $L = mvr_\perp$, where $r_\perp$ is the perpendicular distance from the chosen axis to the particle's line of motion.
Whether a given torque is internal or external depends on what you put in the system. Bearing friction between a wheel and its axle is external if the system is just the wheel, but becomes internal if the system is wheel-plus-axle. The physical torques don't change. The labels do.
A horizontal disk on a frictionless vertical axle has $I_d = 2\,\text{kg}\cdot\text{m}^2$ and spins at $\omega_0 = +4\,\text{rad/s}$. A $m = 1\,\text{kg}$ blob falls straight down and sticks to the disk at radius $r = 1\,\text{m}$. Find the disk's final angular velocity $\omega_f$.
System = disk plus falling mass. With this choice, the impact force between the mass and the disk is internal to the system, so it can't change the system's total $L$.
External forces on the system: gravity (acts at the center of mass along the spin axis, zero torque about the axis) and the axle's normal force on the disk (also along the axis, zero torque). $\tau_{net,ext} = 0$, so $L$ is conserved across the impact.
The mass falls straight down, with velocity parallel to the spin axis. Its angular momentum component about that vertical axis depends only on the tangential part of its velocity, and a straight vertical fall has no tangential component about the axis. So the falling mass contributes $L = 0$ about the spin axis throughout the fall, even though it lands at radius $r$. The system's $L$ before contact is just the disk's:
$$L_i = I_d\,\omega_0 + 0 = (2)(+4) = +8\,\text{kg}\cdot\text{m}^2/\text{s}.$$After sticking, the mass at radius $r$ adds $mr^2$ to the rotational inertia:
$$I_f = I_d + mr^2 = 2 + (1)(1)^2 = 3\,\text{kg}\cdot\text{m}^2.$$Set $L_f = L_i$:
$$L_f = I_f\,\omega_f \;\Rightarrow\; \omega_f = \dfrac{L_i}{I_f} = \dfrac{+8}{3} \approx +2.67\,\text{rad/s}.$$The system choice is the conservation step. Once disk-plus-mass is the system, the impact force is internal and $L$ is conserved across the collision. Pick the system first; classify the torques second; apply $L_f = L_i$ third.
The other classic setup is a system that changes its shape: a skater on frictionless ice pulls her arms in. No external torque acts on her, so $L$ is conserved. But $I$ is not constant. Pulling arms in moves mass closer to the spin axis and lowers $I$. With $L = I\omega$ constant and $I$ shrinking, $\omega$ has to grow. That is the whole pull-in spin-up.
A skater on frictionless ice has $I_i = 4\,\text{kg}\cdot\text{m}^2$ and spins at $\omega_i = 2\,\text{rad/s}$. She pulls her arms in, dropping her rotational inertia to $I_f = 1\,\text{kg}\cdot\text{m}^2$. Find $\omega_f$.
System = the skater. Her muscle forces are internal. The ice's normal force and gravity both act along the spin axis, so they give zero torque about it. $\tau_{net,ext} = 0$, so $L$ is conserved.
$L = I\omega$ for the skater treated as a rigid body at each instant. Set before equal to after:
$$I_i\,\omega_i = I_f\,\omega_f \;\Rightarrow\; \omega_f = \dfrac{I_i\,\omega_i}{I_f} = \dfrac{(4)(2)}{1} = 8\,\text{rad/s}.$$Now compare the energies. Before: $K_{rot,i} = \tfrac{1}{2}(4)(2)^2 = 8\,\text{J}$. After: $K_{rot,f} = \tfrac{1}{2}(1)(8)^2 = 32\,\text{J}$. $K_{rot}$ is four times larger after the pull-in, even though $L$ didn't change. The extra $24\,\text{J}$ came from her muscles, which did work pulling the arm masses inward against the rotational tendency. $L$ conservation is not energy conservation.
$\omega$ changes; $L$ does not. When $I$ changes under zero net external torque, $\omega$ must change the other way. The angular momentum is unaffected; the rotational kinetic energy is not. They are different quantities.
It feels right because in everyday language we say "things keep rotating." But the conserved rotational quantity is $L = I\omega$, not $\omega$ on its own. If $I$ changes for any reason (a skater reshapes, a star collapses, two ice dancers come together), $\omega$ has to change in the opposite direction so the product stays constant.
Fix. Replace the rule with $I_i\omega_i = I_f\omega_f$ and treat $I$ as a variable. When $I$ shrinks, $\omega$ grows. When $I$ grows, $\omega$ shrinks. The ratio $I_i/I_f$ is the scaling factor for $\omega$.
Rotation alone is not the criterion. A wheel on a friction-loaded axle is rotating, but it is slowing down. A child on a merry-go-round being pushed by a parent is rotating, but the parent is delivering angular impulse. The criterion is about external torques, not about whether anything is rotating.
Fix. Before writing $L_f = L_i$, finish the sentence "External torques on the system are zero because ___." If you can't finish it honestly, $L$ is not conserved across the interval, and you need $J_{ext} = \Delta L$ from 6.3 instead.
Whether a torque is internal or external depends on the system boundary, not on the diagram. The same forces look external to a wheel-only system and internal to a wheel-plus-axle system. A skater's muscle forces are external if the system is just her arms but internal if the system is the whole skater. The labels change; the physics does not.
Fix. Draw a literal boundary around the system. A force from inside the boundary to another part inside the boundary is internal. A force from outside the boundary acting on something inside is external. Internal torque pairs cancel; external torques change $L$.
Ten scenarios drill the conservation criterion, the skater pull-in, vertical drops, multi-body engagements, and the $L$-versus-$K_{rot}$ split. Some ask for a number, some ask for an algebraic factor, and some ask for the right concept with no calculation. Pick chips, hit Check answer, read the feedback. Progress saves automatically.