Question 1

**Setting up the motion** Let θ be the angular displacement of the disk's center (measured from the ring's center) from the bottom. The center moves on a circle of radius (R − r), so its speed is v = (R − r)θ̇. **Kinetic energy** (disk rolling without slipping, I = ½mr²): K = ½mv² + ½Iω_spin² = ½mv² + ¼mv² = ¾ m (R − r)² θ̇² **Potential energy** - Gravity: U_g = mg(R − r)(1 − cos θ) ≈ ½ mg(R − r) θ² - Spring (stretched along the arc by the displacement of the disk center, x = (R − r)θ): U_s = ½ k (R − r)² θ² **Equation of motion** Total energy: E = ¾ m (R − r)² θ̇² + ½ [mg(R − r) + k(R − r)²] θ² Differentiating: (3/2) m (R − r)² θ̈ + [mg(R − r) + k(R − r)²] θ = 0 θ̈ + (2/3)[g/(R − r) + k/m] θ = 0 Hence ω = √[(2/3)(g/(R − r) + k/m)] This matches option (A).

No written reasoning was returned for this item.

No written reasoning was returned for this item.