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

2024
Views:

39

Pages:

6
Author:

Jabin F.

# Simple Pendulum: Here, r = l, x = lθ velocity at any instant: v = ẋ = lθ̇ + lθθ̈ = lθ̇ Now P.E. = -mgl cos θ K.E. = (1/2)mv² = (1/2)ml²θ̇² Total Energy: E = (1/2)ml²θ̇² - mgl cos θ Conservation of energy: dE/dt = 0 ⇒ ml²θ̇θ̈ + mgl sin θ = 0 ⇒ θ̈ + (g/l) sin θ = 0 This is not periodic motion. Now we use approximation: for small oscillation, sin θ ≈ θ ⇒ θ̈ + (g/l)θ = 0 This is SHM. ω² = g/l ⇒ ω = √(g/l) T = 2π/ω = 2π√(l/g) Problem 2: A particle of unit mass is dropped under uniform gravity (g = 1). It experiences a drag force F = -v². Its speed after time t will be: Solution: EOM: mg - v² = ma ⇒ 1 - v² = dv/dt ⇒ dv/(1 - v²) = dt ∫dv/(1 - v²) = ∫dt Q-50 In the figure shown, a particle goes from region I to region II. In region I, it has speed v₁, and potential energy u₁, while in region II, it has potential energy u₂. Which of the following is correct? (a) sin θ₂/sin θ₁ = (1 - (2/m)(u₁ - u₂))^(1/2) (b) cos θ₂/cos θ₁ = (1 + (2/m)(u₁ - u₂))^(1/2) (c) sin θ₂/sin θ₁ = (1 + (1/m)(u₁ - u₂))^(1/2) (d) cos θ₂/cos θ₁ = (1 - (2/m)(u₁ - u₂))^(1/2) Solution: PE is given, so we can apply conservation of energy. (KE + PE)₁ = (KE + PE)₂ (1/2)mv₁² + u₁ = (1/2)mv₂² + u₂ ⇒ v₂² = v₁² + (2/m)(u₁ - u₂) ⇒ v₂ = √(v₁² + (2/m)(u₁ - u₂)) PE change in x direction only, not in y direction. Fx = 0 ⇒ px = conserved. Fy = 0 ⇒ py = conserved. Conservation of momentum in y direction: (py)₁ = (py)₂ mv₁ sin θ₁ = mv₂ sin θ₂ sin θ₂/sin θ₁ = v₂/v₁ = √(v₁² + (2/m)(u₁ - u₂)) / v₁ = (1 + (2/m)(u₁ - u₂))^(1/2) P-8 In the previous question, let T₁ be the kinetic energy of particles in region I. Then the ratio sin θ₁/sin θ₂ is given by: Solution: sin θ₁/sin θ₂ = √(1 + (u₁ - u₂)/T₁) P-60 Two particles of masses m₁ and m₂ are placed 'd' distance apart. Due to gravitational attraction they move towards each other. What is speed of m₁ when their separation reduces to d/2? Solution: As all the forces are conservative in nature, apply conservation of energy. (KE + PE)initial = (KE + PE)final (1/2)m₁(0)² + (1/2)m₂(0)² - Gm₁m₂/d = (1/2)m₁v₁² + (1/2)m₂v₂² - Gm₁m₂/(d/2) ⇒ 0 + 0 - Gm₁m₂/d = (1/2)m₁v₁² + (1/2)m₂v₂² - 2Gm₁m₂/d ⇒ 2Gm₁m₂/d = m₁v₁² + m₂v₂² ⇒ 2Gm₁m₂/d = m₁(v₁² + (m₂/m₁)v₂²) Put in v₂ = (m₁/m₂)v₁ from equation (1) ⇒ 2Gm₁m₂/d = m₁(v₁² + (m₁/m₂) * (m₁/m₂)v₁²) ⇒ 2Gm₁m₂/d = m₁(v₁² + m₁²/m₂²) ⇒ 2Gm₁m₂/d = m₁(m₁² + m₂²) / m₂ ⇒ v₁² = (2Gm₂²)/(m₁ + m₂)d ⇒ v₁ = √((2Gm₂²)/(m₁ + m₂)d)