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22Solutions - 'T ENGINEERING MECHANICS Dynamics W.F...

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Unformatted text preview: 'T. ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges [ 13-103* A radar is tracking a rocket. . At some instant of time, the distance, i r, and angle, 9, are measured as 10 mi 0 and 30 , respectively. From successive measurements, the derivatives, r, r, 6, J and 9 are estimated to be 650 ft/s, " 2 2 165 ft/s , 0.031 rad/s, and 0.005 rad/s, { respectively. Determine the velocity and acceleration of the rocket. LI Solution The r— and e-components of velocity are Vr = E = 650 ft/s v6 = r6 = (10)(5280)(0.031) = 1636.8 ft/s 2 2 V = Y 650 + 1636.8 = 1761 ft/s and the velocity of the rocket is W" Then 3 = 1761 ft/s 5 81.660 .......... . ....... . ........... Ans. The r— and G—components of acceleration are u '2 . ar = r — r9 ' 2 = 165 — (10)(5280)(0.031) 2 = 114.26 ft/s a9 = r0 + 2r6 I = (10)(5280)(0.005) + 2(650)(0.03l) 2 = 304.3 ft/s Then 2 2 2 a = V 114.26 + 304.3 = 325 ft/s } and the acceleration of the rocket is _. 2 o a = 325 ft/s 5 80.58 . ....... . ............. ........ Ans. 102 'ING MECHANICS - Dynamics W.F. Riley & L.D. Sturges A particle is following a path given by r(t) = 50 cos 36 where _$3 in radians and r is in millimeters. Given that 9 = 2.5 rad/s 0 when t = 0, determine the velocity and :'ltion of the particle when t = 2 s. I fiflfiEfl the angular velocity 9 = 2.5 rad/s = constant, the angular ban and angular acceleration are 2 fl - 2.5t rad 6 2.5 rad/s 6 = O rad/s - fintiating the radial position r = 50 cos 36 = 50 cos 7.5t mm with :-h to time gives r = —375 sin 7.5t mm/s 2 r = —2812.5 cos 7.5t mm/s zflhen t = 2 s, 6 = 5 rad = 286.48 2 r = -37.984 mm r = -243.858 mm/s 1376.93 mm/S . 2 6 = 2.5 rad/s 0 rad/s } —243.858 mm/s r9 = -94.960 mm/s —243.858 6 — 94.960 6 mm/s r 6 262 mm/s 5 52.24° n I. 2 2 r - r6 = 1614.33 mm/s .. .. 2 r6 + 2:6 = —1219.25 mm/s 5 = 1614.33 at — 1219.25 66 mm/s 2 o = 2023 mm/s Y 69.42 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13—109 A collar that slides along a horizontal rod has a pin that is constrained to move in the slot of arm AB. The arm oscillates with angular position given by 9(t) = 90 - 30 cos wt where w = 1.5 rad/s, t is in seconds, and 9 is in degrees. For t = 5 5, Determine the radial distance r from the pivot A to the pin B. Determine the velocity components Vr and v6 of the collar. Determine the acceleration components ar and .36 of the collar. Verify that the velocity vector V and the acceleration vector 3 are both directed along the horizontal rod. Solution Given the angular position of the arm 9 = 90 - 30 cos l.5t deg = fl/2 — fl/6 cos l.5t rad, the angular velocity and angular acceleration of the arm are 9 = (fl/4) sin l.5t rad/s (3fl/8) cos 1.5t rad/s2 The radial position of the pin is given by r sin 9 = 2 ft Differentiating the radial position with respect to time gives E sin 9 + ré cos 9 = 0 .. 2¢ 2 sin 9 + gié cos 9 + r5 cos 6 — r02 sin 9 = 0 Then at t = 5 s e 79.600 0 = 0.73670 rad/s 2.0334 ft 3 2.03 ft 1‘ o r 2 = —o.27491 ft/s = 1.02553 ft/s = —O.275 ft/s Ve = r6 = 1.498 ft/s 1.523 ft/s B 79.600 to the radial direction or along the horizontal rod .. '2 2 r - r9 = —0.0781 ft/s .. . O 2 r6 + 2r6 = 0.425 ft/s 2 o - 0.432 ft/s B 79.60 (also along the rod) I ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13-112 Arm AB of the cam follower mechanism shown is rotating at a constant angular speed of w = 60 rev/min. A spring holds the pin P against the cam lobes. The equation that ' describes the shape of the cam lobes is r = 20 + 15 cos 9 0 where r is in millimeters. When 9 = 75 , a. Determine the velocity components VI and V9 of the pin P. b. Determine the acceleration components at and a9 of the pin P. c. Verify that the velocity vector V is directed along the cam surface. 0+15WS9 Solution Since the arm rotates at a constant angular rate (6 = 60 rev/min = 2N rad/s), the angular acceleration is zero. Differentiating the 1 expression for the radial position of the pin r = 20 + 15 cos 9 mm —156 sin 6 mm/s '2 2 -l56 cos 6 mm/s O 75 , 23.88229 mm —91.03636 mm/s 2 —153.2665 mm/s L = —91.0 mm/s - 150.1 mm/s _. O I V = 175.5 mm/s F 16.24 00 2 2 r — r9 = —lO96 mm/s r9 + 2&6 = —1144 mm/s2 —. 2 o a = 1584 mm/s E 58.78 (Problem 13—112 continues ...) 112 ...
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