(a) Consider a large stationary cylinder of inner radius R. A smaller solid cylinder of radius r rolls without slipping inside the larger cylinder. Determine the equation of motion of the smaller cylinder. 10

(b) Derive the expression for the gravit

Question

(a) Consider a large stationary cylinder of inner radius R. A smaller solid cylinder of radius r rolls without slipping inside the larger cylinder. Determine the equation of motion of the smaller cylinder. 10

(b) Derive the expression for the gravitational self-energy of a uniform solid sphere of mass M and radius R. 10

(c) A particle of rest mass 1 kg and velocity of magnitude 0·9c collides with a particle of mass 2 kg at rest. After collision the two particles coalesce and form a single particle of mass M and velocity V. Determine M and V. 10

(d) In a double slit Fraunhofer diffraction experiment, the slit width is 0·12 mm and the spacing between the two slits is 0·48 mm. The distance of the screen from the slits is 1·5 m. If the wavelength of the light used is 600 nm, determine (i) the missing orders of the interference maxima, and (ii) the distance between the central maxima and the first minima. 10

(e) A light beam of wavelength 600 nm produced by a 20 mW laser source is incident on a plane mirror. Determine :
(i) number of photons per second striking the surface of the mirror.
(ii) force exerted by the light beam on the mirror. 10

UPSC Answer Check · Accepted Answer

(a)
Let O be the axis of the fixed large cylinder and C the centre of the small rolling cylinder. Let θ be the angle of OC from the downward vertical. The centre C moves on a circle of radius R−r, so its speed is
v_C = (R−r) dθ/dt.

Let φ be the rotation angle of the small cylinder about its own axis. For rolling without slipping inside the fixed cylinder, the contact point is instantaneously at rest. The rolling condition is
(R−r) dθ/dt + r dφ/dt = 0,
so
dφ/dt = −((R−r)/r) dθ/dt.

The moment of inertia of a solid cylinder about its axis is I = (1/2) m r². Therefore the kinetic energy is
T = (1/2) m (R−r)² (dθ/dt)² + (1/2) I (dφ/dt)²
= (1/2) m (R−r)² (dθ/dt)² + (1/2)(1/2 m r²)((R−r)²/r²)(dθ/dt)²
= (3/4) m (R−r)² (dθ/dt)².

Taking zero potential energy at the lowest point,
V = m g (R−r)(1 − cos θ).

Lagrangian:
L = T − V = (3/4) m (R−r)² (dθ/dt)² − m g (R−r)(1 − cos θ).

Using the Euler–Lagrange equation
d/dt(∂L/∂(dθ/dt)) − ∂L/∂θ = 0,
we get
∂L/∂(dθ/dt) = (3/2) m (R−r)² dθ/dt,
so
d/dt(∂L/∂(dθ/dt)) = (3/2) m (R−r)² d²θ/dt².
Also
∂L/∂θ = − m g (R−r) sin θ.
Hence
(3/2) m (R−r)² d²θ/dt² + m g (R−r) sin θ = 0.
Dividing by m(R−r),
d²θ/dt² + (2g/(3(R−r))) sin θ = 0.

This is valid for R > r, no slipping, and the larger cylinder fixed.

(b)
Use the shell-assembly method. For a uniform solid sphere of mass M and radius R, density is
ρ = M/(4/3 π R³) = 3M/(4π R³).

At radius r, the mass inside is
m(r) = (4/3) π r³ ρ = M r³/R³.

Consider a spherical shell of radius r, thickness dr and mass
dm = 4π r² ρ dr = 3M r² dr / R³.

The work done in bringing this shell from infinity to radius r against the gravitational field of the mass already inside is
dU = − G m(r) dm / r.
Substituting,
dU = − G (M r³/R³)(3M r² dr/R³)/r
= − 3G M² r⁴ dr / R⁶.

Integrate from r = 0 to r = R:
U = − (3G M²/R⁶) ∫₀ᴿ r⁴ dr
= − (3G M²/R⁶)(R⁵/5)
= − 3G M²/(5R).

Therefore the gravitational self-energy is
U = − 3G M²/(5R).
The negative sign shows that the sphere is bound; its binding energy is 3G M²/(5R).

(c)
Let m₁ = 1 kg, u = 0.9c, and m₂ = 2 kg at rest. The Lorentz factor for the moving particle is
γ = 1/√(1 − u²/c²)
= 1/√(1 − 0.81)
= 1/√0.19
= 10/√19.

Total energy before collision:
E = γ m₁ c² + m₂ c²
= (10/√19 + 2)c².

Total momentum before collision:
p = γ m₁ u
= (10/√19)(1)(0.9c)
= 9c/√19.

For the composite particle of rest mass M and velocity V,
M²c⁴ = E² − p²c².
Thus
M² = (10/√19 + 2)² − (9/√19)²
= (4 + 40/√19 + 100/19) − 81/19
= 5 + 40/√19.
So
M = √(5 + 40/√19) kg ≈ 3.765 kg.

The velocity is
V = p c²/E
= (9c/√19)/(10/√19 + 2)
= 9c/(10 + 2√19)
= 3(5 − √19)c/4.
Therefore
V = [3(5 − √19)/4] c ≈ 0.4808c ≈ 1.44 × 10⁸ m/s.

Rest mass is not conserved in this inelastic collision; kinetic energy is converted into rest mass of the composite particle.

(d)(i)
For interference maxima:
d sin θ = nλ.
For diffraction minima:
a sin θ = mλ, m = ±1, ±2, ±3, …

A missing interference maximum occurs when an interference maximum coincides with a diffraction minimum:
nλ/d = mλ/a
so
n = (d/a)m.
Here
d/a = 0.48 mm/0.12 mm = 4.
Thus
n = 4m, m = ±1, ±2, ±3, …

Therefore the missing orders are
n = ±4, ±8, ±12, ±16, …
i.e. all non-zero multiples of 4. The central order n = 0 is not missing.

(d)(ii)
For the first interference minimum of the double-slit pattern:
d sin θ = λ/2.
Using the small-angle approximation, y = D tan θ ≈ D sin θ, so
y = Dλ/(2d)
= (1.5 m)(600 × 10⁻⁹ m)/(2 × 0.48 × 10⁻³ m)
= 9.375 × 10⁻⁴ m
= 0.9375 mm = 15/16 mm.

If by “first minima” the first single-slit diffraction minimum is meant, then
a sin θ = λ
and
y = Dλ/a
= (1.5 m)(600 × 10⁻⁹ m)/(0.12 × 10⁻³ m)
= 7.5 × 10⁻³ m
= 7.5 mm = 15/2 mm.

For the complete double-slit pattern, the nearest first minimum is the interference minimum at 0.9375 mm; the first diffraction-envelope minimum is at 7.5 mm.

(e)(i)
The energy of one photon is
Eγ = hc/λ
= (6.626 × 10⁻³⁴ J s)(3 × 10⁸ m/s)/(600 × 10⁻⁹ m)
= 3.313 × 10⁻¹⁹ J.

The laser power is
P = 20 mW = 20 × 10⁻³ W = 0.020 J/s.
Number of photons striking per second is
N = P/Eγ
= 0.020/(3.313 × 10⁻¹⁹)
≈ 6.04 × 10¹⁶ s⁻¹.
Therefore
N ≈ 6.04 × 10¹⁶ photons per second.

(e)(ii)
Assume normal incidence and perfect reflection by the plane mirror. Each photon has incident momentum p = Eγ/c. On reflection, its momentum changes from +p to −p, so the momentum change per photon is 2p. Thus the force on the mirror is
F = 2P/c
= 2(20 × 10⁻³ W)/(3 × 10⁸ m/s)
= 4 × 10⁻²/(3 × 10⁸) N
= 1.33 × 10⁻¹⁰ N.

For a partially reflecting mirror with reflectance R, the force would be F = (1 + R)P/c. For perfect reflection at normal incidence,
F = 4/3 × 10⁻¹⁰ N ≈ 1.33 × 10⁻¹⁰ N.

(a) Consider a large stationary cylinder of inner radius R. A smaller solid cylinder of radius r rolls without slipping inside…

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