Rotation (Calculus-Based)

Moment of Inertia (Integration)

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

For a uniform body with mass density $\lambda = M/L$ (linear) or $\sigma = M/A$ (surface):

Thin rod about center: $I = \int_{-L/2}^{L/2} x^2 \lambda\,dx = \frac{1}{12}ML^2$

Thin rod about end: $I = \int_0^L x^2 \lambda\,dx = \frac{1}{3}ML^2$

Disk about axis: $I = \int_0^R r^2 (2\pi r \sigma)\,dr = \frac{1}{2}MR^2$

Parallel Axis Theorem

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

Torque and Angular Acceleration

$\tau_{\text{net}} = I\alpha = I\frac{d\omega}{dt}$

Angular Momentum

$L = I\omega$ $\tau = \frac{dL}{dt}$

Conservation: if $\tau_{\text{ext}} = 0$, then $L = \text{const}$.

Rotational Kinetic Energy

$KE_{\text{rot}} = \frac{1}{2}I\omega^2$

Rolling Without Slipping

$v_{\text{cm}} = R\omega, \quad a_{\text{cm}} = R\alpha$ $KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}(m + I/R^2)v_{\text{cm}}^2$

Work-Energy Theorem (Rotational)

$W = \int \tau\,d\theta = \Delta KE_{\text{rot}}$

Problem: Derive the moment of inertia of a solid sphere of mass $M$ and radius $R$ about a diameter.

Solution:

Consider the sphere as a stack of thin disks. A disk at height $z$ from the center has radius $r = \sqrt{R^2 - z^2}$ and thickness $dz$.

$dm = \rho \pi r^2\,dz = \rho \pi(R^2 - z^2)\,dz$

Moment of inertia of a disk about its axis: $dI = \frac{1}{2}r^2\,dm = \frac{1}{2}\rho\pi(R^2-z^2)^2\,dz$

$I = \frac{\rho\pi}{2}\int_{-R}^{R}(R^2-z^2)^2\,dz = \frac{\rho\pi}{2}\cdot\frac{16R^5}{15}$

With $\rho = \frac{3M}{4\pi R^3}$: $I = \frac{2}{5}MR^2$

• Moment of inertia via integration: $I = \int r^2\,dm$.
• Parallel axis theorem: $I = I_{\text{cm}} + Md^2$.
• Rotational dynamics: $\tau = I\alpha$, $L = I\omega$, $\tau = dL/dt$.
• Rolling without slipping: total KE = translational + rotational.