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
Lagrangian: L = T − U
Potential energy: U = mgh = mgs cos θ
Kinetic Energy: Tkin = ½ mṡ ²
Rotational kinetic energy: Trot = ½ Iω²
with moment inertia: I = mr² = m(s sin θ)²
→ Trot = ½ m(ωs sin θ)²
L = Tkin + Trot − U = ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ
Answer: E
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: Tkin = ½ mṡ ²
Rotational kinetic energy: Trot = ½ Iω²
with moment inertia: I = mr² = m(s sin θ)²
→ Trot = ½ m(ωs sin θ)²
L = Tkin + Trot − U = ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ
Answer: E
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