Rotational Motion & Moment of Inertia

Rotational motion extends Newton's laws from point particles to rigid bodies spinning about fixed or moving axes. Every translational quantity has a rotational analogue: force maps to torque, mass to moment of inertia, linear momentum to angular momentum, and kinetic energy acquires a rotational term ($\frac{1}{2}$)I*$\omega$².

Moment of Inertia (MOI) quantifies rotational inertia — how hard it is to change an object's angular velocity about a given axis.
For a system of discrete particles, I = sum($m_{i}$ * $r_{i}^{2}$), where $r_{i}$ is the perpendicular distance from the axis.
For continuous bodies, I = integral($r^{2}$ dm). MOI depends on the axis of rotation, not just the body's shape or mass.

Standard MOI results (about the natural symmetry axis unless stated otherwise):

• Thin rod about centre: $\frac{ML^{2}}{12}$; about end: $\frac{ML^{2}}{3}$
• Thin ring about central axis: $MR^{2}$; about diameter: $\frac{MR^{2}}{2}$
• Solid disc/cylinder about central axis: $\frac{MR^{2}}{2}$; about diameter: $\frac{MR^{2}}{4}$
• Solid sphere about diameter: ($\frac{2}{5}$)$MR^{2}$
• Hollow sphere about diameter: ($\frac{2}{3}$)$MR^{2}$

Parallel Axis Theorem: I = $I_{cm}$ + $Md^{2}$, where d is the distance between the parallel axis and the centre-of-mass axis. This always increases I.

Perpendicular Axis Theorem: $I_{z}$ = $I_{x}$ + $I_{y}$, valid only for planar (2D) bodies. The three axes are mutually perpendicular and intersect at the same point.

Torque ($\tau$ = r x F) causes angular acceleration: $\tau_{net}$ = I*$\alpha$. The sign convention (counterclockwise positive) must be consistent throughout the problem.

Angular Momentum: L = I*$\omega$ for rotation about a fixed axis.
Newton's second law for rotation: $\tau_{net}$ = dL/dt.
When $\tau_{net}$ = 0, angular momentum is conserved: $I_{1}$$\omega_{1}$ = $I_{2}$$\omega_{2}$. This explains why a spinning ice skater speeds up when pulling arms inward (I decreases, $\omega$ increases).

Rolling Without Slipping combines translation and rotation.
The constraint is $v_{cm}$ = R*$\omega$, and the contact point has zero velocity.
Total KE = ($\frac{1}{2}$)$Mv_{cm}^{2}$ + ($\frac{1}{2}$)$I_{cm}$$\omega$².
For a solid sphere rolling down an incline, a = gsin($\theta$)/(1 + $I_{cm}$/($MR^{2}$)) = ($\frac{5}{7}$)g*sin($\theta$).

Rotational KE: $K_{rot}$ = ($\frac{1}{2}$)I*$\omega$².
For rolling bodies, $K_{total}$ = ($\frac{1}{2}$)$Mv^{2}$(1 + $\frac{k^{2}}{R^{2}}$), where k is the radius of gyration (I = $Mk^{2}$).

The key problem-solving concept is: identify the axis of rotation, compute MOI about that axis (using parallel/perpendicular axis theorems if needed), apply $\tau$ = I*$\alpha$ or conservation of angular momentum, and use the rolling constraint v = R*$\omega$ for rolling problems.

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