Rotational Dynamics

This lesson covers the equations of rotational motion — the rotational analogues of Newton's laws. This includes torque, angular acceleration, and the rotational analogues of the SUVAT equations.

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1 Newton's Second Law for Rotation

The rotational analogue of $F = ma$F=ma is:

$$\boxed{\tau = I\alpha}$$τ=Iα​

where:

• $\tau$τ = resultant torque (moment) about the axis of rotation (N m),
• $I$I = moment of inertia about the axis (kg m$^2$2),
• $\alpha$α = angular acceleration (rad s$^{-2}$−2).

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2 Torque

The torque (or moment) of a force $F$F about an axis is:

$$\tau = Fr$$τ=Fr

where $r$r is the perpendicular distance from the axis to the line of action of the force. More generally:

$$\tau = Fr\sin\theta$$τ=Frsinθ

where $\theta$θ is the angle between $\mathbf{F}$F and the position vector $\mathbf{r}$r.

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3 Rotational SUVAT Equations

When angular acceleration $\alpha$α is constant:

Equation	Variables
$\omega = \omega_0 + \alpha t$ω=ω0​+αt	$\omega, \omega_0, \alpha, t$ω,ω0​,α,t
$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$θ=ω0​t+21​αt2	$\theta, \omega_0, \alpha, t$θ,ω0​,α,t
$\theta = \omega t - \frac{1}{2}\alpha t^2$θ=ωt−21​αt2	$\theta, \omega, \alpha, t$θ,ω,α,t
$\omega^2 = \omega_0^2 + 2\alpha\theta$ω2=ω02​+2αθ	$\omega, \omega_0, \alpha, \theta$ω,ω0​,α,θ
$\theta = \frac{(\omega_0 + \omega)t}{2}$θ=2(ω0​+ω)t​	$\theta, \omega_0, \omega, t$θ,ω0​,ω,t

These are identical in form to the linear SUVAT equations with the substitutions $s \to \theta$s→θ, $u \to \omega_0$u→ω0​, $v \to \omega$v→ω, $a \to \alpha$a→α.

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Example 1 — Spinning Disc

A torque of 6 N m acts on a disc with moment of inertia 1.5 kg m$^2$2. Find the angular acceleration.

Solution

$\alpha = \tau/I = 6/1.5 = 4$α=τ/I=6/1.5=4 rad s$^{-2}$−2.

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Example 2 — Constant Angular Acceleration

A wheel starts from rest and accelerates at 3 rad s$^{-2}$−2 for 4 seconds. Find: (a) the final angular speed, (b) the total angle turned through.

Solution

(a) $\omega = \omega_0 + \alpha t = 0 + 3(4) = 12$ω=ω0​+αt=0+3(4)=12 rad s$^{-1}$−1.

(b) $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(3)(16) = 24$θ=ω0​t+21​αt2=0+21​(3)(16)=24 rad.

Number of revolutions: $24/(2\pi) \approx 3.82$24/(2π)≈3.82.

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4 Rotational Kinetic Energy

The kinetic energy of a rotating body is:

$$\boxed{KE_{\text{rot}} = \frac{1}{2}I\omega^2}$$KErot​=21​Iω2​

This is the rotational analogue of $\frac{1}{2}mv^2$21​mv2.

For a body that is both translating and rotating (e.g. a rolling wheel):

$$KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$$KEtotal​=21​mvcm2​+21​Icm​ω2

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Example 3 — Rolling Down a Slope

A uniform solid sphere of mass 2 kg and radius 0.1 m rolls without slipping down a slope of height 3 m. Find the speed at the bottom.

Solution

For a solid sphere: $I = \frac{2}{5}MR^2$I=52​MR2. For rolling without slipping: $v = R\omega$v=Rω, so $\omega = v/R$ω=v/R.

Energy conservation:

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5}mR^2 \cdot \frac{v^2}{R^2}$mgh=21​mv2+21​Iω2=21​mv2+21​⋅52​mR2⋅R2v2​

$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$mgh=21​mv2+51​mv2=107​mv2

$v^2 = \frac{10gh}{7} = \frac{10 \times 9.8 \times 3}{7} = 42$v2=710gh​=710×9.8×3​=42

$v = \sqrt{42} \approx 6.48$v=42​≈6.48 m s$^{-1}$−1.