(a) A body moves about a point 'O' under no force, the principal moments of inertia at 'O' being 3A, 5A and 6A. The components of the initial angular velocity about the principal axes are ω₁ = n, ω₂ = 0 and ω₃ = n. Find the components ω₁, ω₂ and ω₃ f

Question

(a) A body moves about a point 'O' under no force, the principal moments of inertia at 'O' being 3A, 5A and 6A. The components of the initial angular velocity about the principal axes are ω₁ = n, ω₂ = 0 and ω₃ = n. Find the components ω₁, ω₂ and ω₃ for large values of time t. 20

(b) A harmonic oscillator is represented by the equation
m d²x/dt² + γ dx/dt + kx = 0;
where m = 0·25 kg, γ = 0·07 kg s⁻¹ and k = 85 Nm⁻¹. Determine (i) the period of oscillation, and (ii) the number of oscillations in which its amplitude will become half of its original value. 15

(c) Show that the electromagnetic wave equation is invariant under Lorentz transformations. 15

UPSC Answer Check · Accepted Answer

(a) For torque-free motion about a fixed point, Euler’s equations are
I₁ dω₁/dt = (I₂ − I₃)ω₂ω₃,
I₂ dω₂/dt = (I₃ − I₁)ω₃ω₁,
I₃ dω₃/dt = (I₁ − I₂)ω₁ω₂.

Here I₁ = 3A, I₂ = 5A, I₃ = 6A. Hence
dω₁/dt = −(1/3)ω₂ω₃,
dω₂/dt = (3/5)ω₁ω₃,
dω₃/dt = −(1/3)ω₁ω₂.

Initially ω₁ = n, ω₂ = 0, ω₃ = n. Since ω₁ = ω₃ initially, and
d(ω₁ − ω₃)/dt = (ω₂/3)(ω₁ − ω₃),
the equality ω₁ = ω₃ is preserved for all t. Put
ω₁ = ω₃ = x, ω₂ = y.
Then
dx/dt = −(1/3)xy, dy/dt = (3/5)x².

Use conservation of kinetic energy. Since the body is torque-free,
T = 1/2(3A x² + 5A y² + 6A x²) = constant.
At t = 0, T = 1/2(3A n² + 6A n²) = (9/2)A n².
Therefore
9x² + 5y² = 9n²,
so x² = n² − (5/9)y².

Substitute this in dy/dt:
dy/dt = (3/5)[n² − (5/9)y²] = (3/5)n² − (1/3)y².
This is separable:
∫₀ʸ dy/((3/5)n² − (1/3)y²) = t.

Using ∫ du/(a − bu²) = (1/√(ab)) artanh(u√(b/a)), with
a = (3/5)n², b = 1/3, √(ab) = n/√5, √(b/a) = √5/(3n),
we get
t = (√5/n) artanh(√5 y/(3n)).
Thus
y = (3n/√5) tanh(n t/√5).
Then
x = √(n² − (5/9)y²) = n sech(n t/√5).

Hence the exact components are
ω₁ = n sech(n t/√5), ω₂ = (3n/√5) tanh(n t/√5), ω₃ = n sech(n t/√5).
For large t, tanh(n t/√5) → 1 and sech(n t/√5) → 0. Therefore
ω₁ → 0, ω₂ → 3n/√5, ω₃ → 0.
The components have units of s⁻¹.

(b) Write the equation as
d²x/dt² + 2β dx/dt + ω₀²x = 0,
where
β = γ/(2m) = 0.07/(2 × 0.25) = 0.14 s⁻¹,
ω₀² = k/m = 85/0.25 = 340 s⁻².
Since β < ω₀, the oscillator is underdamped. The damped angular frequency is
ω_d = √(ω₀² − β²) = √(340 − 0.0196) = √339.9804 = 18.4386 s⁻¹.

(i) The period of oscillation is
T = 2π/ω_d = 2π/18.4386 = 0.3408 s.
Exactly,
T = 100π/√849951 s ≈ 0.3408 s.

(ii) The amplitude decays as
A(t) = A₀ e^−βt.
For half amplitude,
e^−βt = 1/2 ⇒ t = ln2/β = 0.693147/0.14 = 4.9511 s.
Number of oscillations is
N = t/T = 4.9511/0.3408 = 14.53.
Exactly,
N = (√849951 ln2)/(14π) ≈ 14.53.
Thus the amplitude becomes half after about 14.5 oscillations, i.e. during the 15th complete oscillation.

(c) In source-free vacuum, Maxwell’s equations give the electromagnetic wave equation
∇²E − (1/c²) ∂²E/∂t² = 0,
and the same for B. Define the d’Alembertian operator
□ = ∂²/∂x² + ∂²/∂y² + ∂²/∂z² − (1/c²) ∂²/∂t².

Consider a Lorentz boost along x with speed v:
x' = γ(x − vt), y' = y, z' = z, t' = γ(t − vx/c²),
where γ = 1/√(1 − v²/c²).

By the chain rule,
∂/∂x = γ(∂/∂x' − (v/c²) ∂/∂t'),
∂/∂t = γ(∂/∂t' − v ∂/∂x').

Therefore
∂²/∂x² = γ²(∂²/∂x'² − 2(v/c²) ∂²/∂x'∂t' + (v²/c⁴) ∂²/∂t'²),
∂²/∂t² = γ²(∂²/∂t'² − 2v ∂²/∂x'∂t' + v² ∂²/∂x'²).

Now compute
∂²/∂x² − (1/c²) ∂²/∂t²
= γ²[(1 − v²/c²) ∂²/∂x'² − (1/c²)(1 − v²/c²) ∂²/∂t'²]
= ∂²/∂x'² − (1/c²) ∂²/∂t'².

Since y and z coordinates are unchanged,
∂²/∂y² + ∂²/∂z² = ∂²/∂y'² + ∂²/∂z'².
Hence
□ = □'.
Thus
□E = 0 ⇒ □'E' = 0,
and similarly for B. Therefore the electromagnetic wave equation has the same form in all inertial frames; it is invariant under Lorentz transformations.

(a) A body moves about a point 'O' under no force, the principal moments of inertia at 'O' being 3A, 5A and 6A. The components of…

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