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Unformatted text preview: 'T. ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges [ 13103* 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 ecomponents 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 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 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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 Spring '08
 McVay

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