Rotational Motion solutions

For any circular path, the net inward force is m v²/r; “centripetal” describes the resultant, rather than a new interaction.

Why: An inward acceleration needs a real force; giving that resultant a name does not tell us what causes it.

(a) In a level banked turn, the horizontal component of the aerodynamic lift supplies the inward resultant; the vertical component balances weight.

Why: The bank tilts lift, so one component holds the plane up while the other turns it.

(b) The road normal force has an inward horizontal component because the road is banked. Tire–road static friction can add to or oppose that component, depending on speed; on a wet road its available magnitude is reduced.

Why: Tilting the road tilts its normal force; friction adjusts the inward total until the tires reach their traction limit.

(c) The inward component of tension in the vine bends Tarzan’s path. If the vine is tilted, separate the horizontal inward component from the vertical component. Gravity still acts downward.

Why: The vine can only pull along itself, so resolve that pull before deciding how much bends the path.

(d) The Sun’s gravitational attraction supplies the inward force on the planet.

Why: For the planet, the Sun pulls toward the center of the orbit, exactly the direction the acceleration needs.

Convert the values: m = 4.00 × 10⁴ kg and F = 1.00 × 10⁴ N.

Why: Use kilograms and newtons so the stated mass and force match the SI units in G.

Using F = G M_E m/r² gives r = √(G M_E m/F) = 3.9956133 × 10⁷ m from Earth’s center.

Why: Solve the inverse-square force law for the center-to-center distance; the given force fixes that radius.

Subtract the surface radius: h = r − R_E = 3.3576133 × 10⁷ m = 33,576 km.

Why: The force law measures from Earth’s center, so remove Earth’s radius to get height above its surface.

A circular orbit at that radius requires v = √(G M_E/r) = 3160.54 m/s. At only 45 m/s, m v²/r = 2.02722 N, far below the stated 10 kN gravitational force.

Why: At the force-implied radius, gravity must equal the required inward force; the stated speed fails that check.

Conversely, 45 m/s as a circular speed would require r = G M_E/v² = 1.9709785 × 10¹¹ m in an Earth-only model, with force only about 4.06 × 10⁻⁴ N. Thus all three original conditions cannot hold together.

Why: Now test the speed on its own: it demands a completely different radius and force, so the three inputs cannot describe one circular orbit.

Initial translational energy is K_i = ½ m v_i² = ½(9)(40²) = 7200 J.

Why: The pebble’s stated loss applies to the center-of-mass motion, so start with translational kinetic energy only.

Immediately after launch, K_f = 7200 − 2000 = 5200 J, so v_f² = 2 K_f/m = 1155.5556 m²/s² and v_f = 33.9935 m/s.

Why: Remove the specified 2000 J, then solve the remaining translational energy for the launch speed.

Only the vertical translational component can be converted into gravitational energy: v_y = v_f sin 30° = 16.9967 m/s.

Why: Height depends on the upward velocity component; the 30° launch angle picks out that component.

At the highest point v_y = 0; h = v_y²/(2g) = 14.7242 m. The horizontal speed and rotational energy remain present, so neither is available to increase this height.

Why: Gravity stops the upward motion at the peak, while horizontal motion and unchanged spin still retain their energy.

With negligible external torque about the spin axis, I_i ω_i = I_f ω_f, so ω_f = 30(I_i/I_f) rpm.

Why: With no external twisting effect, the skater’s angular momentum stays fixed as her inertia changes.

The total inertia depends on the torso, legs, arms, their masses and their positions. The total mass 75 kg and two hand separations do not determine I_i and I_f.

Why: Hand spacing leaves the rest of her mass distribution unknown, so it cannot give the total inertia ratio.

For the additional idealized model only, take I ∝ r² with r_i = 0.50 m and r_f = 0.20 m. Then I_i/I_f = (0.50/0.20)² = 6.25.

Why: Only the extra model lets us replace the inertia ratio with the squared ratio of the two hand radii.

That model predicts ω_f = 187.5 rpm = 19.635 rad/s. A real skater with a nonzero, unchanged torso inertia would generally have a smaller increase than this arm-distance-only model predicts.

Why: Apply that model’s ratio to the starting spin; a torso that keeps its inertia would reduce the predicted increase.

For a constant-speed circle, a_c = v²/r. Thus r ≥ v²/a_max = 325²/62 = 1703.63 m.

Why: A smaller turn needs a larger inward acceleration at the same speed, so the acceleration cap sets a minimum radius.

The semicircular distance is s = πr. At fixed speed the shortest time uses the smallest permitted radius.

Why: A U-turn covers half a circumference, and reducing that distance reduces the time when speed is fixed.

t_min = πr_min/v = πv/a_max = π(325)/62 = 16.4680 s.

Why: Divide the shortest allowed semicircular path by the plane’s speed to get its minimum turning time.

Start with the bottom beam, since equilibrium of every separate beam is necessary. Its length is 48 cm, so its center is 24 cm from the left end, or 4 cm to the right of its support.

Why: Each beam must balance separately; start at the bottom, where uniform mass places the beam’s weight at its midpoint.

Take counterclockwise torque as positive about that support. The support force has zero lever arm. The left mass contributes +(3 kg)g(0.20 m); the beam’s own weight contributes −(30 kg)g(0.04 m); the star contributes −m_star g(0.28 m).

Why: Taking torques about the support removes its force; left and right downward loads twist the beam in opposite directions.

Setting the sum to zero gives m_star = [3(0.20) − 30(0.04)]/0.28 = −2.142857 kg. That would be a weight of −21.0214 N.

Why: Set those signed torques to zero and solve for the star’s mass, keeping the beam’s own clockwise contribution.

A hanging object cannot have negative weight. Even with no star, the beam already has a net clockwise torque. Therefore the bottom beam cannot be static, and no physical values for the other unknown hanging weights can make the complete pictured mobile static.

Why: The negative result means even a weightless star cannot balance the bottom beam, so the whole mobile fails this model.

If the beams are not uniform, their center-of-mass locations must be supplied. Ignoring their masses would define a different problem and cannot justify an answer to the labeled diagram.

Why: Changing the beam’s mass distribution changes its lever arm; we need that information before trying a different equilibrium model.

The initial circular acceleration is a_i = v_i²/r_i = 16/0.850 = 18.8235 m/s² inward.

Why: The initial path is circular, so its inward acceleration follows from the known speed and radius.

For a force through the central hole, torque about the hole is zero. Therefore angular momentum m r v_t is conserved.

Why: A radial pull has no lever arm about the hole, which makes angular momentum about that hole constant.

At the specified circular initial and final instants, v_f = r_i v_i/r_f = (0.850)(4.0)/0.150 = 22.6667 m/s.

Why: At the two circular instants the velocity is tangential, so the smaller radius requires a larger speed to keep mrv fixed.

The central pull does positive work as the radius decreases. K_i = 32.0 J and K_f = 1027.56 J, so the increase is 995.56 J; conservation of angular momentum does not imply conservation of kinetic energy.

Why: Pulling inward transfers energy to the mass, so angular-momentum conservation does not keep its speed or kinetic energy fixed.

For literal ideal wrapping around a stationary finite cylinder, tension lies along the free string, tangent to the cylinder, so it has torque about the cylinder’s center. In the smooth ideal winding model it is perpendicular to the mass’s instantaneous velocity and does no work, giving constant speed 4.0 m/s. The path is an involute, not a succession of center-based circles; cylinder radius and free-string geometry are needed to find its curvature acceleration.

Why: Wrapping changes the tension’s line of action; in the ideal winding geometry it turns the path without doing work, so the central-pull result does not apply.

For a thin hoop, I = m r². Rim speed gives ω = v/r.

Why: All of a thin hoop’s mass lies at its radius, and its rim speed converts to angular speed using that radius.

Substitute into K_rot = ½Iω²: K_rot = ½m r²(v/r)² = ½mv². The radius cancels when rim speed is fixed.

Why: Substituting the same rim-speed condition cancels the radius, leaving energy dependent on mass and rim speed.

K_1 = ½(0.70)(1.5²) = 0.7875 J; K_2 = ½(0.90)(1.5²) = 1.0125 J.

Why: Evaluate that reduced expression separately for the two given hoop masses.

The heavier hoop stores K_2 − K_1 = 0.2250 J more rotational energy. This is stored mechanical energy; metabolic energy to use a hoop would require additional information.

Why: Subtract the two stored energies; the problem gives no model for the person’s energy use.

Being fixed in longitude requires orbital angular speed 2π/T equal to Earth’s sidereal rotation rate.

Why: To stay over one longitude, the satellite must complete one orbit in the same time Earth turns once.

Equate gravitational and centripetal acceleration: G M_E/r² = (2π/T)²r, so r³ = G M_E T²/(4π²).

Why: Gravity provides the inward acceleration, and the matching rotation period fixes the angular speed in that equation.

Using T = 86164.1 s gives r = 4.2182595 × 10⁷ m. Altitude is h = r − R_E ≈ 3.5802595 × 10⁷ m = 35,803 km.

Why: Take the cube root to obtain orbital radius, then subtract Earth’s radius because the question asks for altitude.

A geostationary orbit must lie in the equatorial plane and travel prograde. An inclined orbit moves north and south in the sky, so it cannot remain directly above one non-equatorial city.

Why: An inclined orbit changes the satellite’s latitude, so matching periods alone cannot keep it over a non-equatorial city.

The disk inertia is I = ½M R² = 0.0810 kg·m². Let a be positive downward for the heavier right block and upward for the left block.

Why: The pulley is a disk about its center; choose each block’s positive direction to follow the same string motion.

The block equations are 7.5g − T_R = 7.5a and T_L − 2.8g = 2.8a. The pulley equation is (T_R − T_L)R = Iα, with α = a/R.

Why: Write one force balance per block and one torque balance for the disk; no slipping connects a to α through R.

Eliminating tensions gives a = (7.5 − 2.8)g/[7.5 + 2.8 + I/R²] = 3.602109 m/s².

Why: Eliminate the two tensions so the mass difference drives both the blocks’ motion and the pulley’s rotation.

The requested RIGHT tension is T_R = 7.5(g − a) = 46.5592 N. For comparison, the left tension is T_L = 2.8(g + a) = 37.5539 N; the difference accelerates the disk.

Why: The right block accelerates downward, so its tension is less than its weight; unequal tensions supply the pulley’s torque.

Angular acceleration has magnitude α = a/0.18 = 20.0117 rad/s², clockwise in the shown front view as the right block descends.

Why: The string’s tangential acceleration equals the rim’s acceleration, fixing both α and its clockwise direction.

Both blocks move equal distances. To close a 3.14 m center separation, each travels s = 1.57 m. From v² = 2as, v = 3.36313 m/s. This also follows from (m_R − m_L)gs = ½(m_R + m_L)v² + ½I(v/R)².

Why: The string makes both blocks travel equally, so each closes half the gap; constant acceleration then gives their passing speed.

With constant α, ω_f = ω_i + αt = 4 + 1.5(8) = 16 rad/s.

Why: Constant angular acceleration adds the same angular-speed change each second.

Angular displacement is θ = ω_it + ½αt² = 4(8) + ½(1.5)(8²) = 80 rad.

Why: The starting spin contributes ω_i t and the steadily increasing spin adds the quadratic term.

Divide by 2π radians per revolution: N = 80/(2π) = 12.7324 revolutions.

Why: One full turn is 2π radians, so divide the angular displacement by that amount.

The target acceleration is a_c = 50(9.81) = 490.5 m/s².

Why: The phrase 50g specifies an acceleration multiplier, so convert it to meters per second squared first.

Use a_c = ω²r: ω = √(490.5/0.120) = 63.9336 rad/s.

Why: The sample’s fixed radius converts the target inward acceleration into the needed angular speed.

Convert to rpm: n = ω(60)/(2π) = 610.521 rpm.

Why: Convert radians to turns and seconds to minutes to express the centrifuge setting in rpm.

Vertical equilibrium gives N = mg.

Why: The level car has no vertical acceleration, so the road supports its full weight.

At the traction limit, m v²/r = μ_s N = μ_s mg.

Why: Static tire friction supplies the inward force and reaches μ_sN at the largest non-skidding speed.

Mass cancels: v_max = √(μ_sgr) = √(0.45 × 9.81 × 80) = 18.7926 m/s.

Why: Cancel the car’s mass and solve for speed; a larger radius or stronger grip raises the speed limit.

The normal force has vertical component N cosθ and inward component N sinθ.

Why: The tilted road makes the normal force point partly upward and partly toward the circle’s center.

The equations are N cosθ = mg and N sinθ = m v²/r. Dividing gives tanθ = v²/(rg).

Why: Vertical balance and inward acceleration use the same N, so dividing the equations removes that unknown force.

Thus v = √(rg tanθ) = √[80(9.81)tan15°] = 14.5013 m/s.

Why: Solve the resulting tangent relation for the speed the banking alone can support.

The circular radius is r = ℓ sinθ = 1.20 sin30° = 0.600 m.

Why: The circular radius is the string’s horizontal projection, rather than its whole length.

Vertical balance gives T cosθ = mg, so T = 0.80(9.81)/cos30° = 9.0621 N.

Why: The bob stays at one height, so the upward tension component must balance its weight.

The inward equation T sinθ = mω²r, together with r = ℓsinθ, gives ω² = g/(ℓcosθ).

Why: The horizontal tension component turns the bob; substitute the radius relation to remove the sine factor.

The period is P = 2π√(ℓcosθ/g) = 2.0450 s.

Why: Convert the resulting angular speed into the time for one revolution.

At the top, both weight and normal force point inward: mg + N = m v_top²/R.

Why: At the loop’s top, both available forces point toward the center, so they add in the inward force balance.

At the limiting contact condition N = 0, v_top² = gR.

Why: The track cannot pull the bead toward itself; the smallest contact-preserving speed is the one with N just zero.

Energy from the release point to the top is mgh = mg(2R) + ½m v_top².

Why: The frictionless climb trades starting gravitational energy for height and the required kinetic energy at the top.

Substitute the limiting speed: h_min = 2R + R/2 = 2.5R = 2.25 m.

Why: Insert the minimum top speed into the energy equation to get the smallest release height.

For the rod, I_rod = mL²/3 = 2(1.20²)/3 = 0.960 kg·m².

Why: The rod turns about its end, so use the end-axis inertia rather than its center-axis value.

The point mass contributes I_point = m_pL² = 0.40(1.20²) = 0.576 kg·m².

Why: Treat the attached mass as a point at distance L from the same pivot.

Inertias about the same axis add: I_total = 1.536 kg·m².

Why: Both objects share the pivot and angular acceleration, so their inertias about that pivot add.

Use τ_net = I_total α = 1.536(2.0) = 3.072 N·m.

Why: Multiply the combined inertia by the requested angular acceleration to find the net torque.

For a uniform annulus about its symmetry axis, I = ½m(R_o² + R_i²).

Why: The inner hole removes near-axis material, so use the annulus formula for the specified symmetry axis.

I = ½(2.0)(0.25² + 0.10²) = 0.0725 kg·m².

Why: Insert both radii and the rotor’s mass to obtain its inertia in kg·m².

The energy is K_rot = ½Iω² = ½(0.0725)(20²) = 14.5 J.

Why: Use that axis-specific inertia with the given angular speed to find the rotor’s stored kinetic energy.

For planar vectors, τ_z = xF_y − yF_x.

Why: The z-component of r × F keeps the sign of each force’s turning effect in the xy-plane.

The first torque is 0.40(12) − 0 = +4.80 N·m.

Why: An upward force to the right of the axle produces a positive, counterclockwise torque.

The second is 0 − 0.30(8) = −2.40 N·m. The center force has r = 0, so its torque is zero.

Why: A rightward force above the axle turns clockwise; a force at the axle has no lever arm at all.

Sum signed torques: τ_net,z = 4.80 − 2.40 = +2.40 N·m.

Why: Add the signed contributions, so the stronger counterclockwise torque leaves a positive net result.

The torque is not constant: α(t) = τ/I = (2.0 rad/s³)t.

Why: Divide the changing torque by the fixed inertia; the angular acceleration therefore grows with time too.

Integrate α: ω(t) = 3.0 + t² in rad/s when t is in seconds. Thus ω(5) = 28.0 rad/s.

Why: Accumulating α over time gives the angular-speed change, then the initial 3 rad/s sets the integration constant.

Integrate ω: θ(t) − θ(0) = 3.0t + t³/3. At 5.0 s, θ = 15 + 125/3 = 56.6667 rad.

Why: Accumulating the changing angular speed gives angle; a constant-acceleration shortcut would miss the cubic term.

As an independent check, angular impulse ∫τdt = 50 N·m·s gives Δω = 50/I = 25 rad/s.

Why: The area under the torque-time graph gives ΔL, which checks the speed change independently of the angle calculation.

Taking the initial rotation as positive, α = −6.0/0.80 = −7.50 rad/s².

Why: The resisting torque opposes the chosen positive rotation, so the angular acceleration is negative.

From 0 = ω_i + αt, t = 30/7.50 = 4.00 s.

Why: Find when the constant negative acceleration reduces the initial angular speed to zero.

From 0 = ω_i² + 2αθ, θ = 30²/(2 × 7.50) = 60.0 rad = 9.549 revolutions.

Why: Use the speed-angle relation with zero final speed to get the angle turned before stopping.

Energy check: the initial K = ½(0.80)(30²) = 360 J equals resisting work magnitude 6.0(60.0) = 360 J.

Why: The negative braking work removes all the initial rotational energy, providing a check on the stopping angle.

Along the incline, mg sinθ − f_s = ma. The torque about the center is f_sR = Iα, with α = a/R.

Why: Gravity drives the center downhill while uphill friction produces the spin; no slipping links the two accelerations.

Thus f_s = I a/R² = ½ma and mg sinθ = 1.5ma, giving a = (2/3)g sin20° = 2.2368 m/s² downhill.

Why: Use the cylinder’s inertia to express friction in terms of a, then solve the downhill force balance.

The required friction is f_s = (1/3)mg sinθ uphill. Since N = mg cosθ and f_s ≤ μ_sN, μ_s,min = (1/3)tan20° = 0.1213.

Why: Compare the required uphill friction with the largest available static friction to find the minimum coefficient.

Static friction supplies the rolling torque but dissipates no energy in the ideal no-slip model.

Why: The contact point stays instantaneously still on the fixed ramp, so ideal static friction produces no thermal loss.

While it slips, a = −μ_kg and α = μ_kmgR/I = (5/2)μ_kg/R in the direction that develops rolling.

Why: Sliding friction slows the center but twists the initially nonspinning sphere in the direction needed for rolling.

Hence v = v_0 − μ_kgt and Rω = (5/2)μ_kgt. Set v = Rω: t = 2v_0/(7μ_kg) = 1.01937 s.

Why: Follow both speeds during slipping and stop the calculation when the contact-point speed v − Rω reaches zero.

The rolling speed is v_f = (5/7)v_0 = 5.00 m/s and ω_f = v_f/R = 25.0 rad/s.

Why: Insert the slip time into the translational speed, then use no slipping to convert that final speed to spin.

Initial energy is ½mv_0² = 24.5 J. Final energy is ½mv_f² + ½I(v_f/R)² = (7/10)mv_f² = 17.5 J.

Why: Count both translation and rotation in the final energy; initially only translation was present.

Dissipated energy is 24.5 − 17.5 = 7.0 J; energy conservation cannot be imposed during the slipping stage.

Why: The missing mechanical energy became internal energy during sliding, so a loss-free energy equation would give the wrong rolling speed.

For the mass, mg − T = ma. For the rotor, TR = Iα = I(a/R), so T = Ia/R².

Why: Weight and tension oppose each other on the descending mass; the same tension twists the rotor, with no slipping tying α to a.

Therefore a = mg/(m + I/R²) = 2(9.81)/(2 + 0.080/0.20²) = 4.905 m/s².

Why: Substitute the rotor’s tension into the mass equation; some of the driving weight must accelerate the rotor.

After descending h = 1.0 m, v = √(2ah) = √9.81 = 3.13209 m/s.

Why: The mass starts from rest and has constant acceleration over the stated downward distance.

Energy check: mgh = ½(m + I/R²)v², which gives the same speed.

Why: Gravity’s energy release becomes both translation and rotor spin, giving a separate check on the final speed.

Initial combined inertia is I_i = 0.45 + 0.20(0.50²) = 0.50 kg·m²; final inertia is I_f = 0.45 kg·m².

Why: The insect contributes mr² when away from the axis and contributes zero point-mass inertia when it reaches the axis.

Conserve angular momentum: ω_f = I_iω_i/I_f = 0.50(4.0)/0.45 = 4.44444 rad/s.

Why: No external torque changes the combined angular momentum, so lowering inertia increases the shared angular speed.

K_i = ½(0.50)(4.0²) = 4.00 J; K_f = ½(0.45)(4.44444²) = 4.44444 J.

Why: Calculate rotational energy with the appropriate inertia and speed at each endpoint.

The increase is 0.44444 J. The insect’s internal energy supplies this work; rotational kinetic energy need not be conserved.

Why: The extra rotational energy comes from the insect’s effort; internal work can change kinetic energy while total angular momentum stays fixed.

Conserve signed angular momentum: L_i = 0.40(10) + 0.60(−2.0) = 2.80 kg·m²/s.

Why: The flywheels spin in opposite directions, so their angular momenta must be added with signs.

The combined inertia is 1.00 kg·m², so ω_f = 2.80/1.00 = +2.80 rad/s.

Why: The locked flywheels share one speed and their inertias add; divide the conserved angular momentum by that total.

K_i = ½(0.40)(10²) + ½(0.60)(2.0²) = 21.20 J.

Why: Kinetic energy depends on squared speeds, so both initial rotations contribute positive energy.

K_f = ½(1.00)(2.80²) = 3.92 J. Internal energy gains 21.20 − 3.92 = 17.28 J due to clutch slipping.

Why: Compare the final shared-spin energy with the initial total; clutch slipping turns the difference into internal energy.

The momentum is p = m v = (−1.20, 0.90) kg·m/s.

Why: Multiply each velocity component by the particle’s mass to obtain the momentum components.

Initially, L_z = x p_y − y p_x = 2(0.90) − 1(−1.20) = +3.00 kg·m²/s.

Why: Use r × p about the origin; subtracting the negative y p_x term gives positive z-angular momentum.

After 8.0 s, r = (2 − 32, 1 + 24) = (−30, 25) m. Then L_z = (−30)(0.90) − 25(−1.20) = +3.00 kg·m²/s.

Why: Advance the position by vt and reuse the unchanged momentum to check the angular momentum at the later time.

With zero force there is zero torque about the fixed origin, so L is constant even though the particle follows a straight line.

Why: Straight-line motion can have angular momentum about an off-path origin; zero torque keeps that quantity fixed.

The impact lever arm is L/2 = 0.60 m. The initial angular momentum about the pivot is L_i = m_bv(0.60) = 1.20 kg·m²/s.

Why: The perpendicular hit occurs half a rod length from the pivot, so that distance is the bullet’s angular-momentum lever arm.

After impact, I = M L²/12 + m_b(0.60)² = 0.144 + 0.0036 = 0.1476 kg·m².

Why: After embedding, the bullet adds point-mass inertia to the rod’s center-pivot inertia.

Angular momentum gives ω_f = 1.20/0.1476 = 8.13008 rad/s.

Why: The brief impact conserves angular momentum about the pivot, so divide the bullet’s initial L by the combined final inertia.

K_f = L_i²/(2I) = 4.87805 J. The initial bullet energy was 200 J; most becomes internal energy in this inelastic impact.

Why: Find the final spin energy using the conserved L; embedding is inelastic, so the much larger bullet energy need not survive as rotation.

The torque magnitude is τ = rF sinθ, where θ is the angle between the handle position vector and the force.

Why: Only the force component perpendicular to the handle turns the bolt, giving the sine factor.

At 30°, F = 18/[0.25 sin30°] = 144 N.

Why: Solve for the angled force needed to produce the specified torque with that wrench length.

For a perpendicular pull, sin90° = 1, so F = 18/0.25 = 72 N.

Why: At a right angle the whole force is perpendicular, so the same torque needs a smaller pull.

The radial component of the angled pull adds no torque; choosing a perpendicular pull uses the whole force effectively.

Why: The component along the handle acts through the bolt and contributes no turning effect.

Gravity produces torque magnitude τ = Mgr = 1.50(9.81)(0.10) = 1.4715 N·m.

Why: The gyroscope’s weight acts at its center of mass, a horizontal lever arm from the support.

Spin angular momentum is L_spin = Iω = 0.020(200) = 4.00 kg·m²/s.

Why: The given spin-axis inertia and spin speed determine the large angular momentum being redirected.

Thus Ω = τ/L_spin = 1.4715/4.00 = 0.367875 rad/s.

Why: Use the supplied slow-precession relation: the gravitational torque changes the direction of that spin angular momentum.

The precession period is P = 2π/Ω ≈ 17.080 s. Since Ω ≪ ω, the supplied slow-precession approximation is self-consistent.

Why: One precession cycle is 2π radians; compare Ω with the spin speed to check the approximation being used.