Angular variables and inertia
Rotation swaps x, v, a for theta, omega, alpha — and adds a new inertia that depends on where the mass sits.
Angular variables
Rotation trades linear variables for angular ones: angle theta in radians, angular velocity omega, angular acceleration alpha. Every constant-acceleration kinematics equation has an exact rotational twin — swap x for theta, v for omega, a for alpha — so there is nothing new to memorize. Angles in radians make the swap work cleanly, which is why radians are the default unit.
\omega = \frac{\Delta\theta}{\Delta t}, \quad \alpha = \frac{\Delta\omega}{\Delta t}Moment of inertia
Moment of inertia I measures how hard it is to spin an object up, and unlike mass it depends on where the mass sits — each bit counts as mr squared, so mass far from the axis matters much more. Point masses add directly, standard shapes have table values: hoop MR^2, disk (1/2)MR^2, rod about its center (1/12)ML^2, solid sphere (2/5)MR^2. The parallel-axis theorem moves the axis: I = I_cm + Md^2.
I = \sum m_i r_i^2, \quad I = I_{\text{cm}} + Md^2Worked example
Two 3.0 kg masses sit at the ends of a light 2.0 m rod. About the center, each mass is 1.0 m from the axis, so I = (3.0)(1.0)^2 + (3.0)(1.0)^2 = 6.0 kg m^2. Now shift the axis to one end. Parallel axis with M = 6.0 kg and d = 1.0 m gives 6.0 + (6.0)(1.0)^2 = 12 kg m^2. Computing directly agrees: the near mass sits on the axis and contributes 0, the far mass is 2.0 m out for (3.0)(2.0)^2 = 12 kg m^2.
I_{\text{end}} = 6.0 + (6.0)(1.0)^2 = 12\ \text{kg}\cdot\text{m}^2Flashcard
Moment of inertia adds mass times radius squared
Check your understanding
A wheel starts at rest and spins up with a constant angular acceleration of 4.0 rad/s^2. Through what angle does it turn in 3.0 s?