Angular Graphs & Calculus

Read a θ–t graph the way you read an x–t graph, with the vertical axis in radians. The angular velocity at an instant is the slope of the tangent there: rise in rad over run in s, so rad s⁻¹. The height of the curve says where the body is; only the slope says how fast it turns. Where the curve is a straight line the slope is fixed, ω is constant and α is zero. Where the curve bends, ω is changing, so α ≠ 0: a graph that steepens has positive α, one that bends over has negative α. The two readings are independent. A θ–t graph can pass through zero with a steep slope, the body turning fast as it crosses its reference line, or sit at a maximum with zero slope while bending sharply, which is the instant a reversing wheel stops and comes back with α still non-zero.

Integrating ω = dθ/dt over an interval gives θ₂ − θ₁ = ∫ ω dt, and on the graph that integral is the area between the curve and the time axis, in rad s⁻¹ × s = rad. For piecewise-straight graphs the area is triangles and rectangles. A fan whose ω rises linearly from 0 to 12 rad s⁻¹ in 4.0 s and then holds 12 rad s⁻¹ for 6.0 s turns through ½ × 4.0 × 12 + 6.0 × 12 = 24 + 72 = 96 rad, which is 96/(2π) = 15.3 revolutions. The area is signed: any part of the graph below the axis, where the body turns the other way, subtracts. Two things the area does not need. It does not need α to be constant, and it does not need θ₁ itself, which is why a graph of ω alone can tell you how much a wheel's angle changed but never where it is.

The same move one rung up: ω₂ − ω₁ = ∫ α dt, the area under the α–t graph, in rad s⁻² × s = rad s⁻¹. Take a motor whose angular acceleration falls linearly from 6.0 rad s⁻² to zero over 10 s as its back-EMF builds. The area is a triangle, ½ × 10 × 6.0 = 30 rad s⁻¹, so a motor starting from rest reaches 30 rad s⁻¹. A student who plugs the initial α into ω = ω₀ + αt gets 60 rad s⁻¹, double the truth, because that formula assumes the α–t graph is a flat line. Averaging α to 3.0 rad s⁻² happens to rescue ω here, but only because a straight α–t line has its mean exactly half-way, and it does not rescue θ: with α changing, the ω–t graph is a parabola, not a line, and ½(ω₀ + ω)t is the area under a chord, not under the curve. Every constant-α relation is an area formula for one particular shape of graph. Change the shape and you must redo the area.

A tracked turntable might report θ(t) = 2.0t³ − 3.0t² + 4.0, in rad with t in s. Differentiate once: ω = 6.0t² − 6.0t. Again: α = 12t − 6.0. Every question about the motion is now a question about these three functions. When is it momentarily at rest? ω = 6.0t(t − 1.0) = 0 at t = 0 and t = 1.0 s. What is the acceleration at those instants? α(0) = −6.0 rad s⁻² and α(1.0) = +6.0 rad s⁻², not zero: the graph is bending while its slope passes through zero. When does the acceleration itself vanish? α = 0 at t = 0.50 s, the inflection of the θ–t curve, where the slope is most negative, ω(0.50) = 1.5 − 3.0 = −1.5 rad s⁻¹. The coefficients carry units, 2.0 rad s⁻³, 3.0 rad s⁻² and 4.0 rad, and a dimensional check of ω = 6.0t² − 6.0t confirms that rad s⁻³ × s² and rad s⁻² × s both give rad s⁻¹.

Run the chain the other way. Suppose a flywheel has α(t) = 12t − 6.0 rad s⁻² and you know ω(0) = 0 and θ(0) = 4.0 rad. Integrating α gives ω = 6.0t² − 6.0t + C₁, and ω(0) = 0 sets C₁ = 0. Integrating ω gives θ = 2.0t³ − 3.0t² + C₂, and θ(0) = 4.0 rad sets C₂ = 4.0. Without those two numbers the integrals return only changes, and the same α(t) describes a whole family of motions differing in starting angle and starting spin. Now ask how far the flywheel turns between t = 0 and 2.0 s. Net displacement is ∫ ω dt from 0 to 2.0 = [2.0t³ − 3.0t²] evaluated from 0 to 2.0 = 16 − 12 = 4.0 rad. But ω is negative until t = 1.0 s: over 0 to 1.0 s the displacement is 2.0 − 3.0 = −1.0 rad, and over 1.0 to 2.0 s it is 4.0 − (−1.0) = 5.0 rad. The wheel backs up 1.0 rad, then turns forward 5.0 rad: 6.0 rad of turning for 4.0 rad of net displacement. Only the second figure is the area under ω; the first is the area under |ω|.

Three consistency checks catch most misread graphs. Units first: a slope of θ–t must come out in rad s⁻¹ and an area under α–t in rad s⁻¹, so if you find yourself holding rad s you have multiplied where you should have divided, and if you are holding rad s⁻³ you have divided where you should have multiplied. Zeros next: every maximum or minimum of θ–t sits directly above a zero crossing of ω–t, and every extremum of ω–t sits above a zero of α–t; three graphs that violate this cannot all describe the same motion. Shapes last: a constant α forces ω–t to be straight and θ–t to be a parabola, so if the ω–t graph you are handed is curved, the constant-α relations are off the table and the area or the integral is the only route. The reverse holds as well. A curved θ–t with a straight ω–t is consistent; a straight θ–t with a sloping ω–t is not, because a straight θ–t means ω is fixed.