Classical Mechanics - Lagrangian

A bead is constrained to slide on a frictionless rod that is fixed at an angle θ with a vertical axis and is rotating with angular frequency ω about the axis, as shown above. Taking the distance s along the rod as the variable, the Lagrangian for the bead is equal to

A. ½ mṡ ² − mgs cos θ 
B. ½ mṡ ² + ½ m(ωs)² − mgs 
C. ½ mṡ ² + ½ m(ωs cos θ)² + mgs cos θ
D. ½ m(ṡ sin θ)² − mgs cos θ 

E. ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

(GR9677 #68)

Solution:

Lagrangian: L = T − U

Potential energy: U = mgh = mgs cos θ 
Kinetic Energy:  T_kin  =  ½ mṡ ²
Rotational kinetic energy:  T_rot =  ½ Iω²
with moment inertia: I = mr²  = m(s sin θ)²
→ T_rot = ½ m(ωs sin θ)²

L = T_kin + T_rot − U =  ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

Answer: E

Optics - Telescope

The angular separation of the two components of a double star is 8 microradians, and the light from the double star has a wavelength of 5500 angstroms. The smallest diameter of a telescope mirror that will resolve the double star is most nearly 

A. 1 mm
B. 1 cm
C. 10 cm
D. 1 m
E. 100 m

(GR9277 #68)

Solution:

Rayleigh telescope resolution limit: sin θ = 1.22 λ/D

D = 1.22 λ/sin θ = 1.22 (5500 × 10⁻¹⁰ m) / sin (8×10⁻⁶)

Since θ ≪, assume sin θ ~ θ

 D = (1.22 × 5.5 / 8) × 10³ × 10⁻¹⁰ × 10⁶ ≈ 10⁻¹ m = 10 cm

Answer: C

Special Relativity - Relativistic Properties

If a newly discovered particle X moves with a speed equal to the speed of light in vacuum, then which of the following must be true?

A. The rest mass of X is zero
B. The spin of X equals the spin of a photon
C. The charge of X is carried on its surface
D. X does not spin
E. X cannot be detected

(GR8677 #68)

Solution:

Relativistic Energy: E² = p²c² + m₀²c⁴

If the particle moves with v = c, de Broglie relation: E = hf = pc

E² = E² + m₀²c⁴ 
m₀²c⁴ = E² − E² = 0
m₀ = 0

Answer: A