Every Linear Motion Formula Has a Twin for Rotation

Almost. Swap displacement for angle, velocity for angular velocity, mass for moment of inertia and force for torque, and nearly every equation of straight-line motion turns into one for rotation.

That means a ceiling fan slowing down, a pulley lifting a bucket and a flywheel storing energy can all be handled with familiar methods.

This part covers rotational kinematics, torque and angular acceleration, rotational work, power and energy, and the full comparison of linear and rotational motion.

**For constant angular acceleration $\alpha$, the rotational equations are $\omega ={\omega }_0+\alpha t$, $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$ and ${\omega }^2={\omega }_0^2+2\alpha \theta$, with angles in radians.

Worked example 1 — speeding up.** A wheel at $10$ rad/s accelerates at $2$ rad/s${}^2$ for $5$ s.

$$\omega =10+2(5)=20\text{ rad/s}\theta =10(5)+\frac{1}{2}(2)(25)=75\text{ rad}\approx 11.9\text{ revolutions}$$

Check: ${\omega }^2=100+2(2)(75)=400$.

Worked example 2 — slowing to rest. A fan turning at $120$ rpm $=4\pi$ rad/s stops uniformly in $10$ s.

$$\alpha =\frac{0-4\pi }{10}\approx -1.26{\text{ rad/s}}^2\theta =\frac{{\omega }_0+\omega }{2}t=2\pi \times 10=20\pi \text{ rad}=10\text{ revolutions}$$

An everyday example. A ceiling fan switched off keeps turning and slows steadily to a stop, following these equations.

The substance. Convert revolutions to radians first — one revolution is $2\pi$ rad, and the equations fail with rpm.

**For rotation about a fixed axis, the net torque equals the moment of inertia times the angular acceleration, $\tau =I\alpha$ — the rotational form of $F=ma$.**

Take $g=10$ m/s${}^2$.

Worked example 1 — a flywheel. A $10$ N m torque acts on a flywheel with $I=2$ kg m${}^2$, starting from rest.

$$\alpha =\frac{10}{2}=5{\text{ rad/s}}^2\omega (4\text{ s})=20\text{ rad/s}$$

Worked example 2 — a pulled rope. A rope wound round a fixed disc pulley ($M=4$ kg, $R=0.2$ m) is pulled with $20$ N.

$$I=\frac{1}{2}(4)(0.2{)}^2=0.08{\text{ kg m}}^2\alpha =\frac{20\times 0.2}{0.08}=50{\text{ rad/s}}^2$$

The rope's linear acceleration is $\alpha R=10$ m/s${}^2$.

Worked example 3 — a hanging bucket. Instead, a $2$ kg bucket hangs from that rope. For the bucket, $mg-T=ma$; for the pulley, $TR=I\frac{a}{R}$, so $T=\frac{M}{2}a$.

$$a=\frac{mg}{m+M/2}=\frac{20}{2+2}=5{\text{ m/s}}^{2}T=2(10-5)=10\text{ N}$$

Check: $TR=2$ N m and $I\alpha =0.08\times 25=2$ N m.

An everyday example. Drawing water from a well with a bucket over a heavy pulley — the pulley's moment of inertia makes the bucket fall more slowly than in free fall.

The substance. A massive pulley means the tensions on its two sides are unequal; only a massless pulley has equal tensions.

**A torque turning a body through angle $\theta$ does work $W=\tau \theta$, delivers power $P=\tau \omega$, and the rotating body has kinetic energy $K=\frac{1}{2}I{\omega }^2$; the net work equals the change in rotational kinetic energy.

Worked example 1 — work-energy check.** In the flywheel example ($\tau =10$ N m, $I=2$ kg m${}^2$, $\alpha =5$ rad/s${}^2$, $4$ s from rest):

$$\theta =\frac{1}{2}(5)(4{)}^2=40\text{ rad}W=10\times 40=400\text{ J}$$

$$K=\frac{1}{2}(2)(20{)}^2=400\text{ J}$$

The work done equals the kinetic energy gained.

Worked example 2 — power. At the moment $\omega =20$ rad/s, the torque delivers $P=10\times 20=200$ W.

Worked example 3 — a motor. A motor giving $50$ N m at $100$ rad/s has output power

$$P=50\times 100=5000\text{ W}=5\text{ kW}$$

An everyday example. A kitchen mixer-grinder runs at very high angular speed, so even a modest torque means substantial power, $P=\tau \omega$.

The substance. Rolling bodies have both kinds of kinetic energy, $\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$.

**Each linear quantity has a rotational partner linked through the radius — $v=\omega r$ and $a_t=\alpha r$ — and each linear law has a rotational form with $I$ in place of $m$ and $\tau$ in place of $F$.

- Displacement** $x$ $\leftrightarrow$ angle $\theta$
- Velocity $v$ $\leftrightarrow$ angular velocity $\omega$
- Acceleration $a$ $\leftrightarrow$ angular acceleration $\alpha$
- Mass $m$ $\leftrightarrow$ moment of inertia $I$
- Force $F=ma$ $\leftrightarrow$ torque $\tau =I\alpha$
- Momentum $p=mv$ $\leftrightarrow$ angular momentum $L=I\omega$
- Work $Fs$ $\leftrightarrow$ $\tau \theta$; power $Fv$ $\leftrightarrow$ $\tau \omega$
- Kinetic energy $\frac{1}{2}m{v}^2$ $\leftrightarrow$ $\frac{1}{2}I{\omega }^2$

Worked example — twin problems. A $6$ N force acts for $3$ s on a $2$ kg block at rest; a $6$ N m torque acts for $3$ s on a wheel with $I=2$ kg m${}^2$ at rest.

$$a=3{\text{ m/s}}^2,v=9\text{ m/s},s=13.5\text{ m},W=6\times 13.5=81\text{ J}=\frac{1}{2}(2)(9{)}^2$$

$$\alpha =3{\text{ rad/s}}^2,\omega =9\text{ rad/s},\theta =13.5\text{ rad},W=6\times 13.5=81\text{ J}=\frac{1}{2}(2)(9{)}^2$$

An everyday example. A moving bicycle shows both at once: the bicycle translates while its wheels rotate.

The substance. **Unlike mass, $I$ changes with the axis**, so the analogy holds only once the axis is fixed.