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

2024
Views:

439

Pages:

7
Author:

Nehemiah Rojas

SOLUTIONS TO HOMEWORK PROBLEMS CHAPTER 16 Chap 16.1-3 HW#5 16-2, 11, 22, Chap 16-5 16-59, 64, 80, 89 Problem 16-2 A flywheel has its angular speed increased uniformly from phi_{2} 1 d_{2} ^ 3 time 1. If the diameter of the wheel is D. determine the magnitudes of the normal and tangential components of acceleration of a point on the rim of the wheel at time 1, and the total distance the point travels during the time period. Given: 2015 rad I = 80 6o_{2} =60 rad s D = 2ft Solution: r = D/2 e_{2} = omega_{2} + omega*l a_{2} = ar a_{M} = e_{2} ^ 2 * r theta= (w_{2} ^ 2 - w_{j} ^ 2) * 2a d = theta*r cr = (phi_{2} - phi_{I})/r a_{I} = 0.563 * (ft)/(s ^ 2) partial H = 3600 * (ft)/(s ^ 2) theta = 3000 rad d = 3000ft alpha = 0.56 * (ind)/(s ^ 2) Problem 16-22 A motor gives gear 4 angular acceleration a A = a * theta ^ 3 +b.If this gear is initially turning with angular velocity C A * 0 deg determine the angular velocity of gear B after A undergoes an angular displacement theta_{U} d l A0 =20 rad s r_{A} = 0.05 r_{B} = 0.15m Given: rev = 2n rad a=0.25 rad s ^ 2 b = 0.5 * (md)/(s ^ 2) theta_{1} = 10 rev Solution: alpha_{A} = a * theta ^ 3 + b e A ^ 2 = omega A dot e ^ 2 +2 int 0 ^ theta I (a dot theta ^ 3 +b)d theta omega A = sqrt alpha A0 ^ 2 +2 int 0 ^ theta I a dot theta ^ 3 + bd*theta omega_{B} = r_{A}/r_{B} * omega_{A} o_{B} =465 rad s O_{A} =1395.94 rad s Problem 16-30 A tape having a thickness s wraps around the wheel which is turning at a constant rate Assuming the unwrapped portion of tape remains horizontal, determine the acceleration of point P of the unwrapped tape when the radius of the wrapped tape is r. Hint: Since r_{\mathcal{P}} = c9P_{e} take the time derivative and note that d/dt (nu) = omega(nu / 2 * pi) Solution v_{P} =c(r^ * ) sigma_{p} = d/dr (v_{p}) = (des)/(dr) * r + co * d/dr (r) since d/dr (phi) = 0 a_{p} = c(d/dt (r)) In one revolution is increased by s, so that (2pi)/(Delta*theta) = x/(Delta*r) Hence, Delta*r = s/(2pi) * Delta*theta a_{p} = s/(2pi) * omega ^ 2 d/dt (r) = s/(2pi) * omega Problem 16-59 The angular velocity of link AB is Cl A * B' Determine the velocity of the collar at Cand the angular velocity of link CB for the given angles and Link CB is horizontal at this instant. Given: alpha AB =4 rad s r AB =500 mm r BC =350 nin phi = 45 deg theta = 60 deg theta_{j} = 30 deg I BC WAR F AB Solution: Guesses Given v_{C} = 1 * m/s o CB =1 rad s ( matrix 0\\ 0\\ 0\\ omega AB matrix )*( matrix r AB cos( phi)\\ r AB sin( phi)\\ 0 matrix )+( matrix 0\\ 0\\ phi CB matrix )*( matrix -r BC \\ 0\\ 0\\ 0 matrix )=( matrix -r C cos( phi)\\ -r C sin( phi)\\ 0 matrix ) VC OCB = Find(VC. @CB) phi CB =7.81 rad s v_{C} = 2.45 * m/5