Rotational Kinematics and Dynamics

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Rotational Kinematics and Dynamics

TL;DR

[58 words]
Rotational motion describes how objects spin around a fixed axis using angular versions of linear concepts: angular displacement (θ), angular velocity (ω), and angular acceleration (α). Torque (τ = r × F) causes angular acceleration just as force causes linear acceleration. Angular momentum (L = Iω) is conserved when no external torque acts, making it the rotational analogue of linear momentum.

1. The Mental Model

[68 words]
Forget trying to track every point on a spinning object separately—that's impossible. Instead, treat rotation as a single motion described by angles instead of distances. Just as linear motion has velocity and acceleration, rotation has angular velocity and angular acceleration. Torque is the "rotational force"—it's what causes spinning, and it depends on how far from the axis you apply the force. The deeper you push this parallel with linear motion, the more intuition you'll build.

2. The Core Material

[~1050 words]

Angular Kinematic Variables

Let's start with the rotational equivalents of position, velocity, and acceleration.

Angular displacement (θ, measured in radians) tells you how far an object has rotated. One full rotation = 2π radians. If a wheel spins 3 full rotations, that's 6π rad.

Angular velocity (ω, rad/s) is how fast the angle changes:
$$\omega = \frac{d\theta}{dt}$$

Angular acceleration (α, rad/s²) is how fast ω changes:
$$\alpha = \frac{d\omega}{dt}$$

Here's the beautiful part: the kinematic equations you already know for linear motion work identically for rotation, just with different symbols:

$$\omega = \omega_0 + \alpha t$$
$$\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$$
$$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$$

The relationship between linear and angular quantities at radius r from the axis is:
$$v = r\omega, \quad a_{\text{tangential}} = r\alpha$$

A point on a spinning object also has centripetal acceleration directed toward the axis:
$$a_c = \frac{v^2}{r} = \omega^2 r$$

Moment of Inertia

Here's where rotation gets its own flavor. In linear motion, mass is the resistance to acceleration (F = ma). In rotation, the resistance to angular acceleration is moment of inertia (I):

$$\tau = I\alpha$$

But I isn't just "mass"—it depends on where that mass sits relative to the axis. A mass far from the axis is harder to spin than the same mass near the axis.

For a point mass m at distance r from the axis:
$$I = mr^2$$

For extended objects, you sum (or integrate) over all mass elements:
$$I = \int r^2 \, dm$$

Common moments of inertia (around the central axis):
- Solid cylinder: $I = \frac{1}{2}MR^2$
- Thin spherical shell: $I = \frac{2}{3}MR^2$
- Solid sphere: $I = \frac{2}{5}MR^2$
- Thin rod (about its center): $I = \frac{1}{12}ML^2$

The parallel axis theorem lets you find I about any axis parallel to the one through the center of mass:
$$I = I_{\text{cm}} + Md^2$$
where d is the distance between the axes.

Torque

Torque is the rotational analogue of force—it's what causes angular acceleration. If you push a door near its hinge, you need a huge force. Push near the handle, a small force works. That's torque:

$$\tau = r F \sin(\phi)$$

where r is the distance from the axis, F is the force magnitude, and φ is the angle between r and F. Equivalently, if you use the perpendicular component of F (call it $F_\perp$):
$$\tau = r F_\perp$$

Direction matters. Torque is a vector perpendicular to both r and F. Use the right-hand rule: fingers curl in the direction of rotation, thumb points in the direction of τ.

The rotational equation of motion is:
$$\tau_{\text{net}} = I\alpha$$

Rotational Energy and Momentum

Rotational kinetic energy:
$$KE_{\text{rot}} = \frac{1}{2}I\omega^2$$

This is structurally identical to $\frac{1}{2}mv^2$—just with I replacing m and ω replacing v.

Angular momentum:
$$L = I\omega$$

This is conserved when no external torque acts on a system. A figure skater pulling in her arms spins faster because I decreases and L stays constant:
$$L_{\text{initial}} = L_{\text{final}} \implies I_1 \omega_1 = I_2 \omega_2$$

If $I_2 < I_1$ (arms in), then $\omega_2 > \omega_1$ (spins faster).

graph TD
    A["Linear Motion Concepts"] -->|"Replace<br/>v with ω<br/>a with α<br/>m with I<br/>F with τ"| B["Rotational Motion Concepts"]
    A -->|"Kinematics"| C["x, v, a"]
    A -->|"Dynamics"| D["F = ma"]
    A -->|"Energy"| E["½mv²"]
    A -->|"Momentum"| F["p = mv"]

    B -->|"Kinematics"| G["θ, ω, α"]
    B -->|"Dynamics"| H["τ = Iα"]
    B -->|"Energy"| I["½Iω²"]
    B -->|"Momentum"| J["L = Iω"]

Rolling Motion

When an object rolls without slipping, linear and angular motion are linked:
$$v_{\text{center}} = R\omega$$

This constraint comes from the fact that the contact point has zero velocity. The total kinetic energy combines translation and rotation:
$$KE_{\text{total}} = \frac{1}{2}Mv_{\text{center}}^2 + \frac{1}{2}I\omega^2$$

Using the constraint to eliminate ω:
$$KE_{\text{total}} = \frac{1}{2}Mv_{\text{center}}^2 + \frac{1}{2}I\frac{v_{\text{center}}^2}{R^2}$$
$$= \frac{1}{2}v_{\text{center}}^2 \left( M + \frac{I}{R^2} \right)$$

The fraction of kinetic energy in rotation depends on the shape (I) of the object.

3. Worked Example

[~320 words]

A solid disk of mass M = 2 kg and radius R = 0.5 m rotates about its central axis. Initially it spins at ω₀ = 4 rad/s. A tangential force F = 3 N is applied at the rim for t = 2 s. Find the final angular velocity.

Step 1: Find the moment of inertia.
For a solid disk: $I = \frac{1}{2}MR^2 = \frac{1}{2}(2)(0.5)^2 = 0.25 \text{ kg·m}^2$

Step 2: Calculate the torque.
Since the force is applied at the rim (r = R) and is tangential (φ = 90°):
$$\tau = rF = 0.5 \times 3 = 1.5 \text{ N·m}$$

Step 3: Find angular acceleration.
$$\alpha = \frac{\tau}{I} = \frac{1.5}{0.25} = 6 \text{ rad/s}^2$$

Step 4: Use kinematics to find final angular velocity.
$$\omega = \omega_0 + \alpha t = 4 + 6(2) = 16 \text{ rad/s}$$

Step 5: Find the angular displacement.
$$\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 4(2) + \frac{1}{2}(6)(2)^2 = 8 + 12 = 20 \text{ rad}$$

That's 20/(2π) ≈ 3.18 full rotations.

Check using energy: Initial rotational KE: $\frac{1}{2}(0.25)(4)^2 = 2$ J. Final KE: $\frac{1}{2}(0.25)(16)^2 = 32$ J. Work done by torque: $\tau \times \theta = 1.5 \times 20 = 30$ J. The difference (32 − 2 = 30 J) matches—good.

4. Key Takeaways

4.1 Most Important Concepts

1. Angular kinematics mirrors linear kinematics. The same equation shapes apply; just replace x→θ, v→ω, a→α, and you're working in the rotational world.

2. Moment of inertia is the rotational mass. It measures resistance to angular acceleration, and it always depends on how far material sits from the axis (the r² factor).

3. Torque causes angular acceleration. τ = Iα is the rotational Newton's second law. Torque depends on force and distance from the axis (the r factor).

4. Angular momentum is conserved. When no external torque acts, L = Iω stays constant. This is why spinning objects speed up when they shrink.

5. Rolling without slipping links v and ω. The constraint v = Rω means a rolling object's motion splits between translation and rotation.

6. Kinetic energy has two parts when rolling. Translational + rotational energy must both be counted; don't forget the I term.

7. Radians are essential. Always use radians, not degrees, when doing calculations with ω and α.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

[~95 words]

Exercise: A 1.5 kg sphere (I = ⅖MR²) of radius 0.3 m starts from rest at the top of an incline. It rolls without slipping down a 4 m frictionless ramp tilted at 20° to the horizontal. Find:
(a) The linear velocity of its center when it reaches the bottom.
(b) The angular velocity at the bottom.
(c) The time it takes to roll down.

Use energy conservation for (a) and (b)—account for both translational and rotational kinetic energy. Then use kinematics for (c).

Success looks like: You obtain three numerical answers with correct units, and you verify that v_center = Rω holds for your answers.

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