5.1 Rotational Kinematics

1. A1.5 rad
   A student who uses the number of revolutions as the number of radians picks this. A radian is much smaller than a revolution: one revolution is 2π ≈ 6.28 rad, so 1.5 revolutions is about 9.4 rad.
2. B540 rad
   A student who works out the angle in degrees, 1.5 × 360° = 540°, and then writes the number with the unit rad picks this. 540° is 540 × (π/180) = 3π rad ≈ 9.4 rad.
3. C3.1 rad
   A student who ignores the complete revolution, because the wheel is back in its starting orientation after it, and counts only the half turn left over (π rad) picks this. Every turn in the same sense adds 2π rad to the angular displacement.
4. D9.4 rad Correct
   One revolution is 2π rad, so 1.5 revolutions is 1.5 × 2π = 3π rad ≈ 9.4 rad.

Working One revolution = 2π rad. Δθ = 1.5 × 2π rad = 3π rad = 9.42 rad ≈ 9.4 rad.

CED 5.1.A.1 · Read this in Fix

1. A3.5 rad/s
   A student who averages the angular velocities of the two segments, (5.0 + 2.0)/2, picks this. The turntable spends 2 s at 5.0 rad/s but 4 s at 2.0 rad/s, so the slower rate lasts longer; dividing the total angular displacement by the total time gives 3.0 rad/s.
2. B3.3 rad/s
   A student who divides the final angular position by the time, 20 rad ÷ 6 s, picks this. The turntable started at θ = 2 rad, not at zero, so its angular displacement is 20 − 2 = 18 rad.
3. C2.0 rad/s
   A student who gives the angular velocity at the end of the interval picks this. 2.0 rad/s is the slope of the last segment, the angular velocity near t = 6 s; the average over the whole 6 s must include the faster first 2 s.
4. D3.0 rad/s Correct
   Average angular velocity is the angular displacement divided by the time taken. From t = 0 to t = 6 s the angular position changes from 2 rad to 20 rad, so ω_avg = 18 rad ÷ 6 s = 3.0 rad/s.

Working ω_avg = Δθ/Δt = (20 − 2) rad ÷ (6 − 0) s = 3.0 rad/s. Segment slopes: (12 − 2)/2 = 5.0 rad/s for 0–2 s and (20 − 12)/4 = 2.0 rad/s for 2–6 s.

CED 5.1.A.2 · Read this in Fix

1. A−4ω₀/t Correct
   With counterclockwise positive, the angular velocity changes from +ω₀ to −3ω₀, a change of −4ω₀. Dividing by the time gives α_avg = −4ω₀/t: an average angular acceleration in the clockwise sense.
2. B+2ω₀/t
   A student who subtracts the two angular speeds, 3ω₀ − ω₀ = 2ω₀, ignoring the reversal, picks this. The wheel first loses all ω₀ of its counterclockwise rotation and then gains 3ω₀ clockwise, so the change is −4ω₀.
3. C−3ω₀/t
   A student who divides the final angular velocity by the time picks this. Angular acceleration is the CHANGE in angular velocity divided by the time, and the wheel did not start from rest.
4. D+4ω₀/t
   A student who takes the change as initial minus final, ω₀ − (−3ω₀) = +4ω₀, picks this. Δω = ω − ω₀, final minus initial, which is −4ω₀, so the average angular acceleration is negative (clockwise).

Working Counterclockwise positive: initial angular velocity +ω₀, final −3ω₀. α_avg = Δω/Δt = (−3ω₀ − ω₀)/t = −4ω₀/t.

CED 5.1.A.3 · Read this in Fix

1. AIts angular velocity has the same value at every instant. Correct
   The angular velocity is the slope of the θ–t graph. The graph is a straight line, so its slope, 20 rad ÷ 5 s = 4 rad/s, is the same at every instant, and the angular acceleration is zero.
2. BIts angular velocity increases, since the graph keeps rising.
   A student who reads the rising height of the graph as a rising angular velocity picks this. The height is the angular position, which does increase; the angular velocity is the slope, which stays the same.
3. CIts angular acceleration is positive, since the turntable keeps turning.
   A student who thinks a system needs an angular acceleration to keep rotating picks this. A turntable turning at a steady rate has zero angular acceleration; an angular acceleration is needed only to change the angular velocity.
4. DIts angular displacement is given by the area under the graph.
   A student who uses the area under every motion graph picks this. The area under an ω–t graph gives an angular displacement; on a θ–t graph the angular displacement is the change in height, here 20 rad − 0 = 20 rad.

Working Slope = (20 − 0) rad ÷ (5 − 0) s = 4 rad/s, constant; α = 0. Δθ = 20 rad (change in height).

CED 5.1.A.4.ii · Read this in Fix

1. AEvery point on the wheel moves in the same direction, since the wheel keeps its shape.
   A student who thinks a system that keeps its shape must move as one piece picks this. That is true of a block sliding without rotating, but in rotation each point follows its own circle, so points on opposite sides of the axle move in opposite directions.
2. BPoints at the top and at the bottom of the wheel move in opposite directions. Correct
   The wheel is a rigid system: it holds its shape, but each point moves around its own circle about the axle. At any instant the top of the rim moves one way and the bottom moves the opposite way, which is why the spinning wheel cannot be modeled as an object with a single velocity.
3. CNo point on the wheel moves, since the wheel's center of mass stays at rest.
   A student who judges whether a system moves only from its center of mass picks this. The center of mass does stay at the axle, but every other point of the wheel moves in a circle about it; the rotation is described with angular quantities.
4. DPoints near the rim turn through larger angles than points near the axle.
   A student who expects points farther out to rotate more picks this. Points near the rim travel farther, along larger circles, but every line drawn from the axle sweeps the same angle in the same time, so all points turn through the same angle.

CED 5.1.A.1.i · Read this in Fix

1. A+4.7 rad Correct
   The counterclockwise revolution gives +1.0 × 2π = +2π rad and the clockwise quarter turn gives −0.25 × 2π = −π/2 rad. Adding them, Δθ = +3π/2 ≈ +4.7 rad: the disk ends three-quarters of a turn counterclockwise from where it started.
2. B+7.9 rad
   A student who adds the two rotations as sizes, 2π + π/2 ≈ 7.9 rad, picks this. That is the total angle turned. Angular displacement takes the sense of rotation into account: with counterclockwise positive, the clockwise quarter turn counts as −π/2 rad.
3. C−1.6 rad
   A student who treats the full counterclockwise revolution as adding nothing, because the disk is back in its starting orientation after it, picks this. That revolution contributes +2π rad; the clockwise quarter turn then subtracts π/2 rad.
4. D+270 rad
   A student who finds the angle in degrees, 360° − 90° = 270°, and writes it with the unit rad picks this. 270° is 3π/2 rad, about 4.7 rad.

Working Counterclockwise positive: Δθ = (+1.0)(2π rad) + (−0.25)(2π rad) = 2π − π/2 = +3π/2 rad = +4.71 rad ≈ +4.7 rad.

CED 5.1.A.1.ii · Read this in Fix

1. AIt is unsuitable for every purpose, since the ball keeps rotating about its own center while it flies.
   A student who thinks a rotating system can never be treated as an object picks this. Earth, for example, spins about its axis but is treated as an object when its revolution around the Sun is described. Whether the model works depends on the question asked.
2. BIt can be used to find how points on the ball's surface move, since the ball keeps its shape.
   A student who thinks a system that keeps its shape moves as one piece picks this. Points on the surface move around the ball's center in different directions; describing them needs the ball's rotation, which is exactly what the object model leaves out.
3. CIt can be used to predict the ball's path, since the spin does not change how the center of mass moves. Correct
   When the question is where the ball goes, the ball's motion is well described by the motion of its center of mass, and with air resistance negligible the spin does not change that motion. The rotation can be neglected and the ball treated as an object, just as Earth's spin is neglected when its orbit around the Sun is described.
4. DIt fails to predict the ball's path correctly, since the spin makes the ball's path curve away to one side.
   A student who carries over the swerve of spinning balls in sports picks this. That swerve comes from the air flowing past the ball. With air resistance negligible, only the gravitational force is exerted on the ball, and its center of mass follows the same path whether or not it spins.

CED 5.1.A.1.iii · Read this in Fix

1. A12 rad, since the wheel turns through ωt = (6.0 rad/s)(2.0 s) in that time.
   A student who multiplies the final angular velocity by the time picks this. The wheel turns at 6.0 rad/s only at the end of the 2.0 s and more slowly before that, so it turns through less than 12 rad.
2. BIt cannot be found from the cart's motion, since points on the wheel move in circles.
   A student who thinks rotation needs different mathematics from motion along a line picks this. About a single fixed axis, θ, ω and α behave exactly like x, v_x and a_x, so the same equations apply with the symbols changed.
3. C8.0 rad, since θ − θ₀ = ω₀t + (1/2)αt² has the same form as the cart's equation. Correct
   Angular displacement, angular velocity and angular acceleration about one axis obey the same relationships as x, v_x and a_x. Here α = (6.0 − 2.0) rad/s ÷ 2.0 s = 2.0 rad/s², and Δθ = (2.0 rad/s)(2.0 s) + (1/2)(2.0 rad/s²)(2.0 s)² = 8.0 rad, matching the cart's 8.0 m.
4. DAbout 50 rad, since by the same analogy the wheel turns through 8.0 revolutions.
   A student who takes revolutions as the rotational counterpart of meters picks this. The counterpart of position in meters is angular position in radians, so the analogy predicts 8.0 rad, a little over one revolution.

Working α = (6.0 − 2.0) rad/s ÷ 2.0 s = 2.0 rad/s². Δθ = ω₀t + (1/2)αt² = (2.0)(2.0) + (1/2)(2.0)(2.0)² = 4.0 + 4.0 = 8.0 rad (cart: a = 2.0 m/s², Δx = 8.0 m). Distractors: 6.0 × 2.0 = 12 rad; 8.0 rev × 2π = 50 rad.

CED 5.1.A.4 · Read this in Fix

1. Aω₀²/α
   A student who multiplies the initial angular velocity by the stopping time, ω₀ × (ω₀/α), picks this. The wheel turns at ω₀ only at the start; its average angular velocity is ω₀/2, so it turns through half this angle.
2. Bω₀²/(2α) Correct
   Using ω² = ω₀² + 2α(θ − θ₀) with ω = 0 and an angular acceleration of −α (opposite to the rotation): 0 = ω₀² − 2αΔθ, so Δθ = ω₀²/(2α). Units: (rad/s)² ÷ (rad/s²) = rad.
3. Cω₀/(2α)
   A student who forgets to square ω₀ when using ω² = ω₀² + 2α(θ − θ₀) picks this. Its units are (rad/s) ÷ (rad/s²) = s, a time, not an angle.
4. D2ω₀²/α
   A student who moves the factor 2 to the wrong side when rearranging ω₀² = 2αΔθ picks this. Dividing both sides by 2α gives Δθ = ω₀²/(2α).

Working ω² = ω₀² + 2(−α)Δθ with ω = 0: Δθ = ω₀²/(2α). Check: stopping time t = ω₀/α; Δθ = (ω₀/2)t = ω₀²/(2α).

CED 5.1.A.4.i · Read this in Fix

1. A2.00
   A student who thinks a greater angular acceleration always means a greater angle picks this. That is true when the fans start from rest and run for the same TIME. Here they reach the same angular velocity, and fan X gets there in half the time, turning through half the angle.
2. B1.00
   A student who writes Δθ = (1/2)αt, forgetting to square t, gets (1/2)α(ω/α) = ω/2 for both fans and picks this. With t squared, Δθ = (1/2)α(ω/α)² = ω²/(2α), which depends on α.
3. C0.25
   A student who notes that fan X takes half as long, and uses Δθ = (1/2)αt² with the time halved but the angular acceleration left unchanged, gets (1/2)² = 1/4 and picks this. Fan X's angular acceleration is also doubled: 2 × 1/4 = 1/2.
4. D0.50 Correct
   Starting from rest, ω² = 2αΔθ, so Δθ = ω²/(2α). Both fans reach the same ω, so the angle is inversely proportional to the angular acceleration: doubling α halves the angle, and Δθ_X/Δθ_Y = 0.50.

Working From rest: ω² = 2αΔθ → Δθ = ω²/(2α). Same ω, α_X = 2α_Y → Δθ_X/Δθ_Y = α_Y/α_X = 0.50. Check with time: t_X = ω/(2α_Y) = t_Y/2; Δθ_X/Δθ_Y = (2α_Y)(t_Y/2)²/(α_Y t_Y²) = 1/2.

CED 5.1.A.4.i · Read this in Fix

1. A24 rad Correct
   The angular acceleration is α = (0 − 12 rad/s) ÷ 4.0 s = −3.0 rad/s². Then θ − θ₀ = ω₀t + (1/2)αt² = (12)(4.0) + (1/2)(−3.0)(4.0)² = 48 − 24 = 24 rad. Equivalently, the average angular velocity is 6.0 rad/s for 4.0 s.
2. B48 rad
   A student who multiplies the initial angular velocity by the whole time, 12 rad/s × 4.0 s, picks this. The fan turns at 12 rad/s only at the instant it is switched off and more slowly afterward, so it turns through less than 48 rad.
3. C72 rad
   A student who takes the angular acceleration as +3.0 rad/s², ignoring that the fan is slowing down, gets 48 + 24 = 72 rad and picks this. When the fan slows, α is opposite in sign to ω₀, so the (1/2)αt² term subtracts.
4. D42 rad
   A student who forgets to square the time in (1/2)αt² gets 48 + (1/2)(−3.0)(4.0) = 42 rad and picks this. With t squared the second term is −24 rad, giving 24 rad.

Working α = (0 − 12) rad/s ÷ 4.0 s = −3.0 rad/s². Δθ = ω₀t + (1/2)αt² = (12)(4.0) + (1/2)(−3.0)(4.0)² = 48 − 24 = 24 rad. Check: ω_avg = (12 + 0)/2 = 6.0 rad/s; 6.0 × 4.0 = 24 rad.

CED 5.1.A.4.i · Read this in Fix

1. AWheel B turns through the greater angle, since it is faster at t = 4 s.
   A student who judges the angle from the height of the graph at the end picks this. B is faster at t = 4 s, but it started from rest; the angle depends on the angular velocity over the whole interval, the area, which is 20 rad for B and 24 rad for A.
2. BWheel B turns through the greater angle, as its graph has the steeper slope.
   A student who reads the angle turned from the slope of the ω–t graph picks this. The slope gives the angular acceleration, which is greater for B; the angle turned is given by the area under the graph.
3. CWheel A turns through the greater angle, since more area lies under its graph. Correct
   The angle turned is the area between the ω–t graph and the time axis. For A, 6 rad/s × 4 s = 24 rad; for B, (1/2)(4 s)(10 rad/s) = 20 rad. A is faster for the first 2.4 s, and B does not make up the difference by t = 4 s.
4. DThe wheels turn through equal angles, since their graphs cross each other.
   A student who takes the crossing of the graphs to mean equal angles turned picks this. Where the graphs cross, at t = 2.4 s, the wheels have equal angular VELOCITIES; A has already turned through more, and it is still ahead at t = 4 s.

Working Area under ω–t = angle turned. A: 6 rad/s × 4 s = 24 rad. B: (1/2)(4 s)(10 rad/s) = 20 rad. A > B. Graphs cross where 2.5t = 6, t = 2.4 s.

CED 5.1.A.4.ii · Read this in Fix

1. AIt slows down throughout, since its angular velocity keeps decreasing.
   A student who thinks a decreasing angular velocity always means slowing down picks this. The value of ω decreases throughout, but after t = 3 s it becomes more and more negative: the wheel turns faster and faster clockwise.
2. BIt rotates clockwise and speeds up between t = 3 s and t = 6 s. Correct
   From t = 3 s to t = 6 s the angular velocity is negative, so the wheel rotates clockwise, and its size grows from 0 to 6 rad/s, so the wheel speeds up. The angular acceleration, −2 rad/s², is in the same (clockwise) sense as the rotation during that time.
3. CIts angular acceleration is zero at t = 3 s, when it is momentarily at rest.
   A student who thinks zero angular velocity means zero angular acceleration picks this. The graph is one straight line with the same slope, −2 rad/s², at t = 3 s as at every other time; the wheel is at rest for only an instant.
4. DIt rotates clockwise from t = 0 to t = 3 s, since the graph slopes down.
   A student who reads the sense of rotation from the slope picks this. The sense of rotation is given by the sign of ω, the height of the graph: from t = 0 to t = 3 s it is positive, so the wheel turns counterclockwise while slowing down.

Working Slope = (−6 − 6) rad/s ÷ 6 s = −2 rad/s² throughout. 0–3 s: ω > 0 (counterclockwise), |ω| falling (slowing). t = 3 s: ω = 0 for an instant. 3–6 s: ω < 0 (clockwise), |ω| rising (speeding up).

CED 5.1.A.4.ii · Read this in Fix

1. AThe hour hand's is 12 times that of the minute hand.
   A student who takes the hand with the longer time per revolution to have the greater angular velocity picks this. A longer time for each revolution means a SMALLER angle each second; the hour hand turns 12 times more slowly.
2. BThey are equal, since both hands turn about one shared axle.
   A student who thinks everything turning about one axle shares one angular velocity picks this. The hands are separate pieces driven at different rates: in one hour the minute hand sweeps 2π rad and the hour hand only π/6 rad.
3. CThey are in the same ratio as the lengths of the two clock hands.
   A student who links angular velocity to distance from the axle picks this. The tip of the longer hand does travel farther, but angular velocity is the angle turned per unit time, which does not depend on the hand's length.
4. DThe minute hand's is 12 times as great as the hour hand's. Correct
   In 12 hours the minute hand turns through 12 × 2π rad and the hour hand through 2π rad. Dividing each angular displacement by the same time, the minute hand's average angular velocity is 12 times the hour hand's.

Working ω_minute = 2π rad ÷ 3600 s = 1.75 × 10⁻³ rad/s; ω_hour = 2π rad ÷ 43 200 s = 1.45 × 10⁻⁴ rad/s; ratio = 43 200/3600 = 12.

CED 5.1.A.2 · Read this in Fix

1. A−1.5 ω²/α
   A student who enters the angular acceleration as +α although the wheel is slowing down picks this: ω² = (2ω)² + 2αΔθ gives a negative Δθ. A wheel that keeps turning counterclockwise has a positive angular displacement; the slowing makes the angular acceleration negative, not Δθ.
2. B−0.5 ω²/α
   A student who sets up ω² = (2ω)² − 2αΔθ correctly but replaces the difference of the squares, ω² − (2ω)², by the square of the change, (ω − 2ω)² = ω², gets ω² = −2αΔθ and so Δθ = −0.5 ω²/α. The difference of the squares is ω² − 4ω² = −3ω², which gives Δθ = +1.5 ω²/α; a negative angular displacement would mean clockwise turning, which this wheel never does.
3. C+1.5 ω²/α Correct
   The angular acceleration is −α, because the wheel slows while turning in the positive sense. ω² = (2ω)² − 2αΔθ gives Δθ = 3ω²/(2α) = +1.5 ω²/α: positive, as the wheel keeps turning counterclockwise.
4. D+2.0 ω²/α
   A student who finds the time to slow down, ω/α, and multiplies it by the initial angular velocity, 2ω, picks this. The angular velocity falls from 2ω to ω during that time; with constant angular acceleration the average is 1.5ω, giving Δθ = 1.5 ω²/α.

Working Counterclockwise positive. The wheel slows while turning counterclockwise, so its angular acceleration is −α. Use ω² = ω₀² + 2α(θ − θ₀) with initial angular velocity 2ω and final angular velocity ω: ω² = (2ω)² − 2αΔθ, so Δθ = 3ω²/(2α) = +1.5 ω²/α. Check with the average angular velocity: slowing takes (2ω − ω)/α = ω/α at an average of 1.5ω, giving 1.5 ω²/α.

CED 5.1.A.4.i · Read this in Fix