9.3 Angular Motion, Torque, and Mass Moment of Inertia

9.3 Angular Motion, Torque, and Mass Moment of Inertia

The NCEES Dynamics specification lists "Angular motion (e.g., torque, inertia, acceleration)" and "Mass moment of inertia" as two of its eight sub-topics — together roughly a quarter of a 9–14 question area. The mathematics parallels linear motion exactly, and building that parallel explicitly is the fastest way to learn it.

The Linear-Rotational Correspondence

Every linear result you know has a rotational twin. Learn the correspondence once and the rotational equations come free.

Angular Kinematics (Constant $\alpha$)

$\omega = \omega_0 + \alpha t, \qquad \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2, \qquad \omega^2 = \omega_0^2 + 2\alpha\theta$

Radians are mandatory. The relationships $s = r\theta$, $v = r\omega$, and $a_t = r\alpha$ are valid only in radians. Convert rev/min to rad/s with $\omega = \dfrac{2\pi N}{60}$, so 1{,}800 rpm $= 188.5$ rad/s.

Mass Moment of Inertia

$I = \int r^2\,dm$

Mass moment of inertia (kg·m², slug·ft²) measures resistance to angular acceleration. It is a different quantity from the area moment of inertia (mm⁴, in⁴) used for bending stress in Chapter 10. They share a name and a symbol and nothing else — check the units to tell them apart.

Parallel Axis Theorem (Mass Version)

$I = I_{\text{cm}} + m d^2$

Structurally identical to the area version $I = \bar{I} + Ad^2$. Verify with the rod: $\frac{mL^2}{12} + m\left(\frac{L}{2}\right)^2 = \frac{mL^2}{12} + \frac{mL^2}{4} = \frac{mL^2}{3}$ ✓ — the tabulated end value.

Radius of Gyration for Mass

$k = \sqrt{\frac{I}{m}} \qquad\Longrightarrow\qquad I = mk^2$

Problems often give you $k$ instead of $I$ (flywheels are commonly specified this way), so recognize that $I = mk^2$ is a one-step substitution rather than a new concept.

Torque and Rotational Dynamics

$\sum T = I\alpha \qquad\text{(about the mass center, or about a fixed axis)}$

For a body rotating about a fixed axis at $O$, use $I_O$ including the transfer term. For a body in general planar motion, take moments about the mass center and use $I_{\text{cm}}$.

Worked Example 1: Flywheel Spin-Down

A flywheel of mass 240 kg and radius of gyration 0.55 m spins at 1{,}200 rpm. A constant braking torque of 85 N·m is applied. How long to stop, and how many revolutions?

$I = mk^2 = 240(0.55)^2 = 240(0.3025) = 72.6\ \text{kg}\cdot\text{m}^2$ $\omega_0 = \frac{2\pi(1{,}200)}{60} = 125.7\ \text{rad/s}$ $\alpha = \frac{-T}{I} = \frac{-85}{72.6} = -1.171\ \text{rad/s}^2$

Time to stop: $t = \frac{0 - 125.7}{-1.171} = \boxed{107.3\ \text{s}}$

Revolutions, from $\omega^2 = \omega_0^2 + 2\alpha\theta$: $\theta = \frac{0 - (125.7)^2}{2(-1.171)} = \frac{-15{,}800}{-2.342} = 6{,}746\ \text{rad} = \frac{6{,}746}{2\pi} = \boxed{1{,}074\ \text{revolutions}}$

Energy cross-check: $KE = \tfrac{1}{2}(72.6)(125.7)^2 = 5.74\times10^5$ J, and work done by the brake $= T\theta = 85(6{,}746) = 5.73\times10^5$ J ✓

Rolling Without Slipping

When a body rolls without slipping, the contact point is the instantaneous center of zero velocity, which locks the rotation to the translation:

$v_{\text{cm}} = R\omega, \qquad a_{\text{cm}} = R\alpha$

Total kinetic energy splits between translation and rotation:

$KE = \tfrac{1}{2}mv_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{cm}}\omega^2 = \tfrac{1}{2}mv^2\left(1 + \frac{I_{\text{cm}}}{mR^2}\right)$

The dimensionless group $\beta = I_{\text{cm}}/(mR^2)$ is the whole story:

The classic result: rolling race order depends only on the shape factor $\beta$ — not on mass and not on radius. A bowling ball and a marble reach the bottom together; a hoop of any size loses to a solid cylinder of any size. Mass and radius cancel out of $a = g\sin\theta/(1+\beta)$ entirely. Candidates who reason "heavier means faster" or "bigger means faster" get this wrong; the physical reason is that a hoop stores a larger share of its energy as rotation, leaving less for translation.

Friction Requirement for Rolling

Rolling without slipping requires enough friction to supply the angular acceleration:

$f_{\text{required}} = \frac{\beta}{1+\beta}\,mg\sin\theta \le \mu_s mg\cos\theta$

$\tan\theta_{\max} = \frac{\mu_s(1+\beta)}{\beta}$

Above that ramp angle the body slips, rolling kinematics break down, and $v = R\omega$ no longer applies.

Worked Example 2: Cylinder Rolling Down a Ramp

A solid cylinder of mass 15 kg and radius 0.20 m rolls without slipping from rest down a 20° incline for 4.0 m. Find its speed at the bottom and the required coefficient of static friction.

By energy, with $\beta = 0.5$:

$h = 4.0\sin 20° = 4.0(0.3420) = 1.368\ \text{m}$

$mgh = \tfrac{1}{2}mv^2(1+\beta) \;\Rightarrow\; v = \sqrt{\frac{2gh}{1+\beta}} = \sqrt{\frac{2(9.81)(1.368)}{1.5}}$

$v = \sqrt{\frac{26.84}{1.5}} = \sqrt{17.89} = \boxed{4.23\ \text{m/s}}$

Note that mass dropped out. A frictionless sliding block would arrive at $\sqrt{2gh} = 5.18$ m/s — faster, because none of its energy goes into rotation.

Required friction:

$f = \frac{0.5}{1.5}mg\sin\theta = \frac{1}{3}(15)(9.81)(0.3420) = 16.8\ \text{N}$ $N = mg\cos\theta = 15(9.81)(0.9397) = 138.3\ \text{N}$ $\mu_s \ge \frac{16.8}{138.3} = \boxed{0.121}$

Any surface with $\mu_s \ge 0.121$ sustains rolling; below that the cylinder slips.

Trap: friction does no work in rolling without slipping, because the contact point has zero velocity — which is exactly why the energy method above is valid despite friction being present and essential. Candidates who subtract a friction loss term get a lower speed and a wrong answer.