University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

371

Pages:

5
Author:

Jabin F.

General Approach to Solving Circular Motion Problems: Resolve all known forces along r ^ and θ ^ directions. Write radial EOM. F_r = m * a_r If a question can't be solved using the above two steps, then use the following steps: Use conservation of energy (if applicable) to find speed. Use tangential EOM (if conservation of energy couldn't be used). F_θ = m * a_θ If the circle is horizontal, then apply the concept of equilibrium in the vertical direction. F_up = F_down Note: Don't use the concept of equilibrium in any other case of circular motion. Angular Momentum: L = r x p = r x (mv) = m(r x v) = m(rvsin90°) L = mvr As r is ⊥ to both r and v Problem: A particle moves in one dimension in a potential U(x) = x²/2 + x³/3 in suitable units. If E is the total energy of the particle, then the motion is: (a) always bounded if 0 ≤ E ≤ 1/6. (b) always unbounded if 0 ≤ E ≤ 1/6. (c) always bounded if E < 0. (d) Bounded if 0 ≤ E ≤ 1/6 and the initial position satisfies -1 < x(0) < 1/6. Solution: U'(x) = x + x² = 0 => x = 0, -1 U''(x) = 1 + 2x At x = 0, U''(x) = 1 > 0: Minimum, stable equilibrium point (SEP) At x = -1, U''(x) = -1 < 0: Maximum, unstable equilibrium point (USEP) Therefore, the particle's motion is bounded if its total energy E is less than or equal to the potential energy at the maximum point (1/6) and its initial position is between the two equilibrium points (-1 and 1/6). Problem: A particle of unit mass is released at x=1 in one-dimensional potential V(x) = x³ - x. Which one of the following statements is incorrect? (a) Particle will come to rest at x=0. (b) Particle's speed is maximum at x=±1/√3. (c) Speed of particle at x=1/2 is √(2/3). (d) Particle will not return to x=1. Solution: Given, V(x) = x³ - x Applying conservation of energy: (KE + PE)ₓ=1 = (KE + PE)ₓ ½m₀² + 0 = 0 + V(x) ⇒ V(x) = ½ Now V'(x) = 3x² - 1 = 0 ⇒ x² = 1/3 ⇒ x = ±1/√3 V''(x) = 6x At x = 1/√3, V''(x) = 6√3 > 0 - min, Vmin, SEP At x = -1/√3, V''(x) = -6√3 < 0 - max, Vmax, USEP