Classical-Mechanics · Unit 20 · Video 2 · Interactive Practice

Rotational Sign Conventions and Tangential Velocity: The Turntable Problem

IKey Formulas

FormulaNameWhat it needs
ωz=dθdt,αz=dωzdt=d2θdt2\omega_z = \dfrac{d\theta}{dt}, \qquad \alpha_z = \dfrac{d\omega_z}{dt} = \dfrac{d^2\theta}{dt^2} Angular velocity and acceleration components A right-handed system: +z+z up, θ\theta increasing counterclockwise seen from above
ωzαz>0\omega_z\alpha_z > 0: speeding up ωzαz<0\qquad \omega_z\alpha_z < 0: slowing down The pair of signs Both components — one sign alone decides nothing
vθ,i=ri ωz,aθ,i=ri αzv_{\theta,i} = r_i\,\omega_z, \qquad a_{\theta,i} = r_i\,\alpha_z Tangential velocity and acceleration of a mass element The perpendicular distance rir_i from the axis
ar,i=−vθ,i 2ri=−ri ωz2a_{r,i} = -\dfrac{v_{\theta,i}^{\,2}}{r_i} = -r_i\,\omega_z^2 Radial (centripetal) acceleration Inward for either sense of rotation — ωz\omega_z appears squared

Key Insight: A single sign fixes nothing. αz<0\alpha_z < 0 says only that α⃗\vec{\alpha} points along −k^-\hat{\mathbf{k}}; whether the body speeds up or slows down is the sign of the product ωzαz\omega_z\alpha_z — and flipping which sense of rotation you call positive changes both components together, leaving that product, and the physics, untouched.

IIVisualization 1 — Two Signs, One Verdict

The sign of αz\alpha_z alone fixes nothing; the sign it shares — or doesn't — with ωz\omega_z decides the motion.

IIIVisualization 2 — Twice as Far, Twice as Fast

One angular velocity for the whole disc, but every element has its own speed: the radius sets it.

IVVisualization 3 — The Turntable Spinning Down

A constant angular acceleration is the slope of ωz\omega_z against time — here a straight fall to zero.

Step 1 — The rate as a frequency
f0=(33 cyclesmin) ⁣(1 min60 s)=0.55 cycless=0.55 Hzf_0 = \left(33\,\frac{\text{cycles}}{\text{min}}\right)\!\left(\frac{1\ \text{min}}{60\ \text{s}}\right) = 0.55\ \frac{\text{cycles}}{\text{s}} = 0.55\ \text{Hz}

💡 The disc's 1.2 kg1.2\ \text{kg} and 13 cm13\ \text{cm} never entered either answer: they fix the torque this deceleration demands, not the kinematics. At the moment of the cut a rim point carried vθ=Rωz=+0.46 m/sv_\theta = R\omega_z = +0.46\ \text{m/s} and ar=−Rωz2=−1.59 m/s2a_r = -R\omega_z^2 = -1.59\ \text{m/s}^2, against a tangential aθ=Rαz=−5.6×10−2 m/s2a_\theta = R\alpha_z = -5.6\times 10^{-2}\ \text{m/s}^2 — smaller by a factor of ωz2/∣αz∣≈28\omega_z^2/|\alpha_z| \approx 28.

VQuiz Questions

Problem 1 · Reading the Pair of Signs

Given: A flywheel seen from above turns clockwise, with ωz=−6.0 rad/s\omega_z = -6.0\ \text{rad/s} and αz=−1.5 rad/s2\alpha_z = -1.5\ \text{rad/s}^2 in the standard right-handed convention (k^\hat{\mathbf{k}} up, θ\theta increasing counterclockwise from above) — find whether it is speeding up or slowing down, and where α⃗\vec{\alpha} points.

✅ Correct! Both components are negative, so both vectors point down the axis — same direction, and the wheel turns clockwise ever faster.
❌ Close — the direction is right, the verdict isn't. αz<0\alpha_z < 0 does put α⃗\vec{\alpha} along −k^-\hat{\mathbf{k}}, but the motion is decided by the product: (−6.0)(−1.5)=+9.0>0(-6.0)(-1.5) = +9.0 > 0.
❌ Not quite. α⃗=αzk^\vec{\alpha} = \alpha_z\hat{\mathbf{k}} with αz<0\alpha_z < 0 points along −k^-\hat{\mathbf{k}}, and ωzαz=+9.0 rad2/s3\omega_z\alpha_z = +9.0\ \text{rad}^2/\text{s}^3.
Show solution

Direction of α⃗\vec{\alpha}: the vector is α⃗=αzk^\vec{\alpha} = \alpha_z\hat{\mathbf{k}}, so a negative component means it points along −k^-\hat{\mathbf{k}}, straight down the axis.

Speeding up or slowing down: compare the two signs, never one alone.

ωzαz=(−6.0)(−1.5)=+9.0>0\omega_z\alpha_z = (-6.0)(-1.5) = +9.0 > 0

Same sign, so ω⃗\vec{\omega} and α⃗\vec{\alpha} point the same way (both down the axis) and the flywheel is speeding up — its clockwise rate is growing, ∣ωz∣|\omega_z| climbing from 6.0 rad/s6.0\ \text{rad/s}.

The trap is reading αz<0\alpha_z < 0 as "decelerating". It only means "down the axis". With a clockwise spin, down the axis is along the motion.

Problem 2 · The Same Disc, the Other Convention

Given: Seen from above a disc turns counterclockwise and its rate is increasing. A second student keeps a right-handed system but calls θ\theta increasing clockwise from above, so that unit vector k^′\hat{\mathbf{k}}' points down — find the signs of that student's ωz′\omega_z' and αz′\alpha_z'.

✅ Correct! Both components flip, their product stays positive, and both students agree the disc is speeding up.
❌ Nothing physical moved — but the measuring stick did. Both vectors are now read against a k^′\hat{\mathbf{k}}' that points down, so both components change sign.
❌ Not quite. Speeding up means ω⃗\vec{\omega} and α⃗\vec{\alpha} are parallel; both point up the axis here, so against a downward k^′\hat{\mathbf{k}}' both components come out with the same sign.
Show solution

The physics first. Counterclockwise from above puts ω⃗\vec{\omega} up the axis. The rate is increasing, so α⃗\vec{\alpha} is parallel to ω⃗\vec{\omega} — also up the axis. Neither vector cares which convention anyone chose.

Now the components. The second student's k^′\hat{\mathbf{k}}' points down, so for any vector A⃗\vec{A} up the axis, Az′=A⃗⋅k^′<0A_z' = \vec{A}\cdot\hat{\mathbf{k}}' < 0:

ωz′<0,αz′<0\omega_z' < 0, \qquad \alpha_z' < 0

The verdict is untouched. Both components changed sign together, so

ωz′αz′=(−)(−)>0\omega_z'\alpha_z' = (-)(-) > 0

which is still "speeding up" — the same conclusion the first student reaches from ωzαz=(+)(+)>0\omega_z\alpha_z = (+)(+) > 0. This is why the product, not either sign, is the physical statement.

Problem 3 · A Speck of Dust on the Rim

Given: At the instant the motor is switched off the turntable has ωz=+3.5 rad/s\omega_z = +3.5\ \text{rad/s} and αz=−0.43 rad/s2\alpha_z = -0.43\ \text{rad/s}^2. A speck of dust rides the rim, R=0.13 mR = 0.13\ \text{m} from the axis — find its two acceleration components.

Tangential component aθa_\theta?

Radial component ara_r?

✅ Correct! The inward pull is 2828 times the tangential drag here — and it would survive even if the motor had been left running.
❌ Check the tangential component. It is aθ=Rαza_\theta = R\alpha_z: one factor of the radius, and the sign of αz\alpha_z carried straight through.
❌ Check the radial component. It is ar=−Rωz2a_r = -R\omega_z^2: the radius once, the angular velocity squared, and a minus sign that never leaves.
Show solution

Tangential. Differentiating vθ=Rωzv_\theta = R\omega_z with RR fixed (the disc is rigid) gives

aθ=Rαz=(0.13 m)(−0.43 rad/s2)=−0.0559≈−5.6×10−2 m/s2a_\theta = R\alpha_z = (0.13\ \text{m})(-0.43\ \text{rad/s}^2) = -0.0559 \approx -5.6\times 10^{-2}\ \text{m/s}^2

Negative: the speck is being dragged backwards along its motion, since vθ=Rωz=+0.455 m/sv_\theta = R\omega_z = +0.455\ \text{m/s} is positive.

Radial. Circular motion always accelerates the speck toward the axis:

ar=−vθ 2R=−Rωz2=−(0.13 m)(3.5 rad/s)2=−(0.13)(12.25)=−1.5925≈−1.6 m/s2a_r = -\frac{v_\theta^{\,2}}{R} = -R\omega_z^2 = -(0.13\ \text{m})(3.5\ \text{rad/s})^2 = -(0.13)(12.25) = -1.5925 \approx -1.6\ \text{m/s}^2

Common slips:

  • Dropping the minus on ara_r — it is inward for either sense of rotation, because ωz\omega_z is squared.
  • Forgetting to square: −Rωz=−0.46 m/s2-R\omega_z = -0.46\ \text{m/s}^2 has the wrong units for an acceleration.
  • Forgetting the radius: −ωz2=−12.25-\omega_z^2 = -12.25 is an acceleration per metre.

Ratio: ∣ar∣/∣aθ∣=ωz2/∣αz∣=12.25/0.43≈28|a_r|/|a_\theta| = \omega_z^2/|\alpha_z| = 12.25/0.43 \approx 28.

Problem 4 · A Rotor Coasting to Rest

Given: Seen from above, a rotor spins clockwise at 45 rev/min45\ \text{rev/min}; the drive is cut and it coasts to rest in 12 s12\ \text{s} at constant angular acceleration — find αz\alpha_z in the standard convention (k^\hat{\mathbf{k}} up, θ\theta increasing counterclockwise from above).

✅ Correct! ωz=−4.7 rad/s\omega_z = -4.7\ \text{rad/s} and αz=+0.39 rad/s2\alpha_z = +0.39\ \text{rad/s}^2: opposite signs, so the rotor slows while α⃗\vec{\alpha} points up the axis.
❌ Slowing down does not mean αz<0\alpha_z < 0. It means ωzαz<0\omega_z\alpha_z < 0 — and here ωz\omega_z is already negative, so αz\alpha_z must come out positive.
❌ Check the conversion. 45 rev/min=0.75 Hz45\ \text{rev/min} = 0.75\ \text{Hz}, and each revolution is 2π2\pi radians, so ∣ω0∣=2π(0.75)=4.7 rad/s|\omega_0| = 2\pi(0.75) = 4.7\ \text{rad/s}.
Show solution

Step 1 — the initial angular velocity, with its sign. Clockwise from above is decreasing θ\theta, so ωz\omega_z starts negative:

f0=4560=0.75 Hz,ωz,0=−2πf0=−4.712…≈−4.7 rad/sf_0 = \frac{45}{60} = 0.75\ \text{Hz}, \qquad \omega_{z,0} = -2\pi f_0 = -4.712\ldots \approx -4.7\ \text{rad/s}

Step 2 — constant angular acceleration.

αz=ωz,f−ωz,0Δt=0−(−4.712… rad/s)12 s=+0.3927≈+3.9×10−1 rad/s2\alpha_z = \frac{\omega_{z,f} - \omega_{z,0}}{\Delta t} = \frac{0 - (-4.712\ldots\ \text{rad/s})}{12\ \text{s}} = +0.3927 \approx +3.9\times 10^{-1}\ \text{rad/s}^2

Step 3 — check it against the sign rule.

ωzαz=(−4.7)(+0.39)<0\omega_z\alpha_z = (-4.7)(+0.39) < 0

Opposite signs, so the rotor is slowing down — which is what "coasts to rest" says. A positive αz\alpha_z here is not an acceleration in the everyday sense; it means α⃗\vec{\alpha} points along +k^+\hat{\mathbf{k}}, up the axis, opposing a ω⃗\vec{\omega} that points down.

Two traps: assuming a braking body must have αz<0\alpha_z < 0, and dropping the factor 2π2\pi (which turns 0.75/12=0.06250.75/12 = 0.0625 into a plausible-looking wrong answer).

Solved: 0 / 4