Unit 8 · Introduction to Physics I

Rotation & Angular Momentum

Everything from Units 2 to 7, restated for spinning objects. Every linear quantity has a rotational twin, every linear equation a rotational copy. Learn the dictionary and the unit is mostly translation.

By the end you can

Translate between linear and rotational quantities.
Compute torque and use $\tau = I\alpha$.
Explain why moment of inertia depends on where the mass sits.
Apply angular momentum conservation to a spinning system.

Why it matters & where it comes from

Why it exists

A rigid body has parts moving at different speeds, so $F = ma$ cannot be applied to it as a whole. Rotational quantities describe the object with a single set of numbers, whatever its parts are doing.

In the real world

Flywheels store energy, gyroscopes hold direction, and a figure skater pulling in her arms speeds up. Every engine, turbine and wheel is a rotational dynamics problem.

Story & open questions

Euler extended Newton's laws to rigid bodies in the 1750s, seventy years after the Principia. Angular momentum conservation turns out to follow from the fact that space has no preferred direction — a connection Emmy Noether proved in 1918.

The dictionary

Rule Every linear quantity has a twin

Position $x$ → angle $\theta$, in radians.

Velocity $v$ → angular velocity $\omega$, with $v = r\omega$.

Acceleration $a$ → angular acceleration $\alpha$, with $a = r\alpha$.

Mass $m$ → moment of inertia $I$.

Force $F$ → torque $\tau$.

$F = ma$ → $\tau = I\alpha$.

$p = mv$ → $L = I\omega$.

$K = \tfrac12mv^2$ → $K = \tfrac12I\omega^2$.

The constant-acceleration equations of Unit 2 carry over unchanged with the substitutions above.

Radians are not optional

The relations $v = r\omega$ and $a = r\alpha$ hold only when angles are in radians. Degrees give answers wrong by a factor of about $57$. Convert first.

Torque

Definition

Torque measures a force's ability to cause rotation:

$\tau = rF\sin\theta$

where $r$ is the distance from the axis to the point of application and $\theta$ the angle between $\vec{r}$ and $\vec{F}$. Equivalently $\tau = r_\perp F$, using the perpendicular distance from the axis to the force's line of action — the lever arm.

Example Why door handles are far from hinges

Pushing at the handle, $0.8$ m from the hinge, with $20$ N perpendicular gives $\tau = 16$ N·m.

The same push $0.1$ m from the hinge gives $\tau = 2$ N·m — eight times less.

Pushing directly toward the hinge gives $\sin\theta = 0$ and no torque at all, however hard you push. Direction matters as much as distance.

Rule Equilibrium needs both

A rigid body is in equilibrium when $\sum F = 0$ and $\sum \tau = 0$. Balanced forces alone are not enough — two equal opposite forces along different lines make a couple, which spins the object while its centre stays put.

Moment of inertia

Definition

For point masses, $I = \sum m_i r_i^2$. It plays the role of mass in rotation, but unlike mass it depends on the axis: the same object has different $I$ about different axes.

Example Standard shapes, all about the centre

Hoop or thin ring: $I = mR^2$ — all the mass at the full radius.

Solid cylinder or disc: $I = \tfrac12 mR^2$.

Solid sphere: $I = \tfrac25 mR^2$.

Rod about its centre: $I = \tfrac{1}{12}mL^2$; about one end, $\tfrac13 mL^2$ — four times larger, because more mass sits far from the axis.

The pattern: mass further from the axis contributes more, and it does so as the square of the distance.

Example The rolling race

A hoop and a solid cylinder of equal mass and radius roll from rest down the same incline. Which wins?

Energy splits two ways: $mgh = \tfrac12mv^2 + \tfrac12I\omega^2$, with $\omega = v/R$ for rolling without slipping.

Substituting $I = cmR^2$ gives $v = \sqrt{\dfrac{2gh}{1+c}}$.

Hoop has $c = 1$ and cylinder $c = \tfrac12$, so the cylinder is faster. Neither mass nor radius appears — only the shape, through $c$. A solid sphere, with $c = \tfrac25$, beats both.

Angular momentum

Definition

$L = I\omega$ for a rigid body, and $L = mvr\sin\theta$ for a point mass about a chosen axis. It is a vector, along the rotation axis.

Rule Conservation

When the net external torque is zero, total angular momentum is conserved:

$I_i\omega_i = I_f\omega_f$

Note this is a separate condition from linear momentum conservation. A system can conserve one and not the other.

Example The spinning skater

A skater spinning with arms out at $2$ rev/s pulls them in, halving her moment of inertia.

$L$ is conserved, so halving $I$ doubles $\omega$: she spins at $4$ rev/s.

Kinetic energy is $\tfrac12I\omega^2$, which doubles. That energy came from her muscles doing work pulling her arms inward against the outward tendency — energy is not conserved here, and it should not be expected to be.

Example Why a neutron star spins fast

A star of radius $700{,}000$ km collapses to $10$ km, keeping its mass.

$I \propto R^2$, so $I$ falls by a factor of about $5 \times 10^9$.

$\omega$ rises by the same factor. A star rotating once a month ends up spinning many times a second — which is exactly what pulsars do.

Zero net force does not mean zero net torque

Two equal opposite forces applied at different points sum to zero force but a nonzero torque, and the object spins up while its centre of mass stays at rest. Check both conditions separately.

Common pitfalls

• Using degrees in $v = r\omega$.
• Quoting a moment of inertia without saying about which axis.
• Forgetting rotational kinetic energy when a body rolls. It is both translating and spinning.
• Expecting kinetic energy to be conserved when angular momentum is.
• Assuming a force through the axis produces torque. Its lever arm is zero.

Practice

Work each fully before opening the answer. Original exercises written for this guide in the style of the course's problem sets.

Practice 1 · Torque

A $30$ N force acts $0.4$ m from an axis, at $60$ degrees to the radius. Find the torque.

Answer

$\tau = (0.4)(30)\sin 60 = 10.4$ N·m.

Practice 2 · Zero torque

When does a force produce no torque?

Answer

When its line of action passes through the axis, so the lever arm is zero — equivalently when $\sin\theta = 0$.

Practice 3 · Rolling race

A hoop, a disc and a sphere roll down an incline. Order them by arrival time.

Answer

Sphere first ($c = 0.4$), then disc ($0.5$), then hoop ($1$). Smaller $c$ means less energy diverted into spinning.

Practice 4 · Skater

A skater at $3$ rev/s reduces $I$ to a third. Find the new rate.

Answer

$9$ rev/s, since $I\omega$ is fixed.

Practice 5 · Rod on an axis

A uniform $2$ kg, $1$ m rod pivots at one end and is released horizontally. Find its initial angular acceleration.

Solution

Torque from weight acting at the centre: $\tau = mg(L/2) = 9.8$ N·m. With $I = \tfrac13mL^2 = 0.667$ kg·m$^2$, $\alpha = 9.8/0.667 = 14.7$ rad/s$^2$.

Practice 6 · Rolling energy

What fraction of a rolling disc's kinetic energy is rotational?

Solution

Translational $\tfrac12mv^2$; rotational $\tfrac12(\tfrac12mR^2)(v/R)^2 = \tfrac14mv^2$. So rotational is a third of the total.

Practice 7 · Both conditions

Give a situation with zero net force but nonzero net torque.

Answer

Two equal opposite forces applied at opposite ends of a rod — a couple. The centre of mass stays still while the rod spins up.

Practice 8 · Merry-go-round

A $40$ kg child runs at $3$ m/s tangentially and jumps onto the rim of a stationary $200$ kg, $2$ m radius merry-go-round ($I = \tfrac12MR^2$). Find the final angular speed.

Solution

Child's angular momentum: $mvr = 40 \times 3 \times 2 = 240$ kg·m$^2$/s. Final moment of inertia: $\tfrac12(200)(4) + (40)(4) = 560$ kg·m$^2$. So $\omega = 240/560 = 0.43$ rad/s. This is a perfectly inelastic collision in rotational form.

Expanded topic checklist

Course coverage. This unit is part of the complete PHY131H1 topic map. It should cover the following items before you move on:

The dictionary. Linear to rotational correspondences and the relations $v = r\omega$, $a = r\alpha$.
Torque. Lever arm, the angle dependence, and rotational equilibrium.
Moment of inertia. Axis dependence and the standard shapes.
Angular momentum. Conservation when net torque is zero, and why energy need not be conserved with it.

Textbook/source alignment

The calendar lists "angular momentum" explicitly. Check the syllabus on Quercus for assigned problems; the rolling-body comparison is a standard exam question.

Core idea map

Rotation reuses every earlier idea with new symbols. Torque drives angular acceleration; moment of inertia resists it; angular momentum is conserved when torque vanishes. The one genuinely new idea is that inertia now depends on geometry, not just on how much mass there is.

Study routine. Write the linear equation first, then translate symbol by symbol. Almost every rotational formula can be reconstructed that way rather than memorised.

Deep practice extensions

Use these after the main practice set. They push on definitions and edge cases rather than routine calculation.

Extension 1 · Axis dependence

Why does moment of inertia depend on the axis?

Guide

Because each mass element contributes $mr^2$, and $r$ is measured from the axis. Move the axis and every distance changes.

Extension 2 · Rolling

Why does a hoop lose a rolling race?

Guide

Its mass is all at the rim, giving the largest moment of inertia, so more of the released potential energy goes into spinning rather than moving.

Extension 3 · Conservation condition

What must vanish for angular momentum to be conserved?

Guide

The net external torque — not the net force. They are separate conditions.

Extension 4 · Energy in a spin-up

Why does a skater's kinetic energy rise when she pulls in her arms?

Guide

She does work pulling them inward. Angular momentum is conserved; energy is supplied by her muscles, so it is not.

← Unit 07Momentum & collisions Unit 09 →Oscillations