Part 1 · Chapter 10

Rotation

Every translational idea reborn for spinning bodies — angle for position, torque for force, rotational inertia for mass

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 45 min
i What you'll learn

Rotation is translational mechanics seen in a mirror — every idea you already own returns with an angular twin. By the end of this chapter you will be able to:

  1. Define the rotational variables — angular position \(\theta\), displacement \(\Delta\theta\), velocity \(\omega=\dfrac{d\theta}{dt}\), and acceleration \(\alpha=\dfrac{d\omega}{dt}\) — as exact analogues of their linear cousins.
  2. Apply the constant-angular-acceleration equations, the term-for-term mirror of the Chapter 2 kinematics — e.g. \(\omega=\omega_0+\alpha t\) and \(\theta-\theta_0=\omega_0 t+\tfrac12\alpha t^2\).
  3. Bridge the linear and angular worlds through the radius \(r\): \(\;s=\theta r,\quad v=\omega r,\quad a_t=\alpha r,\quad a_r=\omega^2 r.\)
  4. Compute rotational kinetic energy \(K=\tfrac12 I\omega^2\) and rotational inertia \(I=\sum_i m_i r_i^{\,2}=\displaystyle\int r^2\,dm\), and relocate the axis with the parallel-axis theorem \(I=I_{\text{com}}+Mh^2\).
  5. Use torque \(\tau=r_\perp F=rF\sin\phi\) and Newton's second law for rotation \(\tau_{\text{net}}=I\alpha\) to predict angular accelerations.
  6. Track work and power in rotation with \(W=\displaystyle\int \tau\,d\theta\) and \(P=\tau\omega\), and confirm the work–energy theorem \(W=\Delta K\).
  7. Reason about direction — why \(\omega\) and \(\alpha\) are vectors fixed by the right-hand rule, while finite angular displacements are not.
Section 10-1

What Is Physics?

Until now we have studied only translation, where an object moves along a line. This chapter turns to rotation, where an object turns about an axis. You meet it everywhere — in every machine, every spinning wheel, the lift on a well-driven golf ball, the curve on a thrown baseball, even metal fatigue in aging aircraft. The strategy is elegant: almost every concept from translational mechanics carries over, provided we swap in the right rotational stand-in. Force becomes torque, mass becomes rotational inertia, and Newton's second law and the work–energy theorem reappear in angular dress.

The rules of the game. We study a rigid body (all parts locked together, no change of shape) rotating about a fixed axis (one that does not move). That rules out a ball of gas like the Sun and a bowling ball rolling down a lane — the latter mixes rotation with translation, which waits for Chapter 11.
Section 10-2

The Rotational Variables

Fix a reference line in the body, perpendicular to the rotation axis. Its angular position \(\theta\) is the angle it makes with a chosen zero direction. From the geometry of a circular arc:

Angular position (radian measure)
\[ \theta = s / r \qquad 1 \mathrm{rev} = 360^{\circ} = 2\pi \mathrm{rad} \]
Here s is the arc length and r the radius. The radian is a ratio of two lengths, so it is dimensionless; 1 rad ≈ 57.3°. We do not reset θ to zero each revolution — two full turns means θ = 4π rad.

The remaining variables are defined exactly as in one-dimensional motion, but for the angle. The angular displacement is \(\Delta \theta = \theta _{2} - \theta _{1}\). Differentiating gives angular velocity, then angular acceleration:

Angular velocity and angular acceleration
\[ \omega = d\theta /dt \qquad \alpha = d\omega /dt \]
Average versions use Δθ/Δt and Δω/Δt over a finite interval. Units: rad/s and rad/s². The magnitude of ω is the angular speed.
↺ Sign convention

Counterclockwise is positive, clockwise is negative. The memory aid: "clocks are negative." A negative Δθ means the reference line ended up clockwise of where it started.

≡ Every particle agrees

Because the body is rigid, \(\theta\), \(\omega\), and \(\alpha\) are the same for every particle in it — they are properties of the whole body, not of one point.

Section 10-3

Are Angular Quantities Vectors?

Angular velocity and angular acceleration are vectors. Their direction is set by a right-hand rule: curl the fingers of your right hand in the sense of rotation, and your extended thumb points along the axis in the direction of \(\omega\). A subtle point — the vector lies along the axis; nothing actually moves in that direction. The vector defines the axis of the spin, not a path.

⚠️
The caution about angular displacements
Finite angular displacements are NOT vectors.

To be a vector, a quantity must add commutatively (order shouldn't matter). Give a book two 90° turns about different axes in one order, then reverse the order: it ends up facing differently. Since the result depends on the order, finite angular displacements fail the test. (Infinitesimally small rotations do behave as vectors — which is why \(\omega\) and \(\alpha\) are fine.)

For rotation about a single fixed axis — the only case in this chapter — we don't need full vector machinery. A plus sign means counterclockwise, a minus sign clockwise. That's all the "direction" we require.

Section 10-4

Rotation with Constant Angular Acceleration

When \(\alpha\) is constant, the angular kinematics are the term-by-term twin of the constant-linear-acceleration equations of Chapter 2 — just replace \(x\to \theta\), \(v\to \omega\), \(a\to \alpha\).

Constant angular-acceleration equations
\[\begin{gathered} \omega = \omega _{0} + \alpha t \\ \theta - \theta _{0} = \omega _{0}t + \tfrac{1}{2}\alpha t^{2} \\ \omega ^{2} = \omega _{0}^{2} + 2\alpha (\theta - \theta _{0}) \\ \theta - \theta _{0} = \tfrac{1}{2}(\omega _{0} + \omega )t \\ \theta - \theta _{0} = \omega t - \tfrac{1}{2}\alpha t^{2} \end{gathered}\]
The first two are the basic pair; the rest follow by eliminating one variable. Pick the equation whose only unknown is the quantity you want — or just keep the first two and solve them simultaneously.
Watch the signs. If ω0 and α have opposite signs, the body first slows, momentarily stops, and reverses — exactly like a ball thrown straight up. The grindstone in the worked examples does precisely this.
Section 10-5

Relating the Linear and Angular Variables

Each particle in a rotating body sweeps its own circle of radius \(r\) (its perpendicular distance from the axis). All share the same \(\omega\) and \(\alpha\), but a particle farther out covers a longer arc, so it moves faster. The link between the two worlds is always the radius:

Linear–angular relations (radian measure)
\[\begin{gathered} s = \theta r \qquad v = \omega r \qquad a_{t} = \alpha r \\ a_{r} = v^{2}/r = \omega ^{2}r \qquad T = 2\pi /\omega \end{gathered}\]
v is tangent to the circle. The acceleration has two parts: a tangential at (present only when α ≠ 0, it changes the speed) and a radial ar (present whenever ω ≠ 0, it changes the direction). T is the period of one revolution.
⟳ Radial (centripetal)

Points inward toward the axis, magnitude \(\omega ^{2}r\). It always exists for a turning body — it's what bends the velocity into a circle.

→ Tangential

Along the direction of motion, magnitude \(\alpha r\). It exists only when the body is speeding up or slowing down.

Section 10-6

Kinetic Energy of Rotation

A spinning saw blade clearly has kinetic energy, yet its center of mass may be at rest, so \(\tfrac{1}{2}Mv_{\text{com}}^{2}\) gives zero. Instead, add up \(\tfrac{1}{2}m_{i}v_{i}^{2}\) over all particles, substitute \(v_{i} = \omega r_{i}\), and factor out the common \(\omega ^{2}\):

Rotational kinetic energy and rotational inertia
\[ K = \tfrac{1}{2}I\omega ^{2} \qquad \text{where} \qquad I = \sum m_{i}r_{i}^{2} \]
The quantity I — the rotational inertia (moment of inertia) — captures how the mass is distributed about the axis. It is the rotational stand-in for mass M, and ω must be in radians. SI unit: kg·m².
Distribution matters, not just amount. Spin a long rod about its length and it turns easily; spin the same rod about its center, perpendicular to its length, and it resists much more. Same mass, very different I — because the mass sits farther from the axis in the second case. Smaller I means easier rotation.
Section 10-7

Calculating the Rotational Inertia

For a handful of particles, sum \(mr^{2}\) directly. For a continuous body, replace the sum with an integral over its mass:

Rotational inertia of a continuous body
\[ I = \int r^{2} dm \]
Standard results (thin rod about its center: I = ¹⁄₁₂ML²; solid disk about its axis: I = ½MR²; solid sphere: I = ⅖MR²; hoop: I = MR²) come from this integral and are tabulated for common shapes.
∥
The parallel-axis theorem
I = Icom + Mh²

If you know the rotational inertia \(I_{\text{com}}\) about an axis through the center of mass, the inertia about any parallel axis a perpendicular distance \(h\) away is just \(I_{\text{com}}\) plus \(Mh^{2}\). Shifting the axis away from the com always increases the rotational inertia.

Example of the theorem at a glance: a thin rod about its center has \(I_{\text{com}} = \tfrac{1}{12}ML^{2}\). Move the axis to one end (\(h = \tfrac{1}{2}L\)) and the theorem gives \(\tfrac{1}{12}ML^{2} + M(\tfrac{1}{2}L)^{2} = \tfrac{1}{3}ML^{2}\) — matching the direct integral, with far less work.

Section 10-8

Torque

A doorknob sits far from the hinges for a reason. Turning a body depends on the force, on where it acts, and on its direction. Combining these, the torque of a force \(F\) applied at position \(r\) from the axis, at angle \(\varphi\) between them, is:

Torque — three equivalent forms
\[ \tau = r F \sin \varphi = r F_{t} = r_{\perp} F \]
Ft = F sin φ is the tangential component (only it rotates the body; the radial part is wasted). r⊥ = r sin φ is the moment arm — the perpendicular distance from the axis to the line of action of F. SI unit: the newton-metre (N·m) — never call it a joule, even though the units match.
± Sign of torque

Positive if it would turn the body counterclockwise from rest, negative if clockwise. Same "clocks are negative" rule as before.

Σ Net torque

Torques superpose: the net torque \(\tau _{\text{net}}\) is the algebraic sum of the individual torques about the axis.

Section 10-9

Newton's Second Law for Rotation

For a single particle of mass \(m\) at radius \(r\), the tangential force gives \(F_{t} = ma_{t}\). Multiply by \(r\) and use \(a_{t} = \alpha r\): the torque becomes \(\tau = (mr^{2})\alpha = I\alpha\). Summing over all particles of a rigid body gives the rotational form of the second law:

🔁
Newton's second law for rotation
τnet = Iα

The net torque equals rotational inertia times angular acceleration — the exact analogue of \(F_{\text{net}} = ma\), with \(F\to \tau\), \(m\to I\), \(a\to \alpha\). As always, \(\alpha\) must be in radian measure.

Section 10-10

Work and Rotational Kinetic Energy

When a torque turns a body, it does work, and the work–kinetic energy theorem reappears with rotational quantities throughout:

Work, power, and the work–energy theorem (rotation)
\[\begin{gathered} W = \int \tau d\theta \qquad (W = \tau \Delta \theta \text{for} \text{constant} \tau ) \\ P = dW/dt = \tau \omega \\ \Delta K = \tfrac{1}{2}I\omega _{f}^{2} - \tfrac{1}{2}I\omega _{i}^{2} = W \end{gathered}\]
Line for line, these mirror the translational W = ∫F dx, P = Fv, and ΔK = ½mv² differences. Torque does work through an angle just as force does work through a distance.
The full analogy in one breath: position \(x\leftrightarrow \theta\), velocity \(v\leftrightarrow \omega\), acceleration \(a\leftrightarrow \alpha\), mass \(m\leftrightarrow I\), force \(F\leftrightarrow \tau\), momentum's law \(F=ma\leftrightarrow \tau =I\alpha\), kinetic energy \(\tfrac{1}{2}mv^{2}\leftrightarrow \tfrac{1}{2}I\omega ^{2}\), power \(Fv\leftrightarrow \tau \omega\). Learn rotation as a translation of mechanics you already know.
Worked Examples

Putting It to Work

1 Angular velocity from angular position

Problem. A disk's reference line obeys \(\theta = -1.00 - 0.600t + 0.250t^{2}\) (rad, s). When does \(\theta\) reach its minimum, and what is that value?

Solution. The angular velocity is the derivative; setting it to zero locates the extremum.

Differentiate, then set ω = 0
\[\begin{gathered} \omega = d\theta /dt = -0.600 + 0.500t \\ 0 = -0.600 + 0.500t \;\Longrightarrow\; t_{\min} = 1.20 s \\ \theta (1.20) = -1.00 - 0.600(1.20) + 0.250(1.20)^{2} \approx -1.36 \mathrm{rad} \end{gathered}\]

At \(t_{\min}\) the disk momentarily stops (ω = 0) at its maximum clockwise position, then reverses to turn counterclockwise.

2 The grindstone — opposite signs of ω and α

Problem. A grindstone has \(\alpha = 0.35 \mathrm{rad}/s^{2}\), with \(\omega _{0} = -4.6 \mathrm{rad}/s\) and \(\theta _{0} = 0\) at \(t = 0\). At what time does it momentarily stop?

Use ω = ω₀ + αt with ω = 0
\[ t = (\omega - \omega _{0})/\alpha = (0 - (-4.6))/0.35 \approx 13 s \]

Velocity and acceleration point oppositely, so the stone slows in the negative direction, stops at 13 s, then spins up the other way — the rotational version of a tossed ball at the top of its flight.

3 Spin-test explosion — rotational kinetic energy

Problem. A solid steel disk, \(M = 272 \mathrm{kg}\), \(R = 0.38 m\), spins at 14 000 rev/min. How much energy is stored (and released if it shatters)?

Solution. For a disk \(I = \tfrac{1}{2}MR^{2}\); convert ω to rad/s, then use \(K = \tfrac{1}{2}I\omega ^{2}\).

Inertia, then energy
\[\begin{gathered} I = \tfrac{1}{2}(272)(0.38)^{2} \approx 19.6 \mathrm{kg}\cdot m^{2} \\ \omega = 14000\cdot (2\pi )/60 \approx 1.47\times 10^{3} \mathrm{rad}/s \\ K = \tfrac{1}{2}(19.6)(1.47\times 10^{3})^{2} \approx 2.1\times 10^{7} J \end{gathered}\]

Twenty-one million joules in a half-metre disk — which is exactly why such tests are run inside lead-brick containment.

4 Falling block turns a disk — coupling the two laws

Problem. A uniform disk (\(M = 2.5 \mathrm{kg}\), \(R = 0.20 m\)) on a fixed axle has a cord over its rim holding a block (\(m = 1.2 \mathrm{kg}\)). Find the block's acceleration, the cord tension, and the disk's angular acceleration. The cord does not slip; the axle is frictionless.

Solution. Block: \(T - mg = -ma\) (taking down as the fall). Disk: \(\tau _{\text{net}} = I\alpha\) gives \(TR = (\tfrac{1}{2}MR^{2})\alpha\), and no-slip means \(a = \alpha R\), so \(T = \tfrac{1}{2}Ma\). Combine:

Solve the coupled equations
\[\begin{gathered} a = g \cdot 2m/(M + 2m) = 9.8\cdot (2\cdot 1.2)/(2.5 + 2.4) \approx 4.8 m/s^{2} \\ T = \tfrac{1}{2}Ma = \tfrac{1}{2}(2.5)(4.8) \approx 6.0 N \\ \alpha = a/R = 4.8/0.20 = 24 \mathrm{rad}/s^{2} \end{gathered}\]

As a sanity check: \(a \lt g\) and \(T \lt mg = 11.8 N\), both as expected. A massless disk (M = 0) recovers free fall (a = g, T = 0).

5 Energy delivered to that disk — rotational work

Problem. Starting from rest, the disk of Example 4 has cord tension 6.0 N and \(|\alpha | = 24 \mathrm{rad}/s^{2}\). What is its rotational kinetic energy at \(t = 2.5 s\)?

Solution. Use \(K = \tfrac{1}{2}I\omega ^{2}\) with \(\omega = \alpha t\) and \(I = \tfrac{1}{2}MR^{2}\).

Kinetic energy from ω = αt
\[\begin{gathered} K = \tfrac{1}{2}(\tfrac{1}{2}MR^{2})(\alpha t)^{2} = \tfrac{1}{4}M(R\alpha t)^{2} \\ = \tfrac{1}{4}(2.5)[(0.20)(24)(2.5)]^{2} \approx 90 J \end{gathered}\]

The same 90 J also equals the work the cord's torque does, \(W = \tau \Delta \theta = TR\cdot (\tfrac{1}{2}\alpha t^{2})\) — the work–energy theorem and direct kinematics agree, as they must.

Review

Chapter Summary

✓ Rotational variables

\(\theta = s/r\), \(\omega = d\theta /dt\), \(\alpha = d\omega /dt\). Counterclockwise positive. Same for every particle of a rigid body.

✓ Constant α

Five equations mirroring linear kinematics, e.g. \(\omega = \omega _{0} + \alpha t\) and \(\theta = \theta _{0} + \omega _{0}t + \tfrac{1}{2}\alpha t^{2}\).

✓ Linear ↔ angular

\(s = \theta r\), \(v = \omega r\), \(a_{t} = \alpha r\), \(a_{r} = \omega ^{2}r\). The radius r is the bridge.

✓ Rotational inertia

\(I = \sum mr^{2}\) or \(\int r^{2} dm\); \(K = \tfrac{1}{2}I\omega ^{2}\). Parallel-axis: \(I = I_{\text{com}} + Mh^{2}\).

✓ Torque

\(\tau = rF \sin \varphi = r_{\perp}F\), in N·m. The moment arm and the tangential component are the two factors that matter.

✓ Rotational dynamics

\(\tau _{\text{net}} = I\alpha\); \(W = \int \tau d\theta\); \(P = \tau \omega\); \(\Delta K = W\). Translation, retold for spin.

Practice

Problems

For each problem, decide which rotational quantity is asked for and pick the matching tool: the constant-α equations for kinematics, \(\tau _{\text{net}} = I\alpha\) for dynamics, or \(K = \tfrac{1}{2}I\omega ^{2}\) and the work–energy theorem for energy. Keep angles in radians, and watch the signs of ω and α.

  1. A baseball is thrown toward home plate at 85 mi/h with a spin of 1800 rev/min. Over a 60 ft straight path, how many revolutions does it make?
  2. Find the angular speed (in rad/s) of the (a) second hand, (b) minute hand, and (c) hour hand of a smoothly running analog watch.
  3. A wheel's angular position is \(\theta = 2.0 + 4.0t^{2} + 2.0t^{3}\) (rad, s). Find (a) θ and (b) ω at t = 0, (c) ω at t = 4.0 s, and (d) α at t = 2.0 s. (e) Is α constant?
  4. A drum spinning at 12.60 rad/s slows at a constant 4.20 rad/s². (a) How long until it stops, and (b) through what angle does it turn while stopping?
  5. A disk initially at 120 rad/s slows at a constant 4.0 rad/s². (a) How long to stop, and (b) through what angle does it turn in that time?
  6. An airplane propeller of tip radius 1.5 m turns at 2000 rev/min while the plane flies at 480 km/h parallel to the rotation axis. Find the tip speed as seen by (a) the pilot and (b) a ground observer.
  7. A flywheel (a thin hoop) of mass 32.0 kg and radius 1.20 m spins at 280 rev/min and must stop in 15.0 s. Find (a) the work needed to stop it and (b) the required average power.
  8. A solid block of mass 0.172 kg has edges a = 3.5 cm, b = 8.4 cm, c = 1.4 cm. Find its rotational inertia about an axis through one corner, perpendicular to the large faces.
  9. A meter stick (mass 0.56 kg, treated as a thin rod) rotates about an axis perpendicular to it at the 20 cm mark. Find its rotational inertia. (Hint: parallel-axis theorem.)
  10. A uniform disk (M = 2.5 kg, R = 0.20 m) on a frictionless axle has a 1.2 kg block hung from a cord over its rim. Using energy methods, find the block's speed after it descends 50 cm from rest.
Tip: the single most common slip is leaving an angle in revolutions or degrees when a formula expects radians. Any relation that mixes linear and angular variables — \(s = \theta r\), \(v = \omega r\), \(a_{t} = \alpha r\), \(K = \tfrac{1}{2}I\omega ^{2}\) — was derived in radian measure. Convert first, compute second.