Rotational Inertia - AP Physics 1

$I = I_{\text{cm}} + m d^2$. Relates inertia about any axis to center of mass axis.

$\tau = I \times \beta$. Rotational analog of Newton's second law ($F = ma$).

$I = m r^2$. All mass concentrated at distance $r$ from axis.

$I = \frac{2}{5} m r^2$. Same as solid sphere about any diameter through center.

$I = \frac{3}{2} m r^2$. Uses parallel axis theorem: $\frac{1}{2}mr^2 + mr^2$.

Distribution of mass relative to axis. Mass farther from axis increases rotational inertia.

Increases rotational inertia. Rotational inertia is directly proportional to mass.

$I = \frac{1}{2} m r^2$. Ring rotating about axis through its diameter.

Sum of individual inertias. Add rotational inertias of each component part.

$I = m r^2$. Cylindrical shell with all mass at outer radius.

$I = \frac{2}{5} m r^2$. Standard formula for uniform solid sphere rotating about center.

$I = \frac{2}{3} m r^2$. Hollow spherical shell about any diameter through center.

$I = \frac{1}{4} m r^2$. Disc rotating about axis through its diameter.

Use $I = m_1 r_1^2 + m_2 r_2^2$. Treat each mass as point mass at its distance.

Use $I = m_1 r_1^2 + m_2 r_2^2$. Apply point mass formula to each mass separately.

Higher inertia, lower angular velocity. Conservation of angular momentum: $L = I\omega = constant$.

$\tau = I \times \beta$. Rotational analog of Newton's second law ($F = ma$).

$I = \frac{1}{12} m L^2$. For uniform rod rotating perpendicular to length at center.

$I = \frac{1}{3} m L^2$. Rod rotating about perpendicular axis at one end.

Increases rotational inertia. Inertia depends on $r^2$, so larger radius increases it.

$I = \frac{7}{5} m r^2$. Uses parallel axis theorem with $d = r$.

$I = \frac{1}{2} m r^2$. Flat circular disc rotating about its center axis.

$I = \frac{2}{3} m r^2$. Thin shell with all mass at surface radius.

Higher inertia, more torque needed. From $\tau = I\alpha$: larger $I$ needs larger $\tau$.

Quadruples rotational inertia. Radius appears squared in inertia formulas.

Further mass increases inertia. Mass farther from rotation axis contributes more.

$I = 2 m r^2$. Uses parallel axis theorem: $I = I_{center} + md^2$.

$I = \frac{1}{2} m r^2$. Hoop rotating about perpendicular axis through diameter.

$I = m r^2$. All mass concentrated at radius $r$ from center.