Rigid body mechanics (HL)

The **angle turned per second** — the rate of change of θ. Unit: **rad s⁻¹**.

The **rate of change of angular velocity** ω. Unit: **rad s⁻²**.

The angle whose **arc length equals the radius**. A full turn = **2π rad = 360°**.

Slope = **angular acceleration** α; area = **angle turned** θ.

$\omega = \omega_0 + \alpha t$.

Multiply by **2π** (one revolution = 2π rad).

The **turning effect** of a force: $\tau = Fr\sin\theta$. Unit: **N m**.

When it acts **through the pivot** (θ = 0, sin θ = 0).

The **total torque about any point is zero** (clockwise = anticlockwise).

A point on an **unknown force's line**, so that force has zero torque.

Bigger **r** → bigger torque for the same force.

Both are N m, but torque is **N m** (a turning effect); energy is the **joule**.

**s = rθ** (and its speed is v = ωr).

When the push's torque about the corner reaches the weight's: **F × H = W × b/2**.

It gives it an **angular acceleration** (its spin changes).

Rotation's version of **mass** — resistance to angular acceleration: $I = \sum m r^{2}$. Unit: **kg m²**.

Because r is **squared** in I = Σmr² — doubling the distance quadruples that part's contribution.

$\tau = I\alpha$ (torque = moment of inertia × angular acceleration).

The **hoop** (I = MR²); the disc is ½MR².

$I = \tfrac{1}{2}MR^{2}$ (given in the question).

$I = MR^{2}$ — all the mass is at radius R.

No — the **exam gives them**; recognise and substitute.

$\alpha = \tau / I$ — rearranged from τ = Iα.

**Torque** τ.

Yes — the same object has different I about different axes.

**kg m²**.

**One eighth**: half the mass and half the length, I ∝ M L².

**1/I**, so I = 1 ÷ gradient.

Their difference gives the **torque** (T₂ − T₁)r = Iα that spins the pulley up.

Rotation's version of momentum: $L = I\omega$. Unit: **kg m² s⁻¹**.

When there is **no external torque** on the system.

$I_1\omega_1 = I_2\omega_2$.

I decreases, so ω increases to keep **L = Iω** constant.

$E_k = \tfrac{1}{2}I\omega^{2}$ (the rotational ½mv²).

$\tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega^{2}$ — translational **plus** rotational.

It **quadruples** (E_k ∝ ω²).

**No** — angular momentum is conserved, but some kinetic energy is lost.

$L = I\omega$.

ω **decreases** (I up, L constant).

**kg m² s⁻¹** (or equivalently N m s).

Torque × time, **ΔL = τΔt = Δ(Iω)**; unit N m s (= kg m² s⁻¹).

**E_k = L²/2I**.

**mgh = ½mv² + ½Iω²** with ω = v/r.