4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A rigid wheel rotates about a fixed axle through its center. Which statement about the motion of the wheel's points is correct?
Answer and reasoning
AAt each instant, every point of the wheel moves in the same direction, as the wheel is rigid. A student who thinks 'rigid' means that all the points move together picks this. That is true only for a rigid system that translates without rotating; in rotation, each point's velocity is tangent to its own circle, so points in different places move in different directions.
BAt each instant, points at opposite ends of a diameter move in opposite directions.Correct The wheel keeps its shape, but each point moves along its own circle around the axle. Two points at opposite ends of a diameter therefore have velocities in opposite directions at every instant. This is why a rigid system cannot be modeled as a single object when its rotation is described.
CIn each second, points near the axle turn through a larger angle than points near the rim. A student who thinks a point on a smaller circle gets around it sooner picks this. All points of a rigid wheel turn through the same angle in the same time; points near the rim travel along longer arcs, not through larger angles.
DModeling the wheel as one object at its center describes how each of its points moves. A student who applies the object model to a rotating system picks this. A single object at the center of a wheel spinning on a fixed axle would not move at all, while the wheel's points do; describing the rotation needs angular quantities.
After its motor is switched off, a flywheel's angular velocity is ω(t) = ω₀e−bt, where ω₀ and b are positive constants. Through what angle does the flywheel turn between t = 0 and t = 1/b?
Answer and reasoning
A0.63 ω₀/bCorrect The angle is the area under the ω–t graph: Δθ = ∫ω₀e−bt dt from 0 to 1/b = (ω₀/b)(1 − e−1) = 0.63 ω₀/b.
B0.37 ω₀/b A student who evaluates the antiderivative, −(ω₀/b)e−bt, only at the upper limit gets a magnitude of (ω₀/b)e−1 = 0.37 ω₀/b. The value at the lower limit, −ω₀/b, must be subtracted, giving (ω₀/b)(1 − e−1).
C1.00 ω₀/b A student who uses Δθ = ωt with the initial angular velocity picks this: ω₀ × (1/b). The flywheel slows throughout the interval, so it turns through less than this.
D0.68 ω₀/b A student who multiplies the constant-α average angular velocity, (ω₀ + ω)/2, by the time picks this: (1 + e−1)/2 = 0.68. ω falls exponentially, not linearly, so that average does not apply; integrating gives 0.63 ω₀/b.
Working Δθ = ∫ω dt from 0 to 1/b = ∫ω₀e−bt dt = (ω₀/b)[−e−bt] from 0 to 1/b = (ω₀/b)(1 − e−1) ≈ 0.63 ω₀/b. (Antiderivative at the upper limit only: (ω₀/b)e−1 ≈ 0.37 ω₀/b. ω₀ taken as constant: ω₀(1/b) = 1.00 ω₀/b. Constant-α average: [(ω₀ + ω₀e−1)/2](1/b) ≈ 0.68 ω₀/b.)
The graph shows the angular velocity ω of a wheel as a function of time t, with counterclockwise taken as positive. What is the wheel's angular acceleration at t = 5.0 s?
Answer and reasoning
A−4.0 rad/s² A student who reads the height of the graph at t = 5.0 s picks this. The height is ω = −4.0 rad/s; α is the slope, which is −2.0 rad/s².
B+2.0 rad/s² A student who thinks a wheel that is speeding up must have a positive angular acceleration picks this. After t = 3 s the angular speed does increase, but in the negative (clockwise) sense, so ω and α are both negative.
C+5.0 rad/s² A student who finds the area under the graph from 0 to 5.0 s picks this: 9 rad − 4 rad = 5.0 rad. That area is the angular displacement; the angular acceleration is the slope.
D−2.0 rad/s²Correct Angular acceleration is the slope of the ω–t graph. The line falls 12 rad/s in 6 s, so α = −2.0 rad/s² at every time shown, including t = 5.0 s; the negative sign means α is clockwise.
Working α is the slope of the ω–t graph. The graph is a straight line from 6 rad/s at t = 0 to −6 rad/s at t = 6 s: α = (−6 rad/s − 6 rad/s)/(6 s) = −2.0 rad/s², at every time shown, including t = 5.0 s. (Height at 5.0 s: −4.0 rad/s. Area from 0 to 5.0 s: 9 rad − 4 rad = 5.0 rad.)
A rotor starts from rest at θ = 0 at time t = 0. Its angular acceleration is α = βt, where β is a positive constant. Through what angle has the rotor turned at time T?
Answer and reasoning
A(1/6)βT³Correct As for linear motion with a = dv/dt and v = dx/dt, integrate twice using the initial values: ω = ∫βt dt = (1/2)βt², then θ = ∫(1/2)βt² dt = (1/6)βT³.
B(1/2)βT³ A student who uses θ = (1/2)αt² with the angular acceleration at time T, α = βT, picks this. That equation holds only for constant α; here α grows from zero, so the rotor turns through less.
C(1/2)βT² A student who stops after one integration picks this. (1/2)βT² is the angular velocity at time T, in rad/s; integrating once more gives the angle.
D(1/4)βT³ A student who uses the average angular acceleration, (1/2)βT, in θ = (1/2)αt² picks this. Averaging α does not make the constant-α equation exact; integrating ω(t) gives (1/6)βT³.
Working ω(t) = ω₀ + ∫₀ᵗ βt′ dt′ = (1/2)βt², with ω₀ = 0. θ(T) = θ₀ + ∫₀ᵀ (1/2)βt² dt = (1/6)βT³, with θ₀ = 0. (θ = (1/2)αt² with α = βT: (1/2)βT³. One integration only: (1/2)βT², which is ω(T). Average α = (1/2)βT in (1/2)αT²: (1/4)βT³.)
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.1.A.1 Angular position, θ Fix
Angular position, θ
The angle, in radians, of a reference line on a rigid system measured from a fixed reference direction, with one sense of rotation (clockwise or counterclockwise) taken as positive. It keeps counting through full turns: one full turn in the positive sense from θ = 0 brings it to θ = 2π rad. SI unit: rad.
Angular displacement, Δθ = θ − θ₀
The change in angular position of a point on a rigid system about a specified axis. Its sign gives the sense of the net rotation; a turn in the positive sense followed by an equal turn back gives Δθ = 0. Unit: rad.
Radian
The angle at the center of a circle that subtends an arc equal in length to the radius. One revolution is 2π rad, or 360°. Angles in the rotational kinematics equations are in radians.
Rigid system
A system that holds its shape but whose points, when it rotates, move in different directions at the same instant (points on opposite sides of an axle move in opposite directions). Its rotation therefore cannot be described by modeling it as a single object.
Positive sense of rotation
A choice, stated or drawn for each problem, that one sense of rotation about the axis, clockwise or counterclockwise, is positive; rotation in the other sense is negative. The signs of Δθ, ω and α are read against this choice.
Object model for a rotating system
A rotating system may be treated as a single object at its center of mass when the motion of interest is well described by the motion of the center of mass, so that its rotation can be neglected; for example, Earth's spin is neglected when describing its revolution about the center of mass of the Earth–Sun system.
Students often think Angular displacement is the total angle a system turns through, so turns in opposite senses add. In fact Not when the rotation reverses. Angular displacement is θ − θ₀, the net change in angular position: turns in the negative sense subtract from turns in the positive sense. The total angle turned, with every turn counted as positive, is larger whenever the system reverses.
Students often think A system's angular displacement over an interval is its angular position at the end of the interval; the starting position need not be subtracted. In fact No. Angular displacement is the change θ − θ₀. It equals the final angular position only when θ₀ = 0.
5.1.A.2 Angular velocity, ω = dθ/dt Fix
Angular velocity, ω = dθ/dt
The rate at which angular position changes with time; the slope of a θ–t graph. Its sign gives the sense of rotation and its magnitude, |ω|, is the angular speed. SI unit: rad/s.
Average angular velocity, ωavg = Δθ/Δt
The angular displacement divided by the time interval. It equals the instantaneous angular velocity at every instant of the interval only when ω is constant.
Students often think Angular position (or angle turned) and angular velocity are treated as the same quantity, so a value or expression for one is used for the other; for example, ω is taken to be zero when θ is zero, or to grow the way θ g… In fact No. ω is the rate of change of θ. θ can be large while ω is zero (at a turning point) and zero while ω is large; on a θ–t graph, ω is the slope, not the height.
Students often think A definite integral can be evaluated from the antiderivative at the upper limit alone, so the value at the lower limit (or the initial condition) can be left out. In fact No. Δθ is the antiderivative evaluated at the upper limit minus its value at the lower limit. For ω = ω₀e−bt, the antiderivative is not zero at t = 0, so dropping it gives the wrong angle.
5.1.A.3 Angular acceleration, α = dω/dt Fix
Angular acceleration, α = dω/dt
The rate at which angular velocity changes with time; the slope of an ω–t graph. When ω and α have the same sign the angular speed increases; when they have opposite signs it decreases. SI unit: rad/s².
Students often think A positive angular acceleration means the rotation is speeding up, and a negative one means it is slowing down. In fact Only if ω is also positive. A system speeds up when ω and α have the same sign and slows down when they have opposite signs; the sign of α alone does not say whether it is speeding up.
Students often think A system rotates in the sense of its angular acceleration, so ω and α have the same sign and α changes sign when the rotation reverses. In fact No. The sense of rotation is given by ω. α gives the sense in which ω is changing, which is opposite to the rotation whenever the system slows down.
5.1.A.4 Rotational–linear analogy Fix
Rotational–linear analogy
For rotation about one axis, θ, ω and α are related exactly as x, v and a are in one dimension: ω = dθ/dt and α = dω/dt, so Δθ = ∫ω dt and Δω = ∫α dt, with initial values supplying the constants.
Constant-angular-acceleration equations
ω = ω₀ + αt, θ = θ₀ + ω₀t + (1/2)αt² and ω² = ω₀² + 2α(θ − θ₀). They hold only when α is constant; when α varies, θ and ω are found by integrating.
Rotational kinematics graphs
On a θ–t graph the slope is ω. On an ω–t graph the slope is α, and the signed area between the graph and the time axis is Δθ. On an α–t graph the signed area is Δω.
Students often think A rate of change is found from the area under a graph (or by integrating) instead of from its slope (or by differentiating). In fact No. A rate of change is a slope, or a derivative: ω is the slope of the θ–t graph and α is the slope of the ω–t graph. The area under an ω–t graph is Δθ, and the area under an α–t graph is Δω.
Students often think The angle turned during an interval equals the angular velocity at one instant (often the start or the end) multiplied by the time, even when ω is changing. In fact No. Δθ = ωt holds only for constant ω. When ω changes, Δθ is the area under the ω–t graph, ∫ω dt, which is not the value of ω at one instant multiplied by the time.
13 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 13
A disk can rotate about a fixed axle, with counterclockwise taken as positive. A mark on the disk starts at angular position θ₀ = 0.400 rad. The disk turns 1.50 revolutions counterclockwise and then 0.250 revolution clockwise. What is the disk's angular displacement?
Answer and reasoning
A11.0 rad A student who adds the two turns as if both were in the same sense picks this: 1.75 rev × 2π rad = 11.0 rad, the total angle turned. The clockwise turn is in the negative sense, so it reduces the angular displacement.
B8.25 rad A student who gives the final angular position as the displacement picks this: 0.400 rad + 7.85 rad = 8.25 rad. Angular displacement is the change θ − θ₀, so the starting value is subtracted again, leaving 7.85 rad.
C1.57 rad A student who thinks a whole revolution adds nothing, because the mark is back where it started, keeps only the extra quarter turn: 0.25 × 2π rad = 1.57 rad. Angular position keeps counting through full turns, so the whole revolution adds 2π rad.
D7.85 radCorrect Give each turn its sign: +1.50 rev − 0.250 rev = +1.25 rev. Converting, Δθ = 1.25 × 2π rad = 7.85 rad. The starting position does not matter, because Δθ = θ − θ₀ is a change in angular position.
Working Net rotation = +1.50 rev − 0.250 rev = +1.25 rev. Δθ = (1.25 rev)(2π rad/rev) = +7.85 rad. θ₀ does not enter, because Δθ = θ − θ₀ is a change. (Total angle turned: 1.75 rev = 11.0 rad. Final angular position: 0.400 rad + 7.85 rad = 8.25 rad. Whole revolution dropped: 0.25 rev = 1.57 rad.)
The figure shows a wheel and the sense of rotation chosen as positive. At one instant the wheel's angular velocity is −3.0 rad/s and its angular acceleration is +1.5 rad/s². Which describes the wheel's motion at that instant?
Answer and reasoning
ATurning clockwise and slowing down A student who takes counterclockwise as positive out of habit picks this, reading ω = −3.0 rad/s as clockwise. The figure makes clockwise positive, so the negative ω means the wheel turns counterclockwise.
BTurning counterclockwise and speeding up A student who thinks a positive angular acceleration means speeding up picks this. Whether the wheel speeds up depends on whether ω and α have the same sign; here they have opposite signs, so it is slowing down.
CTurning counterclockwise and slowing downCorrect The figure makes clockwise positive, so ω = −3.0 rad/s means the wheel is turning counterclockwise. α is positive, opposite in sign to ω, so the angular speed is decreasing: the wheel is slowing down.
DTurning clockwise and speeding up A student who thinks a wheel turns in the sense of its angular acceleration picks this: α is positive, which the figure makes clockwise. The sense of rotation is given by ω, which is negative, so the wheel turns counterclockwise, and with α opposite to ω it is slowing.
A baton is tossed into the air and spins end over end as it flies. Air resistance is negligible. Which statement about modeling the baton is correct?
Answer and reasoning
AIt can be modeled as one object to find the greatest height reached by its center of mass.Correct The center of mass moves like a single object in free fall whatever the spin, so the greatest height it reaches can be found with the object model. The spin does not affect the motion of the center of mass, so it can be ignored for this question.
BIt cannot be modeled as one object for any part of its motion, because it is spinning. A student who thinks any rotation rules out the object model picks this. A rotating system can be treated as one object for any motion that its center of mass describes, such as the height that the baton's center reaches.
CIt can be modeled as one object to find the path of each of its ends, as it is rigid. A student who thinks every point of a rigid system moves the same way picks this. The ends of the spinning baton loop around its center of mass, so their paths differ from the center's parabola and cannot be found from the object model.
DIt can be modeled as one object to find how fast it spins, from the speed of its center. A student who links the spin to the speed of the center of mass picks this. The two motions are independent: the same toss could send the baton up spinning fast or slowly, and a single object at the center of mass carries no information about rotation.
A turntable's angular position is θ(t) = bt² − ct³ for t ≥ 0, where b and c are positive constants. At what time t > 0 is the turntable momentarily at rest?
Answer and reasoning
A1.00 b/c A student who thinks the turntable is at rest when its angular position is zero picks this: bt² − ct³ = 0 at t = b/c. At that time ω = 2b²/c − 3b²/c = −b²/c, so the turntable is turning, not at rest.
B0.67 b/cCorrect At rest means ω = 0. Differentiating, ω = dθ/dt = 2bt − 3ct², which is zero at t = 2b/(3c) = 0.67 b/c. Before this time the turntable turns in the positive sense; afterward it turns back.
C0.33 b/c A student who thinks the turntable is at rest when its angular acceleration is zero picks this: α = 2b − 6ct = 0 at t = b/(3c). At that instant ω = b²/(3c), its greatest value, not zero.
D1.33 b/c A student who integrates θ instead of differentiating it picks this: ∫θ dt = bt³/3 − ct⁴/4, which is zero at t = 4b/(3c). Angular velocity is the slope dθ/dt, not the area under the θ–t graph.
Working At rest means ω = 0. ω = dθ/dt = 2bt − 3ct² = t(2b − 3ct), which is zero for t > 0 at t = 2b/(3c) ≈ 0.67 b/c. (θ = 0: t = b/c. α = dω/dt = 2b − 6ct = 0: t = b/(3c) ≈ 0.33 b/c, where ω is greatest. Integrating θ instead: bt³/3 − ct⁴/4 = 0 at t = 4b/(3c) ≈ 1.33 b/c.)
For θ > 0, a disk's angular velocity depends on its angular position as ω = kθ, where k is a positive constant. What is the disk's angular acceleration when its angular position is θ₁?
Answer and reasoning
Ak A student who differentiates ω with respect to θ picks this: dω/dθ = k. Angular acceleration is dω/dt, so this must be multiplied by dθ/dt = ω; k alone has units of 1/s, not rad/s².
Bkθ₁ A student who takes the angular velocity at θ₁ as the angular acceleration picks this. kθ₁ is ω at that position; α is how fast ω is changing, which is k²θ₁.
Ck²θ₁Correct α is the rate of change of ω with time. By the chain rule, α = (dω/dθ)(dθ/dt) = k × ω = k × kθ₁ = k²θ₁, which has units of rad/s² because k has units of 1/s.
Dk²θ₁/2 A student who uses ω² = ω₀² + 2α(θ − θ₀) from θ = 0 picks this: α = (kθ₁)²/(2θ₁). That equation assumes α is constant, but here α = k²θ grows as the disk turns.
Working α = dω/dt = (dω/dθ)(dθ/dt) = k·ω = k(kθ) = k²θ, so at θ₁, α = k²θ₁ (units: k in 1/s, so rad/s²). (dω/dθ alone: k. ω itself: kθ₁. ω² = 2αθ from θ = 0 with constant α: α = (kθ₁)²/(2θ₁) = k²θ₁/2.)
A bicycle wheel is spinning freely at 16.0 rad/s when a brake gives it a constant angular acceleration of magnitude 8.00 rad/s² until it stops. How many revolutions does the wheel make while stopping?
Answer and reasoning
A5.09 rev A student who multiplies the initial angular velocity by the stopping time picks this: 16.0 rad/s × 2.00 s = 32.0 rad = 5.09 rev. The wheel slows the whole time, so its average angular velocity is only half of 16.0 rad/s.
B7.64 rev A student who enters the angular acceleration as +8.00 rad/s² in θ = ω₀t + (1/2)αt², with t = 2.00 s, gets 32.0 rad + 16.0 rad = 48.0 rad = 7.64 rev. The wheel is slowing, so α is opposite in sign to ω₀ and the (1/2)αt² term subtracts.
C16.0 rev A student who reports the angle in radians as a number of revolutions picks this. The wheel turns through 16.0 rad, which is 16.0/(2π) = 2.55 revolutions.
D2.55 revCorrect With ω = 0 at the end, ω² = ω₀² + 2αΔθ gives Δθ = ω₀²/(2|α|) = (16.0 rad/s)²/(16.0 rad/s²) = 16.0 rad. Dividing by 2π rad per revolution gives 2.55 rev.
Working Take the initial sense as positive: ω₀ = +16.0 rad/s, α = −8.00 rad/s², ω = 0. ω² = ω₀² + 2αΔθ gives Δθ = (0 − (16.0 rad/s)²)/(2 × −8.00 rad/s²) = 16.0 rad. In revolutions: 16.0 rad/(2π rad/rev) = 2.55 rev. (ω₀ kept constant for the stopping time of 2.00 s: 32.0 rad = 5.09 rev. α entered as +8.00 rad/s² in θ = ω₀t + (1/2)αt² with t = 2.00 s: 48.0 rad = 7.64 rev. 16.0 rad reported as revolutions: 16.0 rev.)
A wheel starts from rest at t = 0, and its angular position is θ = kt³, where k is a positive constant. At time T its angular velocity is 1.50 rad/s. What is its angular velocity at time 2T?
Answer and reasoning
A12.0 rad/s A student who scales ω the way θ scales picks this: θ ∝ t³ grows by a factor of 8 when t doubles. ω = dθ/dt ∝ t², so it grows by a factor of 4.
B3.00 rad/s A student who assumes ω is proportional to t, as it would be for constant α from rest, picks this. Here ω = 3kt², so doubling t multiplies ω by 4, not 2.
C6.00 rad/sCorrect Differentiate: ω = 3kt², so ω is proportional to t². Doubling the time multiplies ω by 2² = 4: ω(2T) = 4 × 1.50 rad/s = 6.00 rad/s.
D2.00 rad/s A student who takes θ/t at time 2T as the angular velocity picks this: 8kT³/(2T) = 4kT², which is 4/3 of ω(T) = 3kT². θ/t is the average angular velocity since t = 0, not the instantaneous value.
Working ω = dθ/dt = 3kt², so ω ∝ t²: doubling t multiplies ω by 4. ω(2T) = 4 × 1.50 rad/s = 6.00 rad/s. (Scaled like θ ∝ t³: × 8 = 12.0 rad/s. Direct proportion: × 2 = 3.00 rad/s. θ/t at 2T: 8kT³/(2T) = 4kT² = (4/3)(3kT²), giving 2.00 rad/s.)
The graph shows the angular velocity ω of a disk as a function of time t, with counterclockwise taken as positive. What is the disk's angular displacement from t = 0 to t = 9 s?
Answer and reasoning
A+34 rad A student who adds all the areas as positive picks this: 30 rad + 4 rad = 34 rad, the total angle turned. From 7 s to 9 s the disk turns clockwise, so that area subtracts.
B+26 radCorrect The angular displacement is the signed area under the ω–t graph. Above the axis, from 0 to 7 s: 6 + 18 + 6 = 30 rad. Below the axis, from 7 s to 9 s: −4 rad. So Δθ = 30 rad − 4 rad = +26 rad.
C+30 rad A student who leaves out the area below the time axis picks this. From 7 s to 9 s the disk turns clockwise through 4 rad, which reduces the angular displacement to 26 rad.
D−36 rad A student who multiplies the final angular velocity by the whole time picks this: −4 rad/s × 9 s = −36 rad. ω changes throughout, so Δθ must be found from the area under the graph.
Working Δθ is the signed area between the graph and the t-axis. 0–2 s: (1/2)(2 s)(6 rad/s) = 6 rad. 2–5 s: (3 s)(6 rad/s) = 18 rad. 5–7 s: (1/2)(2 s)(6 rad/s) = 6 rad. 7–9 s: (1/2)(2 s)(−4 rad/s) = −4 rad. Δθ = 6 + 18 + 6 − 4 = +26 rad. (Magnitudes added: 34 rad. Area above the axis only: 30 rad. Final ω × 9 s: −36 rad.)
The graph shows the angular position θ of a wheel as a function of time t, with three points marked on it. Which ranking of the wheel's angular speeds |ω₁|, |ω₂| and |ω₃| at points 1, 2 and 3 is correct?
Answer and reasoning
A|ω₂| > |ω₃| > |ω₁| A student who reads the height of the θ–t graph as the angular speed picks this, ranking 8 rad > 5 rad > 2 rad. The angular speed is the steepness of the graph, not its height.
B|ω₁| = |ω₂| > |ω₃| A student who uses θ/t, the slope of a line from the origin, picks this: 2 rad/1 s = 8 rad/4 s = 2 rad/s, and 5 rad/7 s ≈ 0.7 rad/s. θ/t is the average angular velocity since t = 0; the angular speed at a point is the slope of the tangent there.
C|ω₁| > |ω₂| > |ω₃| A student who compares signed slopes picks this, putting point 3's negative slope last. At point 3 the wheel is turning back, but faster than at point 2: angular speed is |ω|, so the sign does not make it smaller.
D|ω₁| > |ω₃| > |ω₂|Correct Angular speed is the magnitude of the slope of the θ–t graph. The graph is steepest at point 1, falls moderately steeply at point 3 and is nearly flat at point 2, close to the top of the curve, so |ω₁| > |ω₃| > |ω₂|.
Working Angular speed is the magnitude of the slope of the θ–t graph. From tangent lines: about +3 rad/s at point 1, about +0.5 rad/s at point 2 (close to the maximum) and about −1.5 rad/s at point 3. So |ω₁| > |ω₃| > |ω₂|. (Heights: 8 rad > 5 rad > 2 rad gives 2, 3, 1. θ/t: 2 rad/1 s = 8 rad/4 s = 2 rad/s > 5 rad/7 s. Signed slopes: +3 > +0.5 > −1.5.)
Disk 1 starts from rest and turns with constant angular acceleration α for a time 2t. Disk 2 starts from rest and turns with constant angular acceleration 2α for a time t. How do the angles turned by the two disks compare?
Answer and reasoning
AThe disks turn through equal angles, since the two changes cancel. A student who thinks that doubling one variable and halving the other always cancels picks this. The time is squared in θ = (1/2)αt², so doubling the time matters more than doubling α.
BDisk 1 turns through twice the angle that disk 2 turns through.Correct From rest, θ = (1/2)αt². Disk 1: (1/2)α(2t)² = 2αt². Disk 2: (1/2)(2α)t² = αt². Doubling the time quadruples the angle, while doubling α only doubles it, so disk 1 turns through twice the angle.
CDisk 2 turns through twice the angle that disk 1 does. A student who thinks the disk with the greater angular acceleration turns through the greater angle, in proportion to α, picks this. That leaves out the time, which is twice as long for disk 1 and enters squared.
DDisk 1 turns through four times the angle that disk 2 does. A student who compares only the times picks this: θ ∝ t² gives a factor of 4. Disk 2's angular acceleration is twice disk 1's, which halves that factor to 2.
Working From rest, θ = (1/2)αt². Disk 1: (1/2)α(2t)² = 2αt². Disk 2: (1/2)(2α)t² = αt². So θ₁ = 2θ₂. (Changes assumed to cancel: equal. θ ∝ α only: θ₂ = 2θ₁. θ ∝ t² only: θ₁ = 4θ₂.)
A wheel turning counterclockwise (taken as positive) slows down, stops for an instant and then turns clockwise with increasing angular speed; its angular velocity changes at a steady rate throughout. A student claims that the wheel's angular acceleration is zero at the instant it stops. Which statement correctly evaluates the claim?
Answer and reasoning
ACorrect: ω is zero at that instant, and therefore its rate of change, α, is zero as well. A student who thinks zero angular velocity means zero angular acceleration picks this. α depends on how ω is changing, not on its value, and ω is changing as it passes through zero.
BCorrect: α reverses when the rotation reverses, so it passes through zero at that instant. A student who thinks a wheel's angular acceleration is in its sense of rotation picks this. α is negative throughout: opposite to ω while the wheel slows counterclockwise and in the same sense as ω while it speeds up clockwise. It does not change sign when the rotation does.
CIncorrect: ω is still changing as it passes through zero, so α is not zero then.Correct A steady rate of change of ω means a constant, nonzero α: ω goes from positive to negative at the same rate before, at and after the instant it passes through zero. Zero ω at an instant does not mean zero α.
DIncorrect: α must be positive at that instant to start the wheel turning again. A student who thinks speeding up needs a positive angular acceleration picks this. After the instant the wheel speeds up clockwise, in the negative sense, so α is negative, as it was while the wheel slowed.
A wheel starts from rest and turns with constant angular acceleration. It turns through 1.50 rad during the first 1.00 s. Through what angle does it turn during the next 1.00 s?
Answer and reasoning
A4.50 radCorrect From rest, θ = (1/2)αt², so θ ∝ t². In 2.00 s the wheel turns 2² = 4 times the first-second angle, 6.00 rad. The angle turned during the second 1.00 s is 6.00 rad − 1.50 rad = 4.50 rad.
B1.50 rad A student who thinks constant angular acceleration means equal angles in equal times picks this. That describes constant ω; with constant α the wheel turns faster each second, so it covers a larger angle in the second interval.
C3.00 rad A student who multiplies the angular velocity at the start of the second interval by its length picks this: α = 3.00 rad/s², so ω = 3.00 rad/s at 1.00 s, and 3.00 rad/s × 1.00 s = 3.00 rad. ω keeps increasing during the interval, so the angle is larger.
D6.00 rad A student who gives the wheel's angular position at 2.00 s, measured from the start, picks this. The angle turned during the second interval is the change in angular position from 1.00 s to 2.00 s: 6.00 rad − 1.50 rad = 4.50 rad.
Working From rest, θ = (1/2)αt², so θ ∝ t². In 2.00 s the wheel turns 2² = 4 times the angle of the first 1.00 s: 4 × 1.50 rad = 6.00 rad. Angle during the second 1.00 s = 6.00 rad − 1.50 rad = 4.50 rad. (Equal angles in equal times: 1.50 rad. α = 2(1.50 rad)/(1.00 s)² = 3.00 rad/s², so ω = 3.00 rad/s at 1.00 s, kept constant: 3.00 rad. Angular position at 2.00 s: 6.00 rad.)
A turbine starts from rest at t = 0. Its angular acceleration is α = α₀(1 − t/T) for 0 ≤ t ≤ T and zero for t > T, where α₀ and T are positive constants. Through what angle does the turbine turn from t = 0 to t = 2T?
Answer and reasoning
A0.33 α₀T² A student who thinks zero angular acceleration means the turbine is at rest picks this: only the first stage counts, α₀T²/3. After T, α = 0 means only that ω stops changing; the turbine keeps turning at α₀T/2.
B0.83 α₀T²Correct Integrating α from rest gives ω = α₀(t − t²/(2T)), so by t = T the turbine has turned α₀T²/3 and reached ω = α₀T/2. After T, α = 0 means ω stays at α₀T/2, which adds α₀T²/2 over the next interval T. Total: (5/6)α₀T² ≈ 0.83 α₀T².
C1.00 α₀T² A student who multiplies the angular velocity at t = 2T, α₀T/2, by the whole interval 2T picks this: α₀T². During the first stage ω was less than α₀T/2, so the turbine turned through less than (α₀T/2)T then.
D0.75 α₀T² A student who uses the first stage's average angular acceleration, α₀/2, in θ = ½αt² picks this: ¼α₀T² for the first stage plus α₀T²/2 afterward. Because α is largest early, ω grows faster at first than a constant α₀/2 would make it, and the first-stage angle is α₀T²/3, not α₀T²/4.
Working For 0 ≤ t ≤ T: ω = ∫₀ᵗ α₀(1 − t′/T) dt′ = α₀(t − t²/(2T)), so ω(T) = α₀T/2, and θ(T) = ∫₀ᵀ ω dt = α₀(T²/2 − T²/6) = α₀T²/3. For T < t ≤ 2T: α = 0, so ω stays α₀T/2 and the turbine turns a further (α₀T/2)(T) = α₀T²/2. Total: α₀T²/3 + α₀T²/2 = (5/6)α₀T² ≈ 0.83 α₀T². Distractors: α = 0 taken to mean at rest after T, α₀T²/3 ≈ 0.33 α₀T²; ω at t = 2T multiplied by the whole interval, (α₀T/2)(2T) = 1.00 α₀T²; average angular acceleration α₀/2 in θ = ½αt² for the first stage, ¼α₀T², plus α₀T²/2, = 0.75 α₀T². Key and distractors checked with sympy.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
