University:
Angelo State University
Course:
ENGR 2302 | Engineering Mechanics - Dynamics
Academic year:

2024
Views:

368

Pages:

6
Author:

helyettitycx1

SOLUTIONS TO HOMEWORK PROBLEMS HW#3 12-107, 117, 133, 141, 12-152, 163 Problem 12-107 The car travels along the curve having a radius of R. If its speed is uniformly increased from (v_1) to (v_2) in time t, determine the magnitude of its acceleration at the instant its speed is (v_3). 1 1. www.studyblue.com www.studyblue.com Відкриється в новому вікні www.chegg.com curved road with a car on it. The car is shown at a point along the curve, and the radius R is indicated. Given: (v_1 = 15 \frac{m}{s}) (v_2 = 27 \frac{m}{s}) (v_3 = 20 \frac{m}{s}) (t = 3 s) (R = 300 m) Solution: (a_t = \frac{v_2 - v_1}{t}) (a_n = \frac{v_3^2}{R}) (a = \sqrt{a_t^2 + a_n^2}) (a = 4.22 \frac{m}{s^2}) Problem 12-117 Cars move around the "traffic circle" which is in the shape of an ellipse. If the speed limit is posted at v, determine the maximum acceleration experienced by the passengers. Given: (v = 60 \frac{km}{hr}) (\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1) (a = 60 m) (b = 40 m) Solution: Maximum acceleration occurs where the radius of curvature is the smallest. In this case that happens when y=0. (x(y) = a\sqrt{1 - (\frac{y}{b})^2}) (\rho(y) = \frac{\sqrt{(1 + x'(y)^2)^3}}{x''(y)}) (x'(y) = \frac{d}{dy}x(y)) (x''(y) = \frac{d}{dy}x'(y)) (\rho_{min} = \rho(0m)) (\rho_{min} = 26.667 m) (a_{max} = \frac{v^2}{\rho_{min}}) (a_{max} = 10.42 \frac{m}{s^2}) Problem 12-133 The truck travels at speed (v_0) along a circular road that has radius ρ. For a short distance from (s=0), its speed is then increased by (dv/dt = bs). Determine its speed and the magnitude of its acceleration when it has moved a distance (s_1). Given: (v_0 1 = 4 \frac{m}{s}) (\rho = 50 m) (b = \frac{0.05}{s^2}) (s_1 = 10 m) Solution: (a_t = v(\frac{d}{ds}v) = bs) (\int_{v_0}^{v_1} v dv = \int_0^{s_1} bs ds) (\frac{v_1^2}{2} - \frac{v_0^2}{2} = \frac{b}{2}s_1^2) (v_1 = \sqrt{v_0^2 + bs_1^2}) (v_1 = 4.58 \frac{m}{s}) (a_t = bs_1) (a_n = \frac{v_1^2}{\rho}) (a = \sqrt{a_t^2 + a_n^2}) (a = 0.653 \frac{m}{s^2})