University Physics I · Final Examination

Dry sand drains through a circular hole of diameter D in the flat base of a hopper. Experiments show that the mass flow rate ṁ, in kg/s, does not depend on the depth of sand above the hole. Assume that ṁ depends only on the bulk density ρ of the sand, the gravitational field strength g and the diameter D, and use dimensional analysis. If the diameter of the hole is doubled, by what factor does the mass flow rate increase?

1. A2.8
2. B4.0
3. C5.7
4. D8.0

A particle moves in the xy-plane with position vector r(t) = (2.0t² − 8.0t) î + (6.0t − 1.0t²) ĵ, where r is in metres and t is in seconds. At what time is the speed of the particle least?

1. A2.0 s
2. B2.2 s
3. C2.5 s
4. D3.0 s

A 12 kg crate rests on a loading ramp inclined at 20° to the horizontal. The coefficients of friction between the crate and the ramp are μₛ = 0.60 and μₖ = 0.45. Take g = 9.81 m/s². What is the magnitude of the friction force that the ramp exerts on the crate?

1. A40 N
2. B50 N
3. C66 N
4. D71 N

An atom adsorbed on a crystal surface can move only along a straight row of binding sites. Along the row its potential energy is U(x) = −U₀ cos(2πx/a), where U₀ is a positive constant and a is the spacing of the sites, so that the sites lie at x = 0, ±a, ±2a and so on. For positions between x = 0 and x = a/2, where is the force that the surface exerts on the atom along the row greatest in magnitude, and in which direction does that force point there?

1. AAt x = a/2, pointing towards x = 0
2. BAt x = a/4, pointing away from x = 0
3. CAt x = a/2, pointing away from x = 0
4. DAt x = a/4, pointing towards x = 0

A 0.170 kg ice-hockey puck slides at 6.00 m/s directly towards a player, who strikes it straight back along the same line. The force of the stick on the puck, directed away from the player, rises linearly from zero to 800 N in 4.0 ms, stays at 800 N for 3.0 ms, then falls linearly to zero in a further 3.0 ms. Friction with the ice is negligible. What is the speed of the puck just after the strike?

1. A17.5 m/s
2. B24.6 m/s
3. C30.6 m/s
4. D36.6 m/s

A bicycle wheel of radius 0.35 m rolls without slipping along a level road, its centre moving at 7.0 m/s. Relative to the road, what are the speed of the point at the top of the rim and the speed of a point on the rim level with the axle?

1. A14 m/s at the top; 9.9 m/s level with the axle
2. B14 m/s at the top; 7.0 m/s level with the axle
3. C7.0 m/s at the top; 7.0 m/s level with the axle
4. D14 m/s at the top; 14 m/s level with the axle

A spinning bicycle wheel is mounted at one end of a light horizontal axle whose other end rests on a frictionless pivot at the origin. The axle lies along the x-axis with the wheel's centre at x = +0.150 m, and the z-axis points vertically upwards. The wheel has mass 2.00 kg and moment of inertia 0.120 kg m² about the axle, and it spins at 40.0 rad/s with its spin angular momentum pointing in the +x direction. Take g = 9.81 m/s². In the fast-spin approximation, at what rate does the axle precess, and in which sense does it turn when viewed from above, looking down the z-axis?

1. A0.61 rad/s, clockwise
2. B1.6 rad/s, counterclockwise
3. C1.6 rad/s, clockwise
4. D0.61 rad/s, counterclockwise

Transit and radial-velocity observations show that a rocky exoplanet has 4.5 times the mass of Earth and 1.5 times its radius. Treat both planets as spherically symmetric. Compared with the values for Earth, how do the gravitational field strength at the planet's surface and the escape speed from its surface change?

1. AField strength doubles; escape speed increases by a factor of 1.7
2. BField strength doubles; escape speed increases by a factor of 1.4
3. CField strength increases by a factor of 3.0; escape speed increases by a factor of 1.7
4. DField strength doubles; escape speed increases by a factor of 3.0

A block attached to a horizontal spring slides on a frictionless surface in simple harmonic motion with an amplitude of 0.12 m and a period of 0.80 s. What is the speed of the block when its displacement from equilibrium is 0.060 m?

1. A0.13 m/s
2. B0.47 m/s
3. C0.82 m/s
4. D0.94 m/s

Water of density 1000 kg/m³ flows steadily through a horizontal pipe whose internal diameter narrows from 0.040 m to 0.020 m. The speed of the water in the wide section is 1.5 m/s. Treating the water as an ideal incompressible fluid, by how much is the pressure in the narrow section lower than the pressure in the wide section?

1. A3.4 kPa
2. B10 kPa
3. C17 kPa
4. D18 kPa

A laboratory test rig stores braking energy in a flywheel: a uniform steel annular disc of inner radius R₁ = 0.100 m, outer radius R₂ = 0.300 m and thickness 0.0500 m, which spins about its symmetry axis. The density of the steel is 7.80 × 10³ kg/m³. In normal operation the flywheel spins at ω₀ = 314 rad/s.

1. (a)

   Show that Starting from the definition I = ∫r² dm, show that the moment of inertia of a uniform annular disc of mass M about its symmetry axis is I = ½M(R₁² + R₂²).

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   1. 1

      Divides the disc into thin rings of radius r and width dr, each of mass dm = σ(2πr dr), where σ = M/[π(R₂² − R₁²)] is the mass per unit area

   2. 2

      Writes I = ∫ r²σ(2πr) dr from R₁ to R₂ = (πσ/2)(R₂⁴ − R₁⁴)

   3. 3

      Factorises R₂⁴ − R₁⁴ = (R₂² − R₁²)(R₂² + R₁²) and substitutes for σ to obtain I = ½M(R₁² + R₂²)

2. (b)

   Calculate Calculate the mass of the flywheel, its moment of inertia, and the kinetic energy it stores at ω₀.

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   1. 1

      M = ρπ(R₂² − R₁²) × thickness = 7800 × π × 0.0800 × 0.0500 = 98.0 kg

   2. 2

      I = ½ × 98.0 × (0.0100 + 0.0900) = 4.90 kg m²

   3. 3

      K = ½Iω₀² = ½ × 4.90 × 314² = 2.42 × 10⁵ J

3. (c)

   Determine When the flywheel is disconnected at ω₀ it coasts to rest in 1.50 × 10³ s under a constant frictional torque from its bearings. Determine the frictional torque, the number of revolutions the flywheel makes while coasting, and the rate at which friction dissipates energy at the instant the flywheel is disconnected.

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   1. 1

      Constant angular deceleration α = ω₀/t = 314/1500 = 0.209 rad/s²

   2. 2

      Frictional torque τ = Iα = 4.90 × 0.2093 = 1.03 N m (accept 1.02 N m from rounded intermediate values)

   3. 3

      Angle turned θ = ½ω₀t = 2.36 × 10⁵ rad, which is θ/(2π) = 3.75 × 10⁴ revolutions

   4. 4

      Initial dissipation rate P = τω₀ = 1.026 × 314 = 322 W (accept 320 W to 323 W; consistent with τθ = 2.42 × 10⁵ J, the whole of the stored energy)

4. (d)

   Compare Compare the energy this flywheel stores at ω₀ with the energy stored by a uniform solid disc of the same mass and the same outer radius spinning at the same angular speed, and explain the difference in terms of how the mass is distributed.

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   1. 1

      Solid disc: I = ½MR₂² = 4.41 kg m², which stores ½Iω₀² = 2.17 × 10⁵ J, about 10% less than the annular flywheel

   2. 2

      The annulus has all of its mass at 0.100 m or more from the axis, where each kilogram contributes a larger r² to I, so for the same mass and the same ω it has the larger moment of inertia and stores more kinetic energy

An electric car of mass 1500 kg, driver included, starts from rest on a level road. Its motor controller limits the power delivered to the driven wheels to P = 60.0 kW. Ignore air resistance and rolling resistance throughout, and take g = 9.81 m/s².

1. (a)

   Show that Suppose the full power P is delivered from the instant the car starts. Show that the speed of the car at time t is v(t) = √(2Pt/m).

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   1. 1

      Applies the work–kinetic energy theorem: the work done on the car in time t is ∫₀ᵗ P dt = Pt, and this equals the gain in kinetic energy ½mv² from rest

   2. 2

      Rearranges Pt = ½mv² to v = √(2Pt/m)

2. (b)

   Calculate Using this model, calculate the time the car takes to reach 25.0 m/s and its acceleration at that instant.

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   1. 1

      t = mv²/(2P) = (1500 × 25.0²)/(1.20 × 10⁵) = 7.81 s

   2. 2

      The driving force is P/v = 2.40 × 10³ N, so a = P/(mv) = 1.60 m/s²

3. (c)

   Determine Determine, by integrating v(t), the distance the car travels in reaching 25.0 m/s under this model.

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   1. 1

      Sets up x = ∫₀ᵗ √(2P/m) t^(1/2) dt

   2. 2

      Integrates to x = (2/3)√(2P/m) t^(3/2), which equals (2/3)vt

   3. 3

      x = (2/3) × 25.0 × 7.81 = 130 m

4. (d)

   Evaluate The coefficient of static friction between the tyres and the road is 0.80, and all four wheels are driven, so the car's whole weight is available for traction. Evaluate the constant-power model close to t = 0: find the speed below which the car cannot use its full power, and state what limits its motion below that speed.

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   1. 1

      The model gives a = P/(mv), which grows without limit as v → 0, so it demands an arbitrarily large forward friction force from the road and cannot hold at the start

   2. 2

      The largest traction force is μₛmg = 0.80 × 1500 × 9.81 = 1.18 × 10⁴ N, so full power requires P/v ≤ 1.18 × 10⁴ N, that is v ≥ P/(μₛmg) = 5.10 m/s

   3. 3

      Below 5.10 m/s traction, not power, is the limit: the car can accelerate at no more than μₛg = 7.85 m/s², so the constant-power model is valid only above 5.10 m/s

5. (e)

   Sketch The real car accelerates at the traction limit until it can use its full power, then continues at constant power. Sketch a graph of its speed against time from t = 0 until it reaches 25.0 m/s, and mark the point where the two stages join.

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   1. 1

      A straight line from the origin with gradient 7.85 m/s², ending at the marked join at about (0.65 s, 5.1 m/s)

   2. 2

      From the join, a concave-down curve that starts with the same gradient as the line (no corner) and flattens steadily, reaching 25.0 m/s at about 8.1 s, later than the 7.81 s of the full-power model

A uniform horizontal steel beam of length 3.00 m and mass 45.0 kg projects from a vertical wall, to which one end is attached by a frictionless hinge. A steel cable runs from the outer end of the beam to a point on the wall directly above the hinge, making an angle of 30.0° with the beam. A shop sign of mass 80.0 kg hangs from the beam 2.40 m from the wall. The cable is 3.46 m long and 4.00 mm in diameter, and its steel has Young modulus 2.00 × 10¹¹ Pa and yield strength 2.50 × 10⁸ Pa. The weight of the cable is negligible. Take g = 9.81 m/s².

1. (a)

   Explain Explain why taking torques about the hinge is the most efficient first step in finding the tension in the cable.

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   1. 1

      The force exerted by the hinge on the beam is unknown in both magnitude and direction

   2. 2

      That force acts at the hinge, so it has zero lever arm and zero torque about an axis through the hinge, which leaves the cable tension as the only unknown in the torque equation

2. (b)

   Calculate Calculate the tension in the cable.

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   1. 1

      Torques about the hinge: T sin 30.0° × 3.00 m = (45.0)(9.81)(1.50) + (80.0)(9.81)(2.40) = 662 + 1884 = 2546 N m

   2. 2

      T = 2546/(3.00 × sin 30.0°) = 1.70 × 10³ N

3. (c)

   Determine Determine the magnitude and direction of the force exerted by the hinge on the beam.

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   1. 1

      Horizontal equilibrium: H = T cos 30.0° = 1.47 × 10³ N, directed away from the wall

   2. 2

      Vertical equilibrium: V = (125.0 kg)(9.81 m/s²) − T sin 30.0° = 1226.3 N − 848.6 N = 378 N, directed upwards

   3. 3

      Magnitude √(H² + V²) = 1.52 × 10³ N, at 14.4° above the horizontal and pointing away from the wall

4. (d)

   Calculate Calculate the tensile stress in the cable and the extension of the cable.

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   1. 1

      A = π(2.00 × 10⁻³ m)² = 1.26 × 10⁻⁵ m², so the stress is σ = T/A = 1.35 × 10⁸ Pa

   2. 2

      Strain = σ/E = 6.75 × 10⁻⁴, so the extension is 6.75 × 10⁻⁴ × 3.46 m = 2.34 × 10⁻³ m

5. (e)

   Evaluate The building code requires every load-bearing cable to have a safety factor of at least 2.0 against yield. Evaluate whether this cable complies, and recommend the smallest cable diameter that would.

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   1. 1

      Safety factor = (2.50 × 10⁸ Pa)/(1.35 × 10⁸ Pa) = 1.85, which is below 2.0, so the cable does not comply

   2. 2

      Compliance requires σ ≤ 1.25 × 10⁸ Pa, so A ≥ 1.36 × 10⁻⁵ m² and d ≥ 4.16 mm; recommends the next available size above this, for example 4.5 mm or 5 mm

6. (f)

   Describe Describe what the stress–strain curve for the cable steel would show if the cable were loaded beyond its yield point and then unloaded, and what this would mean for the sign.

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   1. 1

      The curve is straight only up to the proportional limit; past the elastic limit and the yield point it lies well off that line and the strain increases substantially for little further increase in stress: the steel deforms plastically

   2. 2

      On unloading, the curve returns along a straight line parallel to the original elastic line and meets the strain axis at a non-zero strain, so the cable keeps a permanent extension and the beam and sign are left sagging

Car A, of mass 1200 kg and travelling due east, and car B, of mass 1800 kg and travelling due north, collide in the middle of an intersection and lock together. Skid marks show that the wreck then slid in a straight line for 8.00 m, directed 35.0° north of east, before stopping. The coefficient of kinetic friction between the tyres and the road is 0.65, and g = 9.81 m/s². The speed limit at the intersection is 13.9 m/s (50 km/h).

1. (a)

   Justify Justify treating the two cars as an isolated system for the duration of the collision, even though friction from the road acts on them.

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   1. 1

      The collision lasts a very short time, of order 0.1 s, and during it the forces between the cars are of order 10⁵ N, several times the largest friction force the road can exert on the wreck, μₖ(3000 kg)g ≈ 1.9 × 10⁴ N

   2. 2

      The external impulse from the road over that brief interval is therefore small compared with the momentum of either car, so the total momentum just after the collision equals the total just before

2. (b)

   Calculate Calculate the speed of the wreck immediately after the collision.

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   1. 1

      The work done by kinetic friction removes all the wreck's kinetic energy: ½(m_A + m_B)V² = μₖ(m_A + m_B)g d, so V = √(2μₖgd)

   2. 2

      V = √(2 × 0.65 × 9.81 × 8.00) = 10.1 m/s

3. (c)

   Determine Determine the speed of each car immediately before the collision.

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   1. 1

      Momentum just after: (3000 kg)(10.10 m/s) = 3.030 × 10⁴ kg m/s, with components 2.482 × 10⁴ kg m/s east and 1.738 × 10⁴ kg m/s north

   2. 2

      The east component is carried only by car A before the collision: v_A = (2.482 × 10⁴)/1200 = 20.7 m/s

   3. 3

      The north component is carried only by car B: v_B = (1.738 × 10⁴)/1800 = 9.66 m/s (accept 9.67 m/s from a rounded component)

4. (d)

   Calculate Calculate the fraction of the cars' kinetic energy that was dissipated in the collision itself.

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   1. 1

      Just before: ½(1200)(20.7)² + ½(1800)(9.66)² = 2.57 × 10⁵ J + 8.40 × 10⁴ J = 3.41 × 10⁵ J; just after: ½(3000)(10.1)² = 1.53 × 10⁵ J

   2. 2

      Fraction dissipated = (3.41 × 10⁵ − 1.53 × 10⁵)/(3.41 × 10⁵) = 0.55

5. (e)

   Estimate Estimate the magnitude of the average force exerted on a 75 kg occupant of car B, held by a seatbelt, during the collision. State the assumptions you make.

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   1. 1

      States the assumptions: the occupant moves with car B, whose velocity changes from 9.66 m/s north to the wreck's velocity of 8.27 m/s east and 5.79 m/s north, over a collision time of about 0.1 s

   2. 2

      |Δv| = √(8.27² + 3.87²) ≈ 9.1 m/s, so F ≈ (75 kg)(9.1 m/s)/(0.1 s) ≈ 7 × 10³ N, roughly nine times the occupant's weight

6. (f)

   Evaluate The driver of car A claims not to have exceeded the speed limit. The investigator's values are uncertain: μₖ lies between 0.55 and 0.75, and the direction of the skid is 35° ± 3° north of east. Evaluate the driver's claim.

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   1. 1

      Writes v_A = (m_A + m_B)√(2μₖgd) cos θ/m_A and identifies that the smallest value comes from the smallest μₖ, 0.55, and the largest angle, 38°

   2. 2

      v_A,min = (3000/1200) × √(2 × 0.55 × 9.81 × 8.00) × cos 38° = 18.3 m/s (66 km/h)

   3. 3

      Even at the least favourable limits of the evidence v_A exceeds 13.9 m/s by about a third, so the evidence contradicts the driver's claim

A uniform rod of mass M = 2.00 kg and length L = 1.20 m hangs vertically at rest from a frictionless horizontal pivot through its upper end. A lump of clay of mass m = 0.400 kg, moving horizontally at 6.00 m/s perpendicular to the pivot axis, strikes the lower end of the rod and sticks to it. Treat the clay as a particle and take g = 9.81 m/s². The moment of inertia of a uniform rod about a perpendicular axis through its centre is ML²/12.

1. (a)

   State State the condition under which the angular momentum of a system about a fixed point is conserved.

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   1. 1

      The net external torque on the system about that point is zero (equivalently, the net external angular impulse about it is zero)

2. (b)

   Calculate Calculate the moment of inertia of the rod and clay together about the pivot, immediately after the collision.

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   1. 1

      Parallel-axis theorem for the rod: I_rod = ML²/12 + M(L/2)² = ML²/3 = 0.960 kg m²

   2. 2

      Adds the clay as a particle at distance L: mL² = 0.576 kg m², giving I = 1.536 kg m² (1.54 kg m²)

3. (c)

   Explain Explain why the linear momentum of the rod–clay system is not conserved in the collision, whereas its angular momentum about the pivot is.

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   1. 1

      The pivot holds the top of the rod in place, so during the collision it exerts a large horizontal impulsive force on the rod; that external force changes the linear momentum of the system

   2. 2

      The pivot force acts at the pivot and has zero torque about it, and the rod's weight acts along the vertical line through the pivot while the rod hangs straight down, with any gravitational angular impulse negligible over the brief collision; so the net external angular impulse about the pivot is zero and angular momentum about it is conserved

4. (d)

   Determine Determine the angular speed of the rod immediately after the collision and the greatest angle from the vertical through which it then swings.

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   1. 1

      Angular momentum about the pivot: mvL = (0.400)(6.00)(1.20) = 2.88 kg m²/s = Iω, so ω = 2.88/1.536 = 1.875 rad/s (1.88 rad/s)

   2. 2

      Mechanical energy is conserved during the swing: ½Iω² = 2.70 J = (MgL/2 + mgL)(1 − cos θ) = (16.5 J)(1 − cos θ)

   3. 3

      1 − cos θ = 0.164, so θ = 33.3° (accept 33.2° from rounded values)

5. (e)

   Determine Determine the horizontal impulse exerted by the pivot on the rod during the collision, including its direction.

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   1. 1

      Linear momentum just after is (ML/2 + mL)ω = (1.20 + 0.480)(1.875) = 3.15 kg m/s, compared with mv = 2.40 kg m/s just before

   2. 2

      The pivot's impulse is 3.15 − 2.40 = 0.750 N s, directed the same way as the clay was moving

6. (f)

   Derive Derive, from conservation of angular momentum about the pivot and the impulse–momentum theorem, the distance x below the pivot at which the clay must strike for the pivot to exert no horizontal impulse during the collision. Show that the result does not depend on the mass of the clay.

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   1. 1

      Angular momentum about the pivot: mvx = (ML²/3 + mx²)ω

   2. 2

      No pivot impulse means linear momentum is conserved: mv = (ML/2 + mx)ω

   3. 3

      Dividing the two equations eliminates v and ω: x(ML/2 + mx) = ML²/3 + mx², so the mx² terms cancel and x = 2L/3 = 0.800 m whatever the value of m

A glider of mass 0.250 kg on a level air track is held between two springs and oscillates with a period of 0.800 s. A card sail fixed to the glider increases the air drag. A motion sensor records the amplitude of the oscillation, the greatest displacement from equilibrium, at the end of every fifth oscillation, with an uncertainty of ±0.3 mm in each reading. Two damping models are proposed. Model V (viscous drag): the resistive force is −bv, so the amplitude decays as A = A₀e^(−bt/(2m)). Model F (sliding friction): a resistive force of constant magnitude acts, which reduces the amplitude by the same amount in every oscillation.

1. (a)

   Describe Describe the pattern that each model predicts in amplitudes recorded at equal intervals of time.

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   1. 1

      Model V: each amplitude is the same fixed fraction of the one before, so successive ratios are constant and ln A falls linearly with t

   2. 2

      Model F: each amplitude is smaller than the one before by the same fixed amount, so successive differences are constant and A falls linearly with t, reaching zero at a finite time

2. (b)

   Justify Justify, using calculations from the table, which model the data support.

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   1. 1

      Successive differences are 22.4, 16.8, 11.2, 8.7, 5.6 and 4.5 mm: they fall steadily by a factor of five, far beyond the ±0.3 mm reading uncertainty, so Model F is rejected

   2. 2

      Successive ratios are 0.720, 0.708, 0.725, 0.706, 0.732 and 0.706: constant at about 0.72, with a scatter of about ±2% that the ±0.3 mm reading uncertainty can account for (combining fractional uncertainties in quadrature, it is ±2.4% in 15.3/20.9 and ±3.4% in 10.8/15.3)

   3. 3

      Concludes that the data support Model V, viscous drag

3. (c)

   Determine Determine the damping constant b by a method that uses all of the readings.

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   1. 1

      Uses ln A = ln A₀ − (b/2m)t: tabulates ln(A/mm) = 4.382, 4.054, 3.709, 3.388, 3.040, 2.728, 2.380 and plots or fits ln A against t

   2. 2

      Best-fit gradient = −0.0833 s⁻¹ (accept −0.080 s⁻¹ to −0.086 s⁻¹)

   3. 3

      b = −2m × gradient = 2 × 0.250 kg × 0.0833 s⁻¹ = 0.0416 kg/s (accept 0.040 kg/s to 0.043 kg/s)

4. (d)

   Calculate Calculate the fraction of the glider's mechanical energy that is lost during one oscillation.

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   1. 1

      Mechanical energy is proportional to A², so over one period it falls by the factor e^(−bT/m) = e^(−0.0416 × 0.800/0.250) = e^(−0.133) = 0.875

   2. 2

      Fraction lost per oscillation = 1 − 0.875 = 0.125, about 12.5% (accept 12% to 13%)

5. (e)

   Predict A 0.250 kg brass block is fixed to the glider, which keeps the same sail and the same springs. Predict the time for the amplitude to fall to half its initial value, before and after the block is added, and give the reason for the change.

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   1. 1

      Before: t½ = 2m ln 2/b = (2 × 0.250 × 0.693)/0.0416 = 8.33 s

   2. 2

      The sail, and so b, is unchanged while m doubles, so the decay constant b/(2m) halves and t½ doubles to 16.7 s (the period also lengthens, by √2 to 1.13 s, but that does not alter the decay time)

A communications satellite is raised from a circular parking orbit of radius r₁ = 6.78 × 10⁶ m to a circular orbit of radius r₂ = 4.22 × 10⁷ m by a two-burn transfer. A short engine burn at radius r₁ places it on an elliptical transfer orbit whose nearest point to Earth's centre (perigee) is at r₁ and whose farthest point (apogee) is at r₂. A second short burn at apogee makes the orbit circular. For Earth, GM = 3.986 × 10¹⁴ m³/s². Treat both burns as instantaneous, treat Earth as a spherically symmetric mass, and ignore every other body.

1. (a)

   Derive Derive, from Newton's law of gravitation and Newton's second law, an expression for the speed v of a satellite in a circular orbit of radius r.

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   1. 1

      The gravitational force provides the centripetal acceleration: GMm/r² = mv²/r

   2. 2

      v = √(GM/r), which is independent of the satellite's mass

2. (b)

   Explain Explain why the satellite's angular momentum about Earth's centre is constant along the transfer orbit, and why its magnitude at perigee and at apogee is simply mrv.

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   1. 1

      The gravitational force is central, directed along the line to Earth's centre, so its torque r × F about the centre is zero and the angular momentum L = r × mv is constant between the burns

   2. 2

      At perigee and apogee r is at its minimum or maximum, so dr/dt = 0 and the velocity is perpendicular to the radius vector; there |r × mv| = mrv sin 90° = mrv, which is not true elsewhere on the ellipse

3. (c)

   Derive Using conservation of angular momentum and conservation of mechanical energy between perigee and apogee, derive an expression for the speed v_p at perigee on the transfer orbit, in terms of GM, r₁ and r₂.

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   1. 1

      Angular momentum: r₁v_p = r₂v_a, so v_a = v_p r₁/r₂

   2. 2

      Energy: ½mv_p² − GMm/r₁ = ½mv_a² − GMm/r₂

   3. 3

      Substituting for v_a: ½v_p²(1 − r₁²/r₂²) = GM(1/r₁ − 1/r₂) = GM(r₂ − r₁)/(r₁r₂)

   4. 4

      Using 1 − r₁²/r₂² = (r₂ − r₁)(r₂ + r₁)/r₂² and cancelling (r₂ − r₁) gives v_p = √(2GMr₂/[r₁(r₁ + r₂)])

4. (d)

   Show that Show that your expression reduces to the circular-orbit speed at r₁ when r₂ = r₁, and to the escape speed from r₁ as r₂ → ∞, and state what each limit means physically.

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   1. 1

      r₂ = r₁ gives v_p = √(2GMr₁/(2r₁²)) = √(GM/r₁): an ellipse whose perigee and apogee coincide is the circular orbit, and no burn is needed

   2. 2

      As r₂ → ∞, r₂/(r₁ + r₂) → 1 and v_p → √(2GM/r₁): an ellipse whose apogee recedes without limit becomes the escape trajectory, with zero total mechanical energy

5. (e)

   Calculate Calculate the change in speed required at each burn, and the total.

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   1. 1

      First burn: v₁ = √(GM/r₁) = 7.67 × 10³ m/s and v_p = 1.007 × 10⁴ m/s, so Δv₁ = 2.40 × 10³ m/s

   2. 2

      Second burn: v_a = v_p r₁/r₂ = 1.62 × 10³ m/s and v₂ = √(GM/r₂) = 3.07 × 10³ m/s, so Δv₂ = 1.46 × 10³ m/s

   3. 3

      Total change in speed = 3.85 × 10³ m/s

6. (f)

   Calculate Calculate the time the satellite spends on the transfer orbit, from perigee to apogee.

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   1. 1

      The semi-major axis is a = (r₁ + r₂)/2 = 2.45 × 10⁷ m, and Kepler's third law gives the period T = 2π√(a³/GM) = 3.81 × 10⁴ s

   2. 2

      The transfer is half an orbit: T/2 = 1.91 × 10⁴ s, about 5.3 hours

A seismic isolation bearing is modelled as a uniform solid cylinder of mass m and radius r that rolls without slipping on the inside of a fixed, concave cylindrical track of radius R, where R > r and both axes are horizontal and parallel. The moment of inertia of the cylinder about its own axis is ½mr². Let φ be the angle between the vertical and the line from the track's axis of curvature to the cylinder's axis, so that φ = 0 at the lowest position. Take g = 9.81 m/s², and ignore rolling resistance and air resistance.

1. (a)

   Explain Explain why the speed of the cylinder's axis is (R − r)|dφ/dt|, and use the rolling condition to obtain the cylinder's angular speed ω about its own axis in terms of dφ/dt.

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   1. 1

      The cylinder's axis stays a fixed distance R − r from the track's axis of curvature, so it moves on a circle of that radius with speed v = (R − r)|dφ/dt|

   2. 2

      Rolling without slipping means the point of the cylinder in contact with the fixed track is instantaneously at rest, so v = rω and ω = (R − r)|dφ/dt|/r

2. (b)

   Derive Derive, from conservation of mechanical energy, the equation of motion for φ.

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   1. 1

      Gravitational potential energy measured from the lowest position: U = mg(R − r)(1 − cos φ)

   2. 2

      Kinetic energy is translational plus rotational: K = ½m(R − r)²(dφ/dt)² + ½(½mr²)ω²

   3. 3

      Substituting ω = (R − r)(dφ/dt)/r gives K = ¾m(R − r)²(dφ/dt)²

   4. 4

      Sets the time derivative of K + U to zero: (3/2)m(R − r)²(dφ/dt)(d²φ/dt²) + mg(R − r) sin φ (dφ/dt) = 0

   5. 5

      Divides by dφ/dt, which is non-zero during the motion, to obtain d²φ/dt² = −[2g/(3(R − r))] sin φ

3. (c)

   Show that Show that, for small oscillations, the motion is simple harmonic with period T = 2π√(3(R − r)/(2g)).

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   1. 1

      For small φ in radians sin φ ≈ φ, so d²φ/dt² = −[2g/(3(R − r))]φ, which has the form of the simple harmonic equation d²φ/dt² = −ω₀²φ

   2. 2

      ω₀ = √(2g/(3(R − r))), so T = 2π/ω₀ = 2π√(3(R − r)/(2g))

4. (d)

   Compare Compare this period with that of a small block sliding without friction whose centre of mass moves on the same circle of radius R − r, and explain the difference.

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   1. 1

      The sliding block is a simple pendulum of length R − r with T = 2π√((R − r)/g); the rolling cylinder's period is longer by a factor of √(3/2) ≈ 1.22

   2. 2

      In both cases the component of gravity along the path, mg sin φ, drives the motion, but the rolling cylinder must also gain rotational kinetic energy: its effective inertia is 3m/2 rather than m, so it accelerates along the path at only two-thirds the rate

5. (e)

   Calculate The engineers want an isolation period of 2.50 s. Calculate the value of R − r required.

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   1. 1

      Rearranges T = 2π√(3(R − r)/(2g)) to R − r = gT²/(6π²)

   2. 2

      R − r = (9.81 × 2.50²)/(6π²) = 1.04 m

6. (f)

   Predict Predict the period of the bearing if the solid cylinder is replaced by a thin-walled tube of the same mass and outer radius, whose moment of inertia about its own axis is mr², with R − r unchanged, and give the reason.

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   1. 1

      For a thin-walled tube I = mr², so K = m(R − r)²(dφ/dt)² and ω₀² = g/(2(R − r)), giving T = 2π√(2(R − r)/g)

   2. 2

      T = 2.50 s × √(4/3) = 2.89 s: a larger share of the energy is rotational, so the effective inertia rises to 2m and the oscillation is slower