Mistake Master
Mistake Master
Topic 5.5 was about when a rigid body's rotation stays steady: $\sum\tau = 0$ about an axis. Topic 5.6 is what happens when it doesn't. When the net torque isn't zero, the body angularly accelerates, and the size of $\alpha$ is set by the rotational version of Newton's second law: $\sum\tau = I\alpha$. Same shape as $\sum F = m a$, with rotational inertia $I$ playing mass.
Last topic was rotational equilibrium: $\sum\tau = 0$ means the body's rotation doesn't change. This topic is the other case. When the net torque isn't zero, $\alpha$ isn't zero, and §2 nails down exactly how big it is. Bigger net torque means bigger $\alpha$; bigger rotational inertia means smaller $\alpha$ for the same torque.
Two warnings before that. First, $\omega$ and $\alpha$ are not the same thing. $\omega$ tells you how fast something is spinning right now; $\alpha$ tells you how that spin is changing. A wheel can spin at $\omega = 1000$ rad/s with $\alpha = 0$ (zero net torque); it can also have $\omega = 0$ for an instant with $\alpha \neq 0$ (a pendulum at the top of its swing). The two behave like position and velocity, not like two flavors of the same thing.
Second, $I$ depends on how mass is arranged, not just how much there is. Same total mass, different layout, different $I$, different $\alpha$. The topic-hub figure shows it on a dumbbell: pull the masses in, $I$ drops, $\alpha$ jumps.
For a rigid body rotating about a chosen axis:
$$\sum\tau = I\alpha.$$Same shape as $\sum F = m a$. Net torque does the work of net force. Angular acceleration $\alpha$ does the work of linear $a$. Rotational inertia $I$ does the work of mass: it measures how hard the body is to spin up or slow down. Sign convention is CCW positive throughout. Solve for $\alpha$:
$$\alpha = \dfrac{\sum\tau}{I}.$$Three things to notice. First, $\omega$ is not in this equation. The current angular velocity doesn't affect the current $\alpha$. A wheel at $\omega = 0$ and a wheel at $\omega = 100$ rad/s get the same $\alpha$ if their $I$ and $\sum\tau$ match. Second, $I$ is about mass distribution, not just total mass. Third, $I$ depends on the axis: same body, different axis, different $I$, different $\alpha$ from the same torque.
A uniform rod of mass $M = 4$ kg and length $L = 2.0$ m is pivoted at one end. It is held horizontal, then released from rest. Find the magnitude of its angular acceleration at the instant of release. Use $g = 10$ N/kg and $I_{\text{end}} = \tfrac{1}{3}ML^2$ for a rod about one end.
The only force is gravity, $Mg$ downward, acting at the rod's center of mass, which is $L/2$ from the pivot. The lever arm is $L/2$ (the rod is horizontal, so the perpendicular distance from the pivot to gravity's line of action equals $L/2$). Torque magnitude: $$\tau = Mg \cdot \dfrac{L}{2} = (4)(10)\dfrac{2.0}{2} = 40 \text{ N·m}.$$
$$I = \tfrac{1}{3}ML^2 = \tfrac{1}{3}(4)(2.0)^2 = \dfrac{16}{3} \approx 5.33 \text{ kg·m}^2.$$
$$\alpha = \dfrac{\tau}{I} = \dfrac{40}{16/3} = \dfrac{40 \cdot 3}{16} = 7.5 \text{ rad/s}^2.$$ Symbolically: $\alpha = (MgL/2)/(ML^2/3) = 3g/(2L)$. The $M$ cancels: at the instant of release, the rod's $\alpha$ depends on $L$ and $g$ only, not on $M$.
7.5 rad/s$^2$. The rod is at rest at this instant ($\omega = 0$) but $\alpha$ is nowhere near zero. Gravity is still pulling on the center of mass with a nonzero lever arm, so the net torque is nonzero, and so is $\alpha$. Notice we never used $\sum F = Ma$. The pivot pushes back on the rod with an unknown force, which makes the linear analysis a mess. When the body is fixed on an axis, $\sum\tau = I\alpha$ is the right tool.
The cleanest place to see $\sum\tau = I\alpha$ next to $\sum F = m a$ is a hanging mass on a pulley. The mass falls; the pulley spins. They're linked by the cord. No slipping on the rim means every centimeter of cord paid out matches the same arc length on the rim, which gives $a = \alpha R$: the constraint that locks the two motions together.
A uniform-disk pulley of mass $M_p = 2$ kg and radius $R = 0.5$ m is mounted on a frictionless axle. Rotational inertia about the center: $I = \tfrac{1}{2}M_pR^2 = \tfrac{1}{2}(2)(0.5)^2 = 0.25$ kg·m$^2$. A massless cord wound on the pulley's rim supports a hanging block of mass $m = 1$ kg. The block is released from rest. Find the angular acceleration of the pulley, the linear acceleration of the block, and the cord tension. Use $g = 10$ N/kg.
Newton's second law on the block (take down as positive): $$mg - T = m a.$$ Two unknowns ($T$ and $a$), one equation. Need another.
The cord pulls down at the rim with tension $T$, producing a torque $TR$ about the pulley's axis (and the axle's reaction goes through the axis, so it contributes nothing). Newton's second law in rotational form: $$T R = I \alpha.$$ A third unknown, $\alpha$, has just entered, but so has a third equation, the no-slip constraint: $$a = \alpha R.$$
Three equations, three unknowns ($T$, $a$, $\alpha$). From the pulley equation, $T = I\alpha/R$. From the constraint, $a = \alpha R$. Substitute both into the block equation: $$mg - \dfrac{I\alpha}{R} = m \alpha R.$$ Solve for $\alpha$: $$\alpha = \dfrac{mg}{I/R + mR} = \dfrac{(1)(10)}{0.25/0.5 + (1)(0.5)} = \dfrac{10}{0.5 + 0.5} = 10 \text{ rad/s}^2.$$ Then $a = \alpha R = (10)(0.5) = 5$ m/s$^2$, and $T = I\alpha/R = (0.25)(10)/0.5 = 5$ N.
The block falls at half of $g$ ($a = 5$ m/s$^2$), and the tension is half of $mg$ ($T = 5$ N). The pulley is taking up the other half of the load and storing it as rotational kinetic energy. Two limiting cases check the algebra. Massless pulley ($I = 0$): $T = 0$, $a = g$, block falls free. Infinitely massive pulley ($I \to \infty$): $\alpha \to 0$, $T \to mg$, block hangs there. Real pulleys land in between.
Each trap here is a misread of $\sum\tau = I\alpha$: letting torque stand in for angular velocity, or mass for how that mass is arranged around the axis.
Constant net torque gives constant angular acceleration, which gives a linearly growing angular velocity. The same trap as the linear case: constant force doesn't give constant velocity; it gives constant acceleration on top of whatever $v_0$ you started with. Rotational version of the same thing.
Fix. $\sum\tau = I\alpha$ ties net torque to $\alpha$, not $\omega$. Constant $\tau$ gives constant $\alpha = \tau/I$, and $\omega(t) = \omega_0 + \alpha t$ marches steadily upward. Constant torque, growing $\omega$.
This one is sticky because at a turnaround it feels like nothing is happening. The body is at rest for an instant, so why would it be accelerating? But $\sum\tau = I\alpha$ has no $\omega$ in it. At a turning point that is not directly above or below the pivot, gravity pulls on the center of mass with a nonzero lever arm, so the net torque is nonzero, and so is $\alpha$.
Fix. "What is $\alpha$ right now?" is the same question as "what is $\sum\tau / I$ right now?" The current $\omega$ doesn't enter. Net torque now, $\alpha$ now.
Mass alone doesn't set rotational response. Two bodies with the same total mass but different layouts have different $I$, so they respond differently to the same torque. Classic example: a solid disk and a hoop of the same mass and radius. The hoop has twice the $I$ (all its mass is at the rim), so it gets half the $\alpha$ from the same torque. The hub figure shows it on a dumbbell: pull the masses in, $I$ drops, $\alpha$ jumps, with no mass added or removed.
Fix. $I$ depends on both how much mass you have and on where it sits. For point masses, $I = \sum m_i r_i^2$. The $r^2$ weighting means outer mass counts way more than inner mass. Compare $\alpha$ values by comparing $I$ values, not masses.