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Topic 10 of 19

Rotational Kinematics

Rotational kinematics is the angular counterpart of 1D kinematics. Every linear kinematic quantity has a rotational analog: displacement → angle, velocity → angular velocity, acceleration → angular acceleration. The kinematic equations take the same form, making this topic straightforward once the 1D version is solid.

10.1 Rotational Variables

When a rigid body rotates about a fixed axis, every particle in the body moves in a circle centered on that axis. We describe the entire rotation with a single angle variable — no need to track individual particle positions separately.

Angular Position

Choose a reference line fixed in the body, perpendicular to the rotation axis. The angular position θ is the angle between that reference line and a fixed external reference direction:

\theta = \frac{s}{r}

where s is the arc length subtended at radius r. The unit is the radian (rad) — a pure ratio with no physical dimensions.

Unit conversions (memorize these):
1 revolution = 2π rad = 360°
1 rad = 360°/(2π) ≈ 57.3°
1 rpm = 2π/60 rad/s ≈ 0.1047 rad/s

Angular Displacement, Velocity, and Acceleration

As the body rotates from θ₁ to θ₂, the angular displacement is Δθ = θ₂ − θ₁. The sign convention follows the right-hand rule: counterclockwise (CCW) is positive, clockwise (CW) is negative.

\omega_{\text{avg}} = \frac{\Delta\theta}{\Delta t}, \qquad \omega = \frac{d\theta}{dt}

\alpha_{\text{avg}} = \frac{\Delta\omega}{\Delta t}, \qquad \alpha = \frac{d\omega}{dt}

The unit of ω is rad/s; α is rad/s². These are the scalar angular velocity and angular acceleration for rotation about a fixed axis. If ω and α share the same sign, the rotation is speeding up; opposite signs mean slowing down.

A rigid body rotating about fixed axis O. The reference line sweeps angle θ from the positive x-axis. Angular velocity ω > 0 (purple arc) means counterclockwise rotation.

10.2 Rotation with Constant Angular Acceleration

When angular acceleration α is constant, the five rotational kinematic equations below follow by direct integration of the definitions of ω and α. They are in exact structural correspondence with the linear kinematic equations from Chapter 2 — substitute θ for x, ω for v, and α for a.

Table 10-1 — Equations for Constant Angular Acceleration

Rotational equation	Missing quantity	Linear analog
ω = ω₀ + αt	Δθ	v = v₀ + at
θ − θ₀ = ω₀t + ½αt²	ω	x − x₀ = v₀t + ½at²
ω² = ω₀² + 2α(θ − θ₀)	t	v² = v₀² + 2a(x − x₀)
θ − θ₀ = ½(ω₀ + ω)t	α	x − x₀ = ½(v₀ + v)t
θ − θ₀ = ωt − ½αt²	ω₀	x − x₀ = vt − ½at²

The same problem-solving strategy as Chapter 2 applies: list known and unknown quantities, identify which equation contains only one unknown, and solve. The structural similarity is not a coincidence — rotation about a fixed axis is mathematically one-dimensional motion.

Grindstone example (Sample Problem 10.01):
A grindstone rotates at ω₀ = 6.0 rad/s. A tool pressed against it produces α = −0.50 rad/s².
(a) How long until ω = 0? Using ω = ω₀ + αt:
0 = 6.0 + (−0.50)t → t = 12 s
(b) Total angle? θ − θ₀ = ½(ω₀ + ω)t = ½(6.0)(12) = 36 rad ≈ 5.7 rev

Once the grindstone reaches ω = 0 it stops — it does not reverse unless a torque acts in the opposite direction. Always check whether the answer makes physical sense before continuing.

10.3 Relating Linear and Angular Variables

A point at radius r from the rotation axis moves along a circular arc. Its linear quantities are directly proportional to the angular quantities through the radius:

s = \theta r \qquad (\theta \text{ must be in radians})

v = \omega r \qquad (\text{tangential speed})

a_t = \alpha r \qquad (\text{tangential acceleration, along velocity})

a_r = \omega^2 r = \frac{v^2}{r} \qquad (\text{centripetal acceleration, toward axis})

These two acceleration components are always perpendicular. The total acceleration magnitude is:

a = \sqrt{a_t^2 + a_r^2}

Point P on a rotating body at radius r from axis O. The tangential component aₜ = αr is parallel to v; the centripetal component aᵣ = ω²r points inward. They are always perpendicular.

Period and frequency:
T = 2π/ω (seconds per revolution)
f = ω/(2π) = 1/T (Hz, revolutions per second)
ω = 2πf (the conversion used most often in problems)

On a merry-go-round, all riders share the same ω. But riders at larger radius r have higher tangential speed v = ωr and higher centripetal acceleration aᵣ = ω²r. The outside moves faster — this is the direct consequence of v = ωr.

10.4 Kinetic Energy of Rotation and Rotational Inertia

A rotating rigid body has kinetic energy even if its center of mass is stationary. Each particle of mass mᵢ at radius rᵢ has speed vᵢ = ωrᵢ and contributes ½mᵢvᵢ² to the total KE. Summing over all particles and factoring out the common ω:

K = \sum \tfrac{1}{2}m_i v_i^2 = \tfrac{1}{2}\!\left(\sum m_i r_i^2\right)\!\omega^2 = \tfrac{1}{2}I\omega^2

where the rotational inertia (moment of inertia) is defined as:

I = \sum m_i r_i^2 \qquad \text{SI unit: kg\,m}^2

I plays the same role in rotation that mass plays in translation. A large I means the object strongly resists angular acceleration, just as large mass resists linear acceleration.

Critical point: I depends on both (1) the mass distribution and (2) the choice of rotation axis. The same object has a different I for a different axis — always specify the axis when stating a value of I.

Translation–Rotation Analogy (so far)

Translation	Rotation
Mass m (kg)	Rotational inertia I (kg·m²)
Linear velocity v (m/s)	Angular velocity ω (rad/s)
Linear acceleration a (m/s²)	Angular acceleration α (rad/s²)
Kinetic energy K = ½mv²	Rotational KE K = ½Iω²

10.5 Calculating Rotational Inertia

For a continuous body the discrete sum becomes an integral:

I = \int r^2 \, dm

where r is the perpendicular distance from mass element dm to the rotation axis. For uniform bodies with standard geometry, these integrals have been evaluated and tabulated:

Table 10-2 — Rotational Inertia of Common Uniform Bodies

Body and axis	I
Thin hoop radius R — central axis (⊥ to plane)	MR²
Annular cylinder, inner R₁ outer R₂ — central axis	½M(R₁² + R₂²)
Solid cylinder (disk), radius R — central axis	½MR²
Thin rod, length L — axis through center ⊥	ML²/12
Thin rod, length L — axis through one end ⊥	ML²/3
Solid sphere, radius R — any diameter	2MR²/5
Thin spherical shell, radius R — any diameter	2MR²/3
Thin hoop, radius R — any diameter	MR²/2

Parallel-Axis Theorem

If I_com is the rotational inertia about an axis through the center of mass, then for any parallel axis displaced a distance h:

I = I_{\text{com}} + Mh^2

I is always larger than I_com for any displaced axis. The center-of-mass axis gives the minimum rotational inertia for a given axis direction.

Example — thin rod, axis at one end:
I_com = ML²/12 (axis through center)
h = L/2 (displacement to one end)
I_end = ML²/12 + M(L/2)² = ML²/12 + 3ML²/12 = ML²/3 ✓

Three common shapes with their moments of inertia. Left: solid disk (½MR²). Center: thin rod about center (ML²/12). Right: solid sphere (2MR²/5). Yellow dashed lines show the rotation axis.

10.6–10.7 Torque and Newton's Second Law for Rotation

Torque τ is the rotational analog of force — it is what causes angular acceleration. For a force F applied at a point P located at distance r from the rotation axis, with angle φ between r and F:

\tau = r F \sin\phi = r_{\perp} F = r F_t

where r⊥ = r sinφ is the moment arm (perpendicular distance from the axis to the line of action of F), and Fₜ = F sinφ is the tangential component of F. All three forms give the same result — use the most convenient.

Sign convention: Torques that would produce CCW rotation are positive; CW torques are negative. This is the same convention used for Δθ, ω, and α.

Force F applied at point P (distance r from axis O). The torque τ = rF sinφ. The moment arm r⊥ = r sinφ is the perpendicular distance from O to the line of action (dashed) of F.

Newton's Second Law for Rotation

Applying Newton's second law to each particle of a rigid body rotating about a fixed axis, multiplying each by the particle's radius, and summing gives the central equation of rotational dynamics:

\tau_{\text{net}} = I\alpha

The net torque about the rotation axis equals the rotational inertia times the angular acceleration. The sign of α matches the sign of τ_net. The axis for I and the axis for τ_net must be the same — mixing axes is a common error.

Door analogy: Pushing near the hinge (small r⊥) requires far more force than pushing at the outer edge (large r⊥) to achieve the same torque τ = r⊥F and thus the same angular acceleration α = τ/I. This is why door handles are placed at the far edge, not near the hinge.

10.8 Work and Rotational Kinetic Energy

As torque rotates a body through angle dθ, it does work dW = τ dθ. For a finite rotation:

W = \int_{\theta_i}^{\theta_f} \tau \, d\theta

W = \tau(\theta_f - \theta_i) \qquad (\text{constant } \tau \text{ only})

The instantaneous power delivered by a torque is:

P = \frac{dW}{dt} = \tau\frac{d\theta}{dt} = \tau\omega

The work–energy theorem for rotation states that the net work done by all external torques equals the change in rotational kinetic energy:

\Delta K = \tfrac{1}{2}I\omega_f^2 - \tfrac{1}{2}I\omega_i^2 = W_{\text{net}}

Motor spinning a flywheel:
A motor exerts constant τ = 10 N·m on a flywheel with I = 2.0 kg·m² starting from rest. After 5 revolutions (Δθ = 10π rad):
W = τ Δθ = (10)(10π) = 100π J
½Iω² = 100π → ω² = 100π → ω ≈ 17.7 rad/s

Table 10-3 — Complete Translation–Rotation Correspondence

Quantity	Translation	Rotation
Displacement	x	θ
Velocity	v = dx/dt	ω = dθ/dt
Acceleration	a = dv/dt	α = dω/dt
Inertia	m (kg)	I = ∫r²dm (kg·m²)
Newton's 2nd law	F_net = ma	τ_net = Iα
Work	W = ∫F dx	W = ∫τ dθ
Kinetic energy	K = ½mv²	K = ½Iω²
Power	P = Fv	P = τω
Work–energy theorem	ΔK = W_net	ΔK = W_net

The translation–rotation correspondence table is the conceptual core of Chapter 10. Every rotational concept mirrors its linear counterpart exactly, with I ↔ m, ω ↔ v, α ↔ a, and τ ↔ F. Master this table and the entire structure of rotational mechanics follows immediately.

Key Concepts

Angular Position

Angle

\theta

(in radians) describing orientation. Arc length and angle relate by

s = r\theta

. One full revolution =

2\pi

rad ≈ 6.28 rad = 360°.

Angular Velocity

Rate of change of angle:

\omega = d\theta/dt

(rad/s). Positive conventionally means counterclockwise. Related to period:

\omega = 2\pi f = 2\pi/T

Angular Acceleration

Rate of change of angular velocity:

\alpha = d\omega/dt

(rad/s²). Constant angular acceleration gives rise to the rotational kinematic equations.

Linear–Angular Relations

For a point at radius

r

from the axis: tangential speed

v = r\omega

, tangential acceleration

a_t = r\alpha

, centripetal acceleration

a_c = r\omega^2 = v^2/r

Rotational Kinematic Equations

Identical in form to the 1D kinematic equations with

\theta \leftrightarrow x

\omega \leftrightarrow v

\alpha \leftrightarrow a

. Valid for constant

\alpha

only.

Period & Frequency

Period

T

is the time for one full rotation. Frequency

f = 1/T

is rotations per second (Hz). Angular frequency

\omega = 2\pi f

Key Equations

Arc length

s = r\theta

θ must be in radians.

Tangential and centripetal acceleration

v = r\omega, \quad a_t = r\alpha, \quad a_c = \frac{v^2}{r} = r\omega^2

Relations between linear and angular quantities for a point at radius r.

Rotational kinematic equations

\omega = \omega_0 + \alpha t \quad\quad \theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2 \quad\quad \omega^2 = \omega_0^2 + 2\alpha\Delta\theta

Direct analogs of the 1D kinematic equations. Valid only for constant α.

Period and angular frequency

T = \frac{2\pi}{\omega}, \quad f = \frac{\omega}{2\pi}, \quad \omega = 2\pi f

Conversions between period, frequency, and angular frequency.

Worked Example

Spinning Up a Wheel

Problem

A wheel starts from rest and reaches 120 rpm in 4 seconds under constant angular acceleration. Find $\alpha$ and the number of revolutions completed.

Solution

Convert 120 rpm to rad/s:

\omega_f = 120 \times \frac{2\pi}{60} = 4\pi \approx 12.6 \text{ rad/s}

Find angular acceleration:

\alpha = \frac{\omega_f - \omega_0}{t} = \frac{4\pi - 0}{4} = \pi \approx 3.14 \text{ rad/s}^2

Find total angle:

\Delta\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = 0 + \tfrac{1}{2}(\pi)(16) = 8\pi \text{ rad}

\text{revolutions} = \frac{8\pi}{2\pi} = 4 \text{ rev}

Answer α = π rad/s² ≈ 3.14 rad/s²; the wheel completes 4 revolutions.

Practice

Exercises

7 problems

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2 free · 5 Pro

Exercise 1 / 7 Free

+10 XP

Watch the wheel spin. Use the Start/Stop Measure buttons to measure Δθ over a time interval Δt. Then calculate ω = Δθ/Δt (in rad/s).

ω = rad/s

Exercise 2 / 7 Free

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The ω-t graph shows angular velocity vs time. Use the yellow triangle to find the slope — which equals the angular acceleration α.

α = rad/s²

3 of 7

A point on a wheel of radius $0.4 \text{ m}$ moves at $2 \text{ m/s}$ . What is the wheel's angular velocity?

Unlock Exercise 3