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Unformatted text preview: ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13—103* A radar is tracking a rocket.
At some instant of time, the distance,
r, and angle, 9, are measured as 10 mi 0
and 30 , respectively. From successive measurements, the derivatives, r, r,
and 5 are estimated to be 650 ft/s, 165 ft/sz, 0.031 rad/s, and 0.005 rad/s?
respectively. Determine the velocity and acceleration of the rocket. Solution The re and G—components of velocity are V = = 650 ft/s 1' 0 v6 = r6 = (10)(5280)(0.03l)
= 1636.8 ft/s 2 2
v = V 650 + 1636.8 = 1761 ft/s velocity of the rocket is V 1761 ft/s 5 81.660 The r— and G—components of acceleration are
I. .2 ar = r  r9
= 165  (10)(5280)(0.03l)2
= 114.26 ft/s2
ra + 2L0
(10)(5280)(0.005) + 2(650)(0.031)
304.3 ft/s2 2 2 2
a = Y 114.26 + 304.3 = 325 ft/s and the acceleration of the rocket is _. 2 o
a = 325 ft/s 5 80.58 102 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13106* A particle is following a path given by r(t) 50 cos 39 where 9(t) is in radians and r is in millimeters. Given that 6 = 2.5 rad/s (constant) and that 9 = 0 when t = 0, determine the velocity and acceleration of the particle when t = 2 5. Solution Given the angular velocity 6 = 2.5 rad/s = constant, the angular position and angular acceleration are 2
9 = 2.5t rad 9 = 2.5 rad/s 6 = O rad/s
Differentiating the radial position I = 50 cos 39 = 50 cos 7.5t mm with respect to time gives r = 375 sin 7.5t mm/s . 2
r = 2812.5 cos 7.5t mm/s Then when t = 2 s, 6 5 rad = 286.480 2
r = —37.984 mm r = —243.858 mm/s = 1376.93 mm/s 6 2.5 rad/s = O rad/s2
L = —243.858 mm/s r6 = 94.960 mm/s V = 243.858 é — 94.960 é mm/s
r 6 = 262 mm/s L 52.24° ’ 2
r — r62 = 1614.33 mm/s I. l 2
r6 + 2r6 = —1219.25 mm/s
5 = 1614.33 é — 1219.25 6 mm/s
r 6 2 o
= 2023 mm/s F 69.42 105 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 WWMMMWWMWNW‘iQF
the pivot A to the pin B. " ' Determine the velocity components vr and we of the collar. Determine the acceleration components at and a9 of the collar. Verify that the velocity vector V and
the acceleration vector 2 are both
directed along the horizontal rod. Solution Given the angular position of the arm 9 = 90  30 cos 1.5t deg =
ﬁ/Z — ﬁ/6 cos 1.5t rad, the angular velocity and angular acceleration
of the arm are 9 = (ﬂ/4) sin 1.5t 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 r sin 9 + ré cos 9 = 0
sin 9 + 2L6 cos 9 + r5 cos 9  réz sin 6 = 0
Then at = 5 s
= 79.600 5 = 0.73670 rad/s _ 0.40837 rad/s2
= 2.0339’ft 3 2.03 ft = O.2749l ft/s = 1.02553 ft/s2 r = r = —O.275 ft/s V6 = :6 = 1.498 ft/s
v 1.523 ft/s L 79.600
to the radial direction or along
the horizontal rod so 2 2
r — I6 = 0.0781 ft/s :6 + 2E6 = 0.425 ft/s2
_. 2 o
a = 0.432 ft/s L 79.60 (also along the rod) 108 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13112 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
D
where r is in millimeters. When 9 = 75 , 3. Determine the velocity components Vr and v9 of the pin P. b. Determine the acceleration components ar and 36 of the pin P. c. Verify that the velocity vector V is directed along the cam surface. r=20+15wse Solution Since the arm rotates at a constant angular rate (9 = 60 rev/min = 2” rad/s), the angular acceleration is zero. Differentiating the
expression for the radial position of the pin r = 20 + 15 cos 9 mm —159 sin 9 mm/s —1592 cos 6 mm/s2 0 75 ,
23.88229 mm = —91.03636 mm/s
2
= —153.2665 mm/s E = 91.0 mm/s = 150.1 mm/s ...o v = 175.5 mm/s P 16.240 2
r — r9 = 1096 mm/s2
.I . I 2
r9 + 2:9 = —1144 mm/s ' 2
a = 1584 mm/s E 58.780 (Problem 13112 continues ...) 112 ...
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This note was uploaded on 02/07/2011 for the course CVEN 501 taught by Professor Beason during the Spring '10 term at Texas A&M.
 Spring '10
 Beason

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