Angle through which object rotates.

Units: radians (rad)

Conversion: $360 degrees = 2\pi$ rad

Positive direction: typically counterclockwise (right-hand rule)

Rate of change of angular displacement.

$\omega = \frac{d\theta}{dt}$

Units: rad/s

Angular Acceleration (alpha):

Rate of change of angular velocity.

$\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$

Units: rad/s2

Rate of change of angular velocity.

$\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$

Units: rad/s2

For point at distance r from rotation axis:

Linear displacement: $x = r\theta$ (for small angles) Linear velocity: $v = r\omega$ (tangential) Tangential acceleration: $a_t = r\alpha$ Centripetal acceleration: $a_c = r\omega^2 = \frac{v^2}{r}$

Constant angular acceleration equations (analogous to linear): $\omega = \omega_0 + \alpha t$ $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$ $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$

Turning effect of force about axis.

$\vec{\tau} = \vec{r} \times \vec{F}$

Magnitude: $\tau = rF\sin\theta = F_{\perp} r$

Where:

• $\vec{r}$ = position vector from axis to force application point
• $\vec{F}$ = force
• $\theta$ = angle between $\vec{r}$ and $\vec{F}$

Units: N - m

Object in rotational equilibrium when net torque = 0.

$\sum \tau = 0$

τ = r × F = rF sinθ:

Torque magnitude formula.

Components:

• $\tau = rF\sin\theta$
• Larger r or F -> larger torque
• Maximum when $\theta = 90 degrees$ (force ⟂ r)
• Zero when $\theta = 0 degrees$ or $180 degrees$ (force ∥ r)

Torque magnitude formula.

Components:

• $\tau = rF\sin\theta$
• Larger r or F -> larger torque
• Maximum when $\theta = 90 degrees$ (force ⟂ r)
• Zero when $\theta = 0 degrees$ or $180 degrees$ (force ∥ r)

Perpendicular distance from axis to line of action of force.

$\text{Lever arm} = r\sin\theta$

Torque = force × lever arm

Two conditions for complete equilibrium:

1. Translational: $\sum \vec{F} = 0$ (or $\sum F_x = 0$, $\sum F_y = 0$)
2. Rotational: $\sum \tau = 0$ about any axis

Procedure:

1. Draw FBD for each object
2. Choose convenient axis (at unknown force to eliminate it from torque equation)
3. Calculate torques (CW = -, CCW = +)
4. Apply equilibrium equations
5. Solve system

Resistance to angular acceleration.

$I = \sum_{i=1}^{N} m_i r_i^2 = \int r^2 \, dm$

Where r = distance from rotation axis.

Units: kg - m2

I = Sigma m_i r_i2:

Discrete system formula.

I = ∫ r2 dm:

Continuous system formula.

I = Sigma m_i r_i2:

Discrete system formula.

I = ∫ r2 dm:

Continuous system formula.

Moment of inertia about any axis.

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

Where:

• $I_{cm}$ = moment about center of mass
• M = total mass
• d = distance between axes

Common shapes (about center of mass):

• Point mass: $mr^2$
• Thin rod (length L, ⟂): $\frac{1}{12}mL^2$
• Solid cylinder/disk (radius R): $\frac{1}{2}mR^2$
• Thin cylindrical shell: $mR^2$
• Solid sphere (radius R): $\frac{2}{5}mR^2$
• Thin spherical shell: $\frac{2}{3}mR^2$

Torque causes angular acceleration.

$\tau_{net} = I\alpha$

Use Sigmaτ = Ialpha to Find Angular Acceleration:

Apply to find rotational motion.

Procedure:

1. Calculate moment of inertia about rotation axis
2. Find net torque (sum of all torques)
3. Calculate angular acceleration: $\alpha = \tau_{net}/I$

Use Sigmaτ = Ialpha to Find Angular Acceleration:

Apply to find rotational motion.

Procedure:

1. Calculate moment of inertia about rotation axis
2. Find net torque (sum of all torques)
3. Calculate angular acceleration: $\alpha = \tau_{net}/I$

Larger torque -> larger angular acceleration Larger inertia -> smaller angular acceleration

Analogy to linear:

• $\tau \leftrightarrow F$
• $I \leftrightarrow m$
• $\alpha \leftrightarrow a$
• $\omega \leftrightarrow v$

Energy of rotation.

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

Units: Joules (J)

Physical meaning:

• Energy required to accelerate object to angular speed ω
• Depends on inertia and angular velocity

Analyzing Combined Motion of Rolling Without Slipping:

Rolling = translation + rotation

Constraint: no slipping at contact point $v_{cm} = R\omega$

Where:

• $v_{cm}$ = center of mass velocity
• R = radius
• $\omega$ = angular velocity

Implications:

• Instantaneous velocity of contact point = 0
• Friction does no work (static friction)
• Both translational and kinetic energy present

Kinematic constraint for pure rolling.

Implications:

• Instantaneous velocity of contact point = 0
• Friction does no work (static friction)
• Both translational and kinetic energy present

$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$

For solid sphere ($I_{cm} = \frac{2}{5}mR^2$): $K_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{5}mv_{cm}^2 = \frac{7}{10}mv_{cm}^2$