Rotational kinematics and moment of inertia

A mark stops, then turns back

Imagine a marked point on an ideal rigid object rotating about a stationary axle. It initially turns counterclockwise, slows, stops for an instant and then turns clockwise. Does that instant of rest mean its acceleration is zero? We will follow one supplied mathematical motion to answer the question.

Rigid means that distances between points in the object stay fixed. Take a fixed inertial reference frame and an axis that is stationary in that frame. View the rotation from the positive end of the axis toward the origin, and choose counterclockwise as positive. The angle history describes the whole rigid object; points at different distances from the axis can still have different linear speeds.

Units, standards, and physical constants prepares the unit reading; Algebra, functions, graphs, and logarithms prepares substitution and graph interpretation. Geometry, trigonometry, and coordinate systems and Vectors and vector spaces help with radius and perpendicular components. The local rate ideas are explained here; Derivatives, Taylor series, and local approximation offers further preparation for their limiting definition.

Keep the angle history

Let θ be the unwrapped angle of the marked radius from a fixed reference ray. Unwrapped means keeping the accumulated turns: an increase from 0 rad to 2π rad is a complete positive revolution even though the final orientation matches the initial one. A single orientation reading cannot reveal how many turns occurred.

The radian measures plane angle. One radian subtends an arc equal in length to the circle’s radius. A full turn is 2π rad. In the SI, rad is a special name for the unit one; retaining it identifies an angle rather than treating every unit-one quantity as the same physical kind. If Δθ is a signed angle change, the corresponding signed circular arc change at fixed radius r is rΔθ/(1 rad). This arc change is different from the straight-line vector between the point’s endpoints.

For our original example, time t is restricted to 0–3 s. Define the dimensionless time τ = t/(1 s), and supply the angle history θ = (1 + 2τ − τ²/2) rad. The initial angle is 1 rad. This smooth function defines the motion throughout the stated interval; the selected values in the table are not a measured record or a fit to four observations.

Read angle and rate together

First compare short angle changes with the time they take. Average angular velocity over an interval is Δθ/Δt, including its sign. Instantaneous angular velocity ω is the limiting value of that ratio around one time: it is the slope of the angle–time curve. Angular acceleration α is the local rate of change of ω, rather than the angular velocity itself. In derivative notation, ω = dθ/dt; α = dω/dt.

For the supplied polynomial, ω = (2 − τ) rad/s; α = −1 rad/s². The table shows ω decreasing steadily. At 2 s its value is zero and the angle curve is locally horizontal, yet α remains negative. Local rates at the two endpoints can be read as the corresponding one-sided limits within our domain.

In the lower panel, signed area adds angle changes: positive area before 2 s contributes +2 rad, and the negative area from 2 to 3 s contributes −0.5 rad. Thus angle height, angle slope and angular-velocity area give three consistent views of the same supplied motion. The two panels share time but have different vertical quantities and units.

Count both parts of the turn

Solving ω = 0 gives τ = 2, hence t = 2 s. Angular velocity is positive before that time and negative afterward. The angle reaches its maximum of 3 rad there. From the start to the turn, θ increases by 2 rad; from the turn to the end, it decreases by 0.5 rad.

The net angle change is 2 rad − 0.5 rad = +1.5 rad. Total angular travel adds the magnitudes of both parts, giving 2.5 rad. Over 3 s, average angular velocity is +0.5 rad/s, while average angular speed is (5/6) rad/s. Angular speed is |ω| and cannot be negative.

A negative α does not always mean slowing down. Before the turn, positive ω decreases toward zero and the angular speed falls. After the turn, ω becomes more negative and the angular speed rises. For this fixed-axis motion the angular-velocity vector can be written ωe_z, with e_z the chosen fixed positive-axis unit vector. Its direction reverses when the sign of ω reverses; the axis itself stays fixed. At ω = 0 the zero vector has no direction. General finite rotations in three dimensions need more than adding signed angles about unrelated axes.

Separate tangent and inward acceleration

Take a point at perpendicular distance r = 0.5 m from our fixed axis. At its current position, choose a unit tangent in the positive counterclockwise direction and a unit radial direction pointing outward. These two perpendicular directions rotate with the point; they are not fixed Cartesian x and y directions. Its signed tangential velocity is vθ = rω/(1 rad), and its speed is r|ω|/(1 rad). All points share ω, but doubling r doubles the speed at the same time.

The acceleration has two components: aθ = rα/(1 rad); ar = −r[ω/(1 rad)]². Tangential acceleration changes the signed rate around the circle. The radial component accounts for the changing tangent direction and is inward whenever ω is nonzero. Dividing by 1 rad makes explicit that these linear formulas use the numerical angle rate in radians; their units are m/s and m/s².

At 2 s, this point has zero speed and zero radial acceleration, but its tangential acceleration is −0.5 m/s². At 3 s, its tangential velocity is −0.5 m/s, tangential acceleration is −0.5 m/s² and radial acceleration is −0.5 m/s². The last two perpendicular components give total acceleration magnitude √(0.5² + 0.5²) m/s² = (√2/2) m/s². A point exactly on the stationary axis has r = 0 and stays at rest in this model.

For the 0.5 m radius, the net signed arc change is 0.75 m and the total circular path length is 1.25 m. Neither quantity is the Cartesian endpoint-displacement vector. The motion is stipulated here; determining which forces produce its acceleration is a separate dynamics question.

Measure mass spread from an axis

Moment of inertia I records how mass is distributed in perpendicular distance from a chosen axis. For point masses, I = Σ mᵢr⊥,ᵢ²; r⊥,ᵢ is the perpendicular distance of mass mᵢ from that axis. Moving the same mass farther out increases its contribution with distance squared. I is neither a force nor another name for the total mass; its unit is kg m².

For a continuous body, the corresponding sum becomes I = ∫r⊥² dm. This requires a specified nonnegative mass distribution with a finite squared-distance moment. I is nonnegative and is positive when a nonzero amount of mass lies off the axis. For a fixed rigid mass distribution and chosen axis, changing angular velocity alone does not change I.

Compare a stipulated thin ring and thin uniform disk, each with mass M = 4 kg and radius R = 0.5 m, about its central perpendicular axis. Every ring mass element is at distance R, so Iring = MR² = 1 kg m². Disk mass occupies radii between 0 and R, giving Idisk = MR²/2 = 0.5 kg m². These formulas depend on the stated distribution and axis, not just on the outline’s radius.

For the disk formula, a narrow annulus at radius r has mass dm = (2M/R²)r dr because the mass per area is uniform. Multiplying by r² and adding from 0 to R gives I = (2M/R²)∫₀ᴿr³ dr = MR²/2, for M > 0 and R > 0. This explains the disk’s smaller inertia: some of its mass lies closer to the axis than the ring’s mass.

Compare rotational kinetic energy

For classical pure rotation about our stationary axis, each mass element has speed r⊥|ω|/(1 rad). Adding its nonnegative kinetic energy, (1/2)mv², gives K = (1/2)I[ω/(1 rad)]². The formula assumes a common angular velocity and the chosen fixed axis in the same inertial frame. It describes this rotational kinetic energy; translation, deformation and other internal motions would need their own accounting.

In a separate comparison, give the 4 kg ring and disk the same angular velocity +3 rad/s about their central perpendicular axes. The ring has K = 4.5 J and the disk K = 2.25 J. Same mass, outer radius and angular velocity do not imply equal rotational kinetic energy. Reversing ω while keeping its magnitude and I leaves K unchanged: kinetic energy does not identify the direction of rotation.

Choose the axis before calculating

For a different exact body, connect two 1 kg point masses at x = −0.5 m and x = +0.5 m with a massless rigid connector. About the perpendicular z-axis through x = 0, both distances are 0.5 m and I = 0.5 kg m². About a parallel stationary axis through x = 0.25 m, the distances are 0.75 m and 0.25 m, giving I = 0.625 kg m². These are two alternative axis choices for the same mass arrangement, not simultaneous motions about both axes.

The fixed-axis model lets us use one signed angular coordinate and one axis-dependent scalar I. A translating axis, an axis that changes orientation, elastic deformation or relativistic motion can require a different description. Kinematics specifies the motion; inertia and kinetic energy describe the chosen body and motion under these assumptions. Torque, angular momentum and gyroscopes develop further questions about rotational interactions.

Worked examples

Try it yourself

1. At 3 s in the main model, is the object slowing down or speeding up? Give ω, α and the rotation direction.

   Show the answer and reasoning

   ω = −1 rad/s, α = −1 rad/s²; the object rotates clockwise and is speeding up.

   After the zero crossing at 2 s, ω becomes more negative. Its magnitude grows even though α is negative. Direction and speed change must both be considered.

2. A point at radius 0.5 m makes exactly one positive full turn without reversing. Compare its angle change, circular path length and Cartesian endpoint displacement.

   Show the answer and reasoning

   Angle change is +2π rad, path length is π m and Cartesian endpoint displacement is the zero vector.

   A full revolution returns to the same spatial endpoint. The path still has length (0.5 m)(2π rad)/(1 rad), and the unwrapped angle retains the turn.

3. A second point is 0.25 m from the main model’s axis. At 1 s, find its tangent velocity, two acceleration components and total acceleration magnitude.

   Show the answer and reasoning

   vθ = +0.25 m/s, aθ = −0.25 m/s², ar = −0.25 m/s² and |a| = (√2/4) m/s².

   At 1 s, ω = +1 rad/s and α = −1 rad/s². Apply the fixed-radius formulas, then add squared perpendicular acceleration components. The signed tangent component and inward radial component are different quantities.

4. Recalculate I for the two 1 kg point masses at x = ±0.5 m about the z-axis through x = 0 and the parallel axis through x = 0.25 m.

   Show the answer and reasoning

   I = 0.5 kg m² for the first axis and 5/8 kg m² for the second.

   The first distances are 0.5 m each. The second distances are 0.75 m and 0.25 m, so I = (1 kg)[(0.75 m)² + (0.25 m)²] = 0.625 kg m². Axis choice changes the distances; the masses are unchanged.

5. A separate thin uniform disk has mass 4 kg and radius 1 m. About its stationary central perpendicular axis at +3 rad/s, find I and K. Compare it with the 0.5 m disk.

   Show the answer and reasoning

   I = 2 kg m² and K = 9 J; each is four times the corresponding 0.5 m disk value.

   At fixed mass and the same ideal uniform-disk distribution, I = MR²/2 scales with radius squared. K scales with I when the angular-speed magnitude is fixed.

6. The original thin ring rotates in one trial at +3 rad/s and in another at −3 rad/s about the same stationary axis. Do the kinetic energies differ, and does either rate alone specify α?

   Show the answer and reasoning

   Both kinetic energies are 4.5 J. The directions are opposite; neither instantaneous angular velocity alone determines angular acceleration.

   The square in K removes the sign. To determine α, a local change of ω with time is needed. A positive or negative instantaneous rate can belong to a speeding-up, slowing-down or constant-rate motion.

What to carry forward

• Unwrapped angle keeps turns and reversals; signed angular velocity gives direction and angular speed is its magnitude.
• At a reversal, zero angular velocity can coexist with nonzero angular and tangential acceleration.
• At fixed radius, tangential acceleration follows α and inward acceleration follows ω squared; their directions are perpendicular.
• Moment of inertia needs both a mass distribution and a chosen axis. Its unit is kg m².
• For pure classical rotation about a stationary axis, rotational kinetic energy is (1/2)I times the squared angular rate in radians.