University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2024
Views:

366

Pages:

16
Author:

floribund1qn

**Physics 101, Dr. Moneeb Shatnawi** **Physics Dept., The University of Jordan** **Solutions Chp. 9/5** **Example 9.6:** The angular position of a swinging door is described by: θ(t) = (5 + 10t + 2t²) rad. Determine the angular position, angular speed, and angular acceleration of the door: a) at t = 0 b) at t = 3 sec. **Solution:** θ(t) = (5 + 10t + 2t²) rad. So, θ(t = 0) = 5 + 10(0) + 2(0)² = 5 rad. ω(t) = dθ(t)/dt = d/dt [5 + 10t + 2t²] ω(t) = 0 + 10 + 4t So, ω(t = 0) = 10 + 4(0) = 10 rad/s α(t) = dω(t)/dt = d/dt [10 + 4t] = 0 + 4 = 4 rad/s² α(t = 0) = 4 rad/s² b) θ(t = 3 sec) = [5 + 10(3) + 2(3)²] = 53 rad. ω(t = 3 sec) = [10 + 4(3)] = 22 rad/s. α(t = 3 sec) = 4 rad/s². **Example 9.7:** An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angular speed of 2000 rad/s. The engine's rotation slows down at a constant rate of 20 rad/s². Determine the time it takes for the rotor to come to rest. **with an angular acceleration of magnitude 80 rad/s².** **a) Determine the angular speed after 10 sec.** **b) How long does it take for the rotor to come to rest?** **Solution:** **a)** * **wᵢ = -2000 rad/s** (since it's clockwise) * **α = +80 rad/s²** (opposite to w as the speed is decreasing) Since α is constant, then we can use the equations of motion under constant acceleration. ω = ωᵢ + αt ⇒ ω = -2000 + 80(10) ⇒ ω = -1200 rad/s. **b)** ω = 0 ⇒ 0 = ωᵢ + αt ⇒ 0 = -2000 + (80)t ⇒ 80t = 2000 ⇒ t = 2000/80 = 25 sec. **Example 9.8** A centrifuge in a medical laboratory rotates at a rotational speed of 3600 rev/min. When switched off, it rotates 50 times before coming to rest. Find the constant angular acceleration of the centrifuge. **Solution:** Since we are given that the acceleration is constant, then we can use the equations of motion under constant acceleration. ω² = ωᵢ² + 2αΔθ **Given:** ωᵢ = 3600 rev/min Δθ = 50 rev ω = 0, find α. **a) How long does the wheel take to reach its final rotational speed of 1200 rev/min?** **b) Through how many revolutions does it turn while accelerating?** **Solution:** Uniform solid disk. R = 7 cm = 0.07 m. m = 2 kg. (T) = 0.6 N.m (I) disk = 1/2 MR² = 1/2 (2)(0.07)² = 4.9*10⁻³ kg.m² Στ = Iα ⇒ α = 0.6 / 4.9*10⁻³ = 122 rad/s² (α constant) ω = ωᵢ + αt ⇒ t = (ω - ωᵢ) / α = (1200 rev/min - 0 rev/min) / 122 rad/s² Note that ω = 1200 rev/min = (1200 rev/min)(2π rad/1 rev)(1 min/60 sec) = 125.6 rad/sec. t = (125.6 rad/sec) / 122 rad/s² = 1.03 sec. **b)** Δθ = ωᵢt + 1/2 αt² = 0 + 1/2 (122) (1.03)² = 64.7 rad. Δθ = (64.7 rad)(1 rev/2π rad) = 10.3 rev. **Example 9.16:** A block of mass m₁ = 2 kg and a block of mass m₂ = 6 kg are connected by a massless string over a pulley in the shape of a disk having radius R = 0.25 m and mass M = 10 kg. These blocks are allowed to move on a fixed block-wedge of angle θ = 30°, as shown in the adjacent figure. The coefficient of kinetic friction for both blocks is 0.360. **a) Draw free-body diagrams of both blocks and of the pulley.** [Image of free-body diagrams for m₁, m₂, and the pulley] **b) Determine the acceleration of the two blocks.** **c) Determine the tensions in the string on both sides of the pulley.** **Solution:** **a)** [See the provided image for the free-body diagrams] **b)** For mass m₁: ΣFx = ma ΣFy = ma So, T₁ - fk₁ = m₁a ... (1) N₁ - m₁g = 0 ... (2) fk₁ = μk * N₁ = μk * m₁g ... (3) For mass m₂: ΣFx = ma ΣFy = ma So, m₂g sinθ - T₂ - fk₂ = m₂a ... (4) N₂ - m₂g cosθ = 0 ... (5) fk₂ = μk * N₂ = μk * m₂g cosθ ... (6) For the pulley: Στ = Iα T₂R - T₁R = (1/2)MR²α ... (7)