(a) (i) Define moment of inertia and radius of gyration of a body of mass M rotating about an axis. State and prove Parallel Axis…

(a)(i) Moment of inertia, radius of gyration and the parallel axis theorem

The moment of inertia of a body about an axis is I = Σmᵢrᵢ² = ∫r² dm, where r is the perpendicular distance of the mass element from the axis. The radius of gyration k is defined by I = Mk², so k = √(I/M): the distance from the axis at which the whole mass could be concentrated without changing I.

Parallel axis theorem: the moment of inertia I about any axis equals the moment of inertia I_cm about a parallel axis through the centre of mass plus Md², where d is the distance between the axes:

I = I_cm + Md².

Proof: take the centre of mass as origin and the axis through it along z. Let the parallel axis pass through (a, b) in the xy-plane, so d² = a² + b². For an element dm at (x, y), the squared distance from the new axis is (x − a)² + (y − b)². Then

I = ∫[(x − a)² + (y − b)²]dm = ∫(x² + y²)dm − 2a∫x dm − 2b∫y dm + (a² + b²)∫dm.

For the centre of mass, ∫x dm = ∫y dm = 0, ∫(x² + y²)dm = I_cm and ∫dm = M. Hence I = I_cm + Md².

(a)(ii) Kinetic energy of a rolling sphere

For a uniform solid sphere, I = (2/5)MR² and ω = v/R, so the total kinetic energy is translational plus rotational:

E = ½Mv² + ½Iω² = ½Mv² + (1/5)Mv² = (7/10)Mv² = 0.7 × 0.5 × 9 = 3.15 J.

(For a thin hollow spherical shell, I = (2/3)MR² and E = (5/6)Mv² = 3.75 J.)

(b) Gravitational potential at the Earth's surface

V = −GM/R, with M = (4/3)πR³ρ, so V = −(4/3)πGρR².

V = −(4/3)π × 6.66×10⁻¹¹ × 5.5×10³ × (6.4×10⁶)² = −6.28×10⁷ J/kg,

taking the potential to be zero at infinity. (Check: V/R gives g ≈ 9.8 m/s².)

(c) Damped harmonic oscillation

Damped harmonic oscillation is oscillation in which a resistive force −b(dx/dt) removes energy, so the amplitude decays with time. The equation of motion is

m d²x/dt² = −kx − b dx/dt, or d²x/dt² + 2β dx/dt + ω₀²x = 0,

where β = b/2m and ω₀² = k/m. Trying x = e^(rt) gives r² + 2βr + ω₀² = 0, so r = −β ± √(β² − ω₀²), and the general solution is x = C₁e^(r₁t) + C₂e^(r₂t).

Cases: Dead beat (overdamped), β > ω₀: both roots are real, x = e^(−βt)(C₁e^(qt) + C₂e^(−qt)) with q = √(β² − ω₀²). The body returns slowly to equilibrium without oscillating. Critical damping, β = ω₀: x = (C₁ + C₂t)e^(−βt). This is the quickest return without oscillation. Oscillatory (underdamped), β < ω₀: x = Ae^(−βt) cos(ωt + φ), with ω = √(ω₀² − β²). The amplitude decays exponentially.

Logarithmic decrement: λ = ln(A_n/A_(n+1)) = βT. After 20 oscillations A₂₀ = A₀e^(−20λ), so

λ = (1/20) ln(30/3) = (ln 10)/20 = 0.115.