Physics course of the year 1 of the natural sciences course of the University of Cambridge (Part IA)

Solution

Let

m_{p} = 0.250 kg

and let

m_{w}

be the hanging mass. Taking up-slope and downward respectively as positive,

T - m_{p} g sin θ - μ m_{p} g cos θ = m_{p} a, m_{w} g - T = m_{w} a .

Eliminate the common tension to obtain

m_{w} = \frac{m _{p} { a + g ( sin θ + μ cos θ )}}{g - a} = 0.130 kg .

Here

θ = 1 0^{\circ}

μ = 0.1

and

a = 1.60 m s^{- 2}

. The pulley changes the direction of the string force; in the ideal model it does not change its magnitude.

d

Solution

The required constant deceleration follows from

v^{2} = u^{2} + 2 a s

a = - \frac{4. 8 ^{2}}{2 ( 2 )} = - 5.76 m s^{- 2} .

Both gravity and sliding friction oppose ascent, so

g (sin θ + 0.1 cos θ) = 5.76

. Write

sin θ + 0.1 cos θ = 1.01 sin (θ + arctan 0.1) .

The physical incline solution is $θ = 30. 0^{\circ}$ more precisely

30.038 9^{\circ}

. The combined mass cancels from the equation. The top of the track is consequently

H = 2 sin θ = 1.00117 m

above its base.

e

Solution

Let

x = 0.400 m

be the compression and

m_{t} = 0.550 kg

the combined mass. Starting from rest at the top, it falls through

H + x

before maximum compression. Hence

\frac{1}{2} k x^{2} = m_{t} g (H + x), k = 94.5 N m^{- 1} .

The gravitational drop during compression must be included. Equating

k x

to the weight would instead describe a static equilibrium, not the turning point after a fall.

f

Solution

Immediately after the parcel leaves, the spring still stores

\frac{1}{2} k x^{2}

, but only the

m_{s} = 0.300 kg

sledge is launched. Choose gravitational potential zero at the ramp base. Its initial position is

- x

, so conservation of energy to the apex gives

\frac{1}{2} k x^{2} - m_{s} gx = m_{s} g H_{s} .

Therefore

H_{s} = \frac{k x ^{2}}{2 m _{s} g} - x = 2.17 m

above the ramp base. Its upward travel from the compressed position is

H_{s} + x = 2.56882 m

, a different distance from the height above the base.

g

Solution

For actual complete bounce distances

D_{1} = 4 m

and

D_{n + 1} = r D_{n}

, the geometric series gives

D_{total} = \frac{4}{1 - r} m, 0 \leq r < 1.

A natural intended estimate compares the observed first round trip with the loss-free round trip from the initially compressed position,

D_{ideal} = 2 (H_{s} + x) = 5.13764 m .

If the reduction factor is taken to be

r = D_{1} / D_{ideal} = 0.77857

, the answer is $D_{total} ≃ 18.1 m$ . The first term is the actual 4 m, not the hypothetical loss-free distance.

This numerical inference needs the stated comparison assumption. Strictly, knowing the first actual bounce distance and saying subsequent distances share a ratio does not independently specify that ratio. A change in spring compression also means that an energy-loss fraction is not automatically the same as a vertical-distance ratio.

7

a

Solution

The collinear relativistic velocity addition formula is

u = \frac{v + u ^{'}}{1 + v u ^{'} / c ^{2}} .

With both velocities directed toward Earth,

u = \frac{0.6 c + 0.8 c}{1 + 0.48} = \frac{35}{37} c = 0.946 c .

The denominator keeps the result below

c

. Adding the two speeds by the Galilean rule would give an inadmissible

1.4 c

b

Solution

The separation simultaneous in the mothership frame at launch is the length contraction of the Earth-frame simultaneous separation:

D^{'} = \frac{2 ly}{γ _{m}} = 1.60 ly, γ_{m} = \frac{1}{1 - 0. 6 ^{2}} = 1.25.

In that frame Earth approaches the launch position at

0.6 c

while the lander travels toward Earth at

0.8 c

. Their coordinate separation closes at

1.4 c

, so

t_{arrival}^{'} = \frac{1.60 ly}{1.4 c} = \frac{8}{7} yr = 1.14 yr .

A closing rate between two objects can exceed

c

; it is not either object's speed in the other's rest frame. Simultaneity matters: simply transforming the Earth event at Earth-frame launch time would not give the simultaneous mothership-frame separation.

c

Solution

For the lander,

u / c = 35/37

, so

γ_{u} = \frac{1}{1 - ( 35/37 ) ^{2}} = \frac{37}{12} .

The relativistic kinetic energy is

K = (γ_{u} - 1) m c^{2} = \frac{25}{12} (4 \times 1 0^{5}) c^{2} = 7.49 \times 1 0^{22} J .

The rest energy

m c^{2}

is excluded from kinetic energy.

d

Solution

Set the launch event to

(t, x) = (0, 0)

and measure

x

toward Earth in light-years and

t

in years, so light has coordinate slope magnitude one. Earth's worldline is

x = 2

, and the lander's is

x = (35/37) t

. The outgoing signal is

x = t

and reaches Earth at

B = (2, 2)

. Earth's reply is sent at

C = (25/12, 2)

The return signal has

x = 2 - (t - 25/12)

. Its intersection with the lander gives

D = (1813/864, 1715/864)

. Both signal lines are null; the lander's worldline lies inside the light cone. The drawing exaggerates the small gap near Earth in an inset so that B, C and D remain distinguishable.

Figure 1.
Earth-frame spacetime diagram of the outgoing signal, delayed reply and lander worldline
. **Details**.

e

Solution

Equating the lander and return-signal positions gives

\frac{35}{37} t_{D} = 2 - (t_{D} - \frac{25}{12}), t_{D} = \frac{1813}{864} yr = 2.09838 yr .

This is measured from launch, so about 25.18 months pass on Earth-frame synchronized clocks. D occurs slightly before the lander reaches Earth at

t = 74/35 yr

The two events A and D lie on the inertial lander's own worldline, and their proper time difference is

Δ τ = \frac{t _{D}}{γ _{u}} = \frac{49}{72} yr = 0.68056 yr ≃ 8.17 months .

Equivalently, applying the Lorentz transformation with

Δ x = u Δ t

gives

Δ t^{'} = γ_{u} Δ t (1 - u^{2} / c^{2}) = Δ t / γ_{u}

8

a

Solution

The central rod occupies the holes in the discs, so the disc masses must be annular rather than counted as solid discs overlapping the rod. Define

C = L_{1} ρ_{1} = 2 L_{2} ρ_{2}

. Then

M = π r_{1}^{2} L_{1} ρ_{1} + 2 π (r_{2}^{2} - r_{1}^{2}) L_{2} ρ_{2} = π C r_{2}^{2} .

The axial moment of inertia is

I = \frac{π}{2} L_{1} ρ_{1} r_{1}^{4} + π L_{2} ρ_{2} (r_{2}^{4} - r_{1}^{4}) = \frac{π}{2} C r_{2}^{4} .

Thus $I = \frac{1}{2} M r_{2}^{2}$ , exactly the axial inertia of a uniform disc of outer radius

r_{2}

and mass

M

. The cancellation uses the given density–length relation.

b

Solution

The contact radius is

r_{1}

, so rolling without slipping gives

v = r_{1} ω

. With

I = M r_{2}^{2} /2

, the energy at the minimum of the stated parabolic motion model is

M g h_{0} = \frac{1}{2} M v_{m a x}^{2} + \frac{1}{2} \frac{M r _{2}^{2}}{2} \frac{v _{m a x}^{2}}{r _{1}^{2}} .

Consequently

v_{m a x} = \frac{2 g h _{0}}{( r _{2} / r _{1} ) ^{2} + 2} = 0.148 m s^{- 1} .

The corresponding spin angular frequency is $ω_{m a x} = v_{m a x} / r_{1} = 14.8 rad s^{- 1}$ . It is the rod radius, not the disc radius, that relates translation and rotation; the discs determine the larger rotational inertia.

c

Solution

In the small-amplitude approximation,

\overset{x}{˙}

is the speed along the track, and

y = B x^{2}

supplies the gravitational potential. The conserved energy is

E = \frac{M}{2} (1 + \frac{r _{2}^{2}}{2 r _{1}^{2}}) \overset{x}{˙}^{2} + M g B x^{2} .

Differentiating and extending through turning points by continuity gives

(1 + \frac{r _{2}^{2}}{2 r _{1}^{2}}) \overset{x}{¨} + 2 g B x = 0.

This is simple harmonic motion with

ω_{0}^{2} = 4 g B / {(r_{2} / r_{1})^{2} + 2}

. Hence

f_{0} = \frac{g B}{π ( r _{2} / r _{1} ) ^{2} + 2} = 0.235 Hz .

The frequency is lower than for a sliding point mass because energy is also stored in rotation.

d

Solution

For weak linear damping, write the amplitude envelope as

A (t) = A_{0} e^{- ω_{0} t / (2 Q)}

. The amplitude e-folding time is

2 Q / ω_{0}

, and one period is approximately

2 π / ω_{0}

. Thus

N = \frac{Q}{π} = 31.8 oscillations .

Energy, being proportional to amplitude squared, decays twice as rapidly in its exponent. Using its e-folding time instead would introduce a factor-of-two error.

e

Solution

At the initial turning point the oscillation energy is

E = M g h_{0} = 0.04905 J

. In the high-quality factor regime,

Q = 2 π E /Δ E

, so the required input per period is

Δ E ≃ \frac{2 π}{100} (0.04905) = 3.08 \times 1 0^{- 3} J .

This replaces the small dissipated energy each cycle, rather than supplying the entire stored energy again.

9

a

Solution

V (x)

is the particle's potential energy as a function of position, and

E

is its total energy eigenvalue in this time-independent Schrödinger equation. Solving it with the boundary conditions determines the allowed stationary states and energy levels.

The wavefunction is a probability amplitude: the Born rule gives position probability density

∣ ψ (x) ∣^{2}

, normalized to unit integral for one particle.

ψ

itself need not be real or positive and is not a classical matter-density profile. The full stationary-state time dependence is

Ψ (x, t) = ψ (x) e^{- i Et /ℏ}

, leaving

∣Ψ ∣^{2}

time independent.

b

Solution

For a nonzero integer mode, normalize the wavefunction:

1 = ∣ A ∣^{2} \int_{0}^{L} sin^{2} \frac{nπ x}{L} d x = ∣ A ∣^{2} \frac{L}{2} .

Thus $∣ A ∣ = 2/ L$ . Choosing an arbitrary overall phase makes

A = 2/ L

real and positive. The physical modes are conventionally indexed

n = 1, 2, \dots

;

n = 0

would be the zero function and cannot be normalized.

c

Solution

Substitution into the Schrödinger equation gives

E_{n} = V_{0} + n^{2} π^{2} ℏ^{2} / (2 m L^{2})

. For the stated electron well,

E_{2} = 37.6 eV = 6.02 \times 1 0^{- 18} J .

The transition emits the energy difference,

Δ E = E_{2} - E_{1} = 3 E_{1} = 28.2 eV

, so

λ = \frac{h c}{Δ E} = 44.0 nm .

Using the full excited-state energy as the photon energy would neglect the nonzero ground-state energy.

d

Solution

The position probability density is

p (x) = \frac{2}{L} sin^{2} \frac{2 π x}{L} (0 \leq x \leq L),

and zero outside. It has zeros at

0, L /2, L

and equal maxima

2/ L

L /4, 3 L /4

. The two lobes have equal integrated probability. The vertical axis is a density with units of inverse length, not a dimensionless probability at a point.

Figure 1.
Second-state position densities for an infinite well and an illustrative finite well, including evanescent tails
. **Details**.

e

Solution

Use the dimensionless coordinate

u = x / L

P = \int_{0.45}^{0.55} 2 sin^{2} (2 π u) d u = [u - \frac{s i n 4 π u}{4 π}]_{0.45}^{0.55} .

Therefore

P = 0.1 - \frac{sin ( π /5 )}{2 π} = 0.00645 ≃ 0.645%.

The central interval surrounds a node, which explains why the probability is much smaller than its ten-percent share of the well's length.

f

Solution

The density is symmetric about

L /2

p (L - x) = p (x)

. Denote the expectation by

μ = \int_{0}^{L} x p (x) d x

. Substitution

u = L - x

gives

μ = \int_{0}^{L} (L - u) p (u) d u = L \int_{0}^{L} p (u) d u - μ = L - μ .

Thus $⟨ x ⟩ = L /2$ . An expectation need not coincide with a high-density position: the centre is a node of this particular state, but the two equally weighted lobes balance about it.

g

Solution

For a symmetric finite well supporting this excited bound state, the wavefunction and its derivative match continuously at each finite potential step. Outside, where

V_{wall} > E

, the solution decays exponentially with

κ = \frac{2 m ( V _{wall} - E )}{ℏ} .

The density therefore extends into the classically forbidden region instead of dropping to zero at the walls. Its interior lobes and normalization change, and the energy is lower than for the corresponding infinite-wall state at the same well bottom. Symmetry preserves the central node for the first excited state. The diagram shows an explicitly illustrative barrier height, not an extra numerical value supplied by the question.

10

a

Solution

In magnetostatics, Ampère's circuital law is

\oint_{C} B \cdot d ℓ = μ_{0} I_{enclosed} .

It relates magnetic circulation around a closed curve to the steady current through a spanning surface. It is especially useful when symmetry makes the tangential field constant on a chosen path: an infinite straight wire gives

B = μ_{0} I / (2 π r)

, while an ideal long solenoid gives approximately

B = μ_{0} n I

inside.

The Biot-Savart law gives the field contribution from the actual current geometry:

B (r) = \frac{μ _{0} I}{4 π} \int \frac{d ℓ ^{'} \times ( r - r ^{'} )}{∣ r - r ^{'} ∣ ^{3}} .

The cross product determines the direction by the right-hand rule. Integrating around a circular loop of radius

a

gives

B = μ_{0} I / (2 a)

at its centre. A finite wire segment can also be integrated directly, even though it lacks enough symmetry for an Ampère-loop shortcut. The full steady-current circuit includes a return path; a segment calculation specifies that segment's contribution.

b

Solution

Place the wire on the

y

axis, its endpoints at

- a

and

b

, and the observation point at

(- d, 0)

. With current along

+ y

d ℓ^{'} \times r

points out of the page. Its magnitude gives

B = \frac{μ _{0} I}{4 π} \int_{- a}^{b} \frac{d d y}{( d ^{2} + y ^{2} ) ^{3/2}} = \frac{μ _{0} I}{4 π d} [\frac{y}{d ^{2} + y ^{2}}]_{- a}^{b} .

In the question's angle convention,

cos β = b / d^{2} + b^{2}

and

cos α = - a / d^{2} + a^{2}

. Therefore

B = \frac{μ _{0} I}{4 π d} (cos β - cos α) .

The minus sign in

cos α

matters because

α

is the exterior obtuse angle.

c

Solution

For a regular hexagon of side

a

, the perpendicular centre-to-side distance is

d = 3 a /2

. Each side subtends endpoint angles with

cos β = 1/2

and

cos α = - 1/2

. Its centre field is consequently

μ_{0} I / (4 π d)

All six contributions have the same normal direction, so

B_{centre} = 6 \frac{μ _{0} I}{4 π d} = \frac{3 μ _{0} I}{πa} .

The direction follows the current's right-hand rule. Equal side lengths alone would not fix an arbitrary distorted hexagon's field; the usual regular-coil geometry is used here.

d

Solution

The regular hexagon has area

A = 6 (3 a^{2} /4) = 33 a^{2} /2

. Choose

t = 0

with its positive normal parallel to the imposed field. The external magnetic flux is

Φ = B_{0} A cos ω t

, so Faraday's law of induction gives

E = - \frac{d Φ}{d t} = B_{0} A ω sin ω t .

Taking the stated resistance to determine the loop response,

I (t) = \frac{3 3 B _{0} a ^{2} ω}{2 R} sin ω t .

A different initial orientation adds a phase. The sign is with respect to the chosen loop orientation and obeys Lenz's law; reversal of that orientation reverses the signed current. This expression neglects self-inductance, as in the resistive-loop model.

2022

Original paper, released PDF pages 53–60

The released examiner commentary is not a complete worked solution set. It records an accepted ambiguity in question 8(d); question 7(e) also requires care about its unstated drag model.^[ref]

Section A

1

Solution

By dimensional analysis, power has dimensions

M L^{2} T^{- 3}

, mass density has dimensions

M L^{- 3}

, and angular velocity has dimensions

T^{- 1}

; radians are dimensionless. Thus

M L^{2} T^{- 3} = M^{α} L^{γ - 3 α} T^{- β} .

Matching the three independent dimensions gives

α = 1

β = 3

and

γ - 3 α = 2

, hence

α = 1, β = 3, γ = 5, P \propto ρ ω^{3} D^{5} .

With unchanged mass density, angular velocity and the same performance coefficient (a dimensionless quantity), the pump power affinity law gives

\frac{P _{new}}{P _{old}} = (\frac{D _{new}}{D _{old}})^{5} = 1. 1^{5} = 1.61051.

The required power therefore increases by $61.1%$ . This uses the stipulated dependence on only these three quantities, with comparable pump geometry and performance; dimensional analysis alone does not guarantee that unrelated pumps have the same performance coefficient.

2

Solution

The relativistic kinetic energy is the total energy minus the rest energy:

K = (γ - 1) M c^{2}, γ = \frac{1}{1 - v ^{2} / c ^{2}} .

Here

γ

is the Lorentz factor and

c

the speed of light. The binomial series, with

b = v / c

, gives

(1 - b^{2})^{- 1/2} = 1 + \frac{1}{2} b^{2} + \frac{3}{8} b^{4} + O (b^{6}), K = \frac{1}{2} M v^{2} + \frac{3}{8} M \frac{v ^{4}}{c ^{2}} + O (M \frac{v ^{6}}{c ^{4}}) .

For

v ≪ c

, the leading term is the classical kinetic energy

M v^{2} /2

. The ratio of the fourth-order term to that leading term is

3 v^{2} / (4 c^{2})

. Consequently

\frac{3}{4} \frac{v ^{2}}{c ^{2}} = 0.01 ⟹ \frac{v}{c} = \frac{0.04}{3} = 0.11547 ≃ 11.5%.

The condition concerns one term of the binomial series; it is not the condition that the entire relativistic correction be exactly

1%

3

Solution

Let

x

be signed displacement along the rails, measured from the centre. The diagram in the original paper shows a guide extending across the centre, so the signed coordinate can pass through zero. Hooke's law gives the radial force

- k x

. In the rotating reference frame, the radial centrifugal force is

+ m Ω^{2} x

. The Coriolis force is perpendicular to the rails and is balanced by their reaction; it has no radial component. Therefore Newton's second law gives

m \overset{x}{¨} = - k x + m Ω^{2} x, \overset{x}{¨} + (\frac{k}{m} - Ω^{2}) x = 0.

Equivalently, the radial component of acceleration in polar coordinates in the inertial frame is

\overset{x}{¨} - Ω^{2} x

. The radial spring oscillator on a rotating guide has simple harmonic motion only when

k / m - Ω^{2} > 0

, with

ω_{osc} = \frac{k}{m} - Ω^{2}, f_{osc} = \frac{1}{2 π} \frac{k}{m} - Ω^{2} .

The angular frequency is in

rad s^{- 1}

and the frequency in

Hz

. The critical angular speed is

Ω_{crit} = k / m, 0 \leq Ω < Ω_{crit} .

Strictly this is an upper bound, not an attained maximum for oscillations: at equality

\overset{x}{¨} = 0

, so a displaced object released from rest stays at its displaced position relative to the rails; above it, the central equilibrium is unstable.

4

Solution

Take the thin lens at

x = 0

, with the object on the left and light rays travelling to the right. The positive object distance is

u = 12 cm

and the focal length is

f = 8 cm

. The thin lens equation gives

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = (\frac{1}{8} - \frac{1}{12}) c m^{- 1} = \frac{1}{24} c m^{- 1}, v = 24 cm .

The transverse magnification and signed image height are

M = - \frac{v}{u} = - 2, h_{i} = M h_{o} = - 2 (5 cm) = - 10 cm .

Thus the real image is $24 cm$ beyond the lens, inverted and $10 cm$ high. For the ray tracing, a light ray parallel to the optical axis emerges through the far optical focus; a light ray through the optical centre continues straight in the ideal thin lens model. They intersect at

(24, - 10) cm

, independently reading off both answers from the diagram.

Figure 1.
Ray tracing for the converging thin lens: object and image heights and both focal points
. **Details**.

The calculation and diagram use the ideal paraxial approximation implicit in the thin lens equation. The comparatively large object height does not supply enough information to calculate optical aberrations of a real lens.

5

Solution

Write

a = 0.020 m

for the side length. In an equilateral triangle, the distance from centre to a vertex is

a / 3

; it is two thirds of the altitude

3 a /2

. By the superposition principle, the electric potential at the centre due to the three fixed electrons is

Φ (0) = - \frac{3 e}{4 π ε _{0} ( a / 3 )} .

Here

e > 0

is the elementary charge and

ε_{0}

the vacuum permittivity. The approaching electron has electric charge

- e

, so its potential energy, measured relative to infinity, is positive:

U (0) = (- e) Φ (0) = \frac{3 3 e ^{2}}{4 π ε _{0} a} .

The fixed sources do not move, and radiation is excluded. Hence conservation of mechanical energy gives

\frac{1}{2} m_{e} v_{\infty}^{2} = U (0), v_{\infty} = \frac{6 3 e ^{2}}{4 π ε _{0} m _{e} a} .

Using

e = 1.60218 \times 1 0^{- 19} C

m_{e} = 9.10938 \times 1 0^{- 31} kg

for the electron mass, and

(4 π ε_{0})^{- 1} = 8.98755 \times 1 0^{9} N m^{2} C^{- 2}

gives

U (0) = 5.99 \times 1 0^{- 26} J = 3.74 \times 1 0^{- 7} eV, v_{\infty} ≃ 3.63 \times 1 0^{2} m s^{- 1} .

This speed is small enough to justify classical kinetic energy. There is an idealization in the word “stops”: along the line normal to the triangle through its centre, this threshold energy makes the electron approach the central equilibrium asymptotically. A smooth, conservative trajectory cannot reach an equilibrium with exactly zero velocity in finite time unless it was already there. The energy threshold, and therefore the requested initial speed, is nevertheless unambiguous.

Section B

6

a

Solution

Choose three mutually perpendicular axes through the same point, and label coordinates

x, y, z

. The moments of inertia of any rigid body are

I_{x} = \int (y^{2} + z^{2}) d m, I_{y} = \int (z^{2} + x^{2}) d m, I_{z} = \int (x^{2} + y^{2}) d m .

Subtracting gives

I_{x} + I_{y} - I_{z} = 2 \int z^{2} d m \geq 0.

For a thin planar lamina lying in

z = 0

, the right-hand side vanishes, proving the perpendicular axis theorem:

I_{z} = I_{x} + I_{y} .

The two in-plane axes and the perpendicular axis must intersect at one point; this need not be the centre of mass. For an extended three-dimensional mass distribution, the equality generally fails. The general moment-of-inertia triangle inequalities are

I_{x} + I_{y} \geq I_{z}, I_{y} + I_{z} \geq I_{x}, I_{z} + I_{x} \geq I_{y},

with equality in the first precisely when all mass lies in the plane perpendicular to its

z

axis, and cyclically for the others. None of these identities requires principal axes.

b

Solution

Let

I_{⊥}

denote the moment of inertia normal to the disc and

I_{d}

that about any central diameter. Rotational symmetry makes the two perpendicular diameter values equal. By the perpendicular axis theorem,

I_{⊥} = 2 I_{d}

. The intended angular impulse has magnitude

J = I_{⊥} ω_{0}

; delivering that magnitude about a diameter instead gives

J = I_{d} ω

. Thus

I_{d} ω = 2 I_{d} ω_{0}, ω = 2 ω_{0} .

These are principal axes, so angular momentum is parallel to angular velocity in both cases. Equal delivered angular impulse means equal magnitude of the change in angular momentum, not equal rotational kinetic energy; the actual rotational kinetic energy is twice the intended value.

c

Solution

Take

+ x

along the upper ship's initial velocity, and take

+ y

from the lower ship to the upper ship at closest approach. The upper and lower ships then have initial velocities

v_{1} x

and

- v_{2} x

. Their equal masses put each at distance

l /2

from the centre of mass, so

I_{CM} = 2 m (l /2)^{2} = \frac{1}{2} m l^{2} .

By momentum conservation, the centre of mass keeps moving with constant velocity

V_{CM} = \frac{v _{1} - v _{2}}{2} x .

In that frame the ships have opposite velocities of magnitude

u = (v_{1} + v_{2}) /2

. Their initial angular momentum about the centre of mass is nonzero even though their trajectories were straight:

L = - m l \frac{v _{1} + v _{2}}{2} z, L = \frac{m l ( v _{1} + v _{2} )}{2} .

The tether exerts central internal forces, so conservation of angular momentum gives

ω = \frac{L}{I _{CM}} = \frac{v _{1} + v _{2}}{l} .

Each ship follows uniform circular motion of radius

l /2

around the translating centre of mass, clockwise viewed from

+ z

. More explicitly, put the initial centre at the origin and define

n (t) = sin (ω t) x + cos (ω t) y, r_{1, 2} (t) = V_{CM} t \pm \frac{l}{2} n (t) .

(5)

The kinetic-energy decomposition about the centre of mass verifies conservation of mechanical energy directly:

E_{after} = \frac{1}{2} (2 m) V_{CM}^{2} + \frac{1}{2} I_{CM} ω^{2} = \frac{m}{4} (v_{1} - v_{2})^{2} + \frac{m}{4} (v_{1} + v_{2})^{2} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = E_{before} .

(6)

At closest approach the relative radial velocity is already zero, so connecting the tether requires no dissipative radial impulse. Subsequently its tension is perpendicular to each ship's velocity relative to the centre of mass and does no net work on the pair at fixed

l

d

Solution

Each ship needs centripetal acceleration

ω^{2} (l /2)

, supplied entirely by the tension. Thus

T = m ω^{2} \frac{l}{2} = \frac{m ( v _{1} + v _{2} ) ^{2}}{2 l} = \frac{2 L ^{2}}{m l ^{3}} .

The result has units of force,

N

e

Solution

During quasi-static tether contraction, momentum conservation keeps

V_{CM}

constant and conservation of angular momentum keeps

L

constant. To leading order in the slow radial motion, the kinetic energy is

E = m V_{CM}^{2} + \frac{L ^{2}}{2 I _{CM}} = m V_{CM}^{2} + \frac{L ^{2}}{m l ^{2}} .

Its derivative at fixed

L

, not fixed

ω

, is

\frac{d E}{d t} = - \frac{2 L ^{2}}{m l ^{3}} \dot{l} > 0 (\dot{l} < 0) .

To compute the power supplied by the tension, use the instantaneous unit vector

n

from the lower to the upper ship. Their radial velocities relative to the centre of mass are

\pm (\dot{l} /2) n

, and the tension forces are

\mp T n

. Consequently

\dot{W} = (- T n) \cdot (\frac{l ˙}{2} n) + (T n) \cdot (- \frac{l ˙}{2} n) = - T \dot{l} = - \frac{2 L ^{2}}{m l ^{3}} \dot{l} = \frac{d E}{d t} .

The translational contributions cancel between the two ships and the tangential contributions are zero. The contraction mechanism supplies positive work, increasing the rotational kinetic energy.

The slow-contraction approximation matters: for a prescribed smooth

l (t)

with a taut tether, the exact kinetic energy and tension include radial terms,

E_{exact} = m V_{CM}^{2} + \frac{L ^{2}}{m l ^{2}} + \frac{m}{4} \dot{l}^{2}, T_{exact} = \frac{2 L ^{2}}{m l ^{3}} - \frac{m}{2} \ddot{l} .