This preview shows pages 1–18. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ENGINEERING MECHANICS  Dynamics WF. Riley & L.D. Sturges 132* The position of a particle moving along the x—axis is given as a function of time by x(t) 15  4t m. Determine the velocity of the particle as a function of time.
Determine the acceleration of the particle as a function of time. Evaluate the position, velocity, and acceleration of the particle at ‘
5 s. t Determine the total distance traveled by the particle between 0 and t 5 5. Sketch x(t), v(t), and a(t); O < t < 8 5. Solution Given x(t) = 15  4t m, take two time derivatives to get v(t) —4 m/s 2
0 m/s a(t) Evaluating these expressions at x = 5 m V 4 m/s 2
0 m/s 3 Since x(t) is uniformly decreasing x(5) — x(0) 20 S: m Problem 132 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 133* The position of a particle moving along the x—axis is given as a t/3
function of time by x(t) = 3e ft. Determine the velocity of the particle as a function of time. Determine the acceleration of the particle as a function of time. Evaluate the position, velocity, and acceleration of the particle at
= 5 5. Determine the total distance traveled by the particle between
and t = 5 5. Sketch x(t), v(t), and a(t); O < t < 8 5. Solution t/3 . . .
Given x(t) = 3e ft, take two time derivatives to get t./3
v(t) = —e ft/s a(t) = (1/3)e—U3 ft/s2 Evaluating these expressions at
x = 0.567 ft
V = O.1888 ft/s
a = 0.0630 ft/s2 Since x(t) is uniformly decreasing s = lx(5)  x(0) = 2.43 ft Problem 133 Time. t (s) ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 134 The position of a particle moving along the x—axis is given as a function of time by x(t) = 4 sin wt m (w = 3/2 rad/s). a. Determine the velocity of the particle as a function of time.
b. Determine the acceleration of the particle as a function of time. Evaluate the position, velocity, and acceleration of the particle at
= 5 5. Determine the total distance traveled by the particle between t = O
and t = S 5. Sketch x(t), v(t), and a(t); 0 < t < 8 5. Solution
Given x(t) = 4 sin 3t/2 m, take two time derivatives to get
V(t) = 6 cos 3t/2 m/s
a(t) = 9 sin 3t/2 m/s2 Evaluating these expressions at
x = 3.75 m
V = 2.08 m/s 2
a = —8.44 m/s The velocity is zero (and the particle reverses its direction of motion)
when t ﬁ/3 s, H s, Sﬂ/B s .... Therefore, the distance traveled is
= x(1r/3) — x(0) + Ixur) — x(1I/3) + x(5)  X(N) = 15.75 m Problem 134 Time, t (s) ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 139* The velocity of a particle moving along the x—axis is given as a —t/3
function of time by v(t) = 30e ft/s and x(3) = 20 ft. Determine the position of the particle as a function of time.
Determine the acceleration of the particle as a function of time. Evaluate the position, velocity, and acceleration of the particle at
t = 8 5. Determine the total distance traveled by the particle between t = 5
and t = 8 5. Sketch x(t), V(t), and a(t), O < t < 10 5. Solution
The position is obtained by integrating the velocity
t/3
V(t) 30e ft/s —’L/3
x(t) 53.109 — 909 ft where the constant of integration is chosen so that x(3) = 20
acceleration is obtained by differentiating the velocity to get t/3 2
a(t) = —lOe ft/s Evaluating these expressions at
x 46.9 ft
v 2.08 ft/s 2
a —O.695 ft/s since x(t) is monotonically increasing, the distance traveled is s = lx(8)  x(5) = 10.75 ft Problem 139 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1325 The jet shown is
catapulted from the deck of an
aircraft carrier by a hydraulic
ram. Determine the acceleration
of the jet if it accelerates uniformly from rest to 160 mi/h in 300 ft. Solution Use the chain—rule of differentiation to write the acceleration dv
dx f a dx to get 2
V 2 where the constant of integration C O is chosen so that v(0) =
Then, when x = 300 ft, v = 160 mi/h 234.67 ft/s, and therefore 2
a = 91.8 ft/s E 2.85 g 23 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1351* Train A is traveling eastward at 80 mi/h while train 8 is
traveling westward at 60 mi/h.
Determine The velocity of train A
relative to train 8. The velocity of train 8
relative to train A. 80 mi/h —+ = +80 mi/h 60 mi/h F~ 60 mi/h VA/B = VA  VB = (80) — (60)
= 140 mi/h % . . . . . ... . . . . . . . . . ... . . . . . . . . . . . . . .. Ans. b. VB/A = vB  VA = (—60)  (80) —l4O mi/h = 140 mi/h 6— . . . . . . . . . . . . . . . . . . . . .. Ans. 13—52* Boat A travels down a straight
river at 20 m/s while boat 8 travels up the river at 15 m/s. Determine The velocity of boat A relative to
boat 8. The velocity of boat 8 relative to
boat A. Solution —20 m/s 15 m/s (—20) — (15)
= 35 m/s l <15) 48 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13—72* In Fig. P13—72 block 3 has
a constant downward acceleration of
0.8 m/sz. Determine the speed and
acceleration of block A 5 s after the system starts from rest. Solution
For block 3 2
aB = 0.8 m/s ¢ 00 SB = constant 0.8t m/s ¢ 5
B But the length of the rope is constant
K = +
25A 53 and taking two time derivatives gives = zét + A
A B 25 + s
A 66 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13103* 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 I
o on 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 r— and Gcomponents of velocity are v = = 650 ft/s r 0 v6 = r6 = (10)(5280)(0.031) = 1636.8 ft/s 2
V = V 650 + 1636.82 = 1761 ft/s and the velocity of the rocket is v = 1761 ft/s a 81.660 The r— and G—components of acceleration are ar=;r92
z 165 — (10)(5280)(o.o31)2
= 114.26 ft/s2
r5 + 2L6
(10)(5280)(0.005) + 2(650)(0.031)
304.3 ft/s2 2
a = Y 114.26 + 304.32 = 325 ft/s2 and the acceleration of the rocket is .—. _ 2 o
a = 325 ft/s 1 80.58 102 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13—106* A particle is following a path given by r(t) 50 cos 36 where
6(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 9 2.5 rad/s = constant, the angular position and angular acceleration are 2
9 = 2.5t rad 6 = 2.5 rad/s 9 = 0 rad/s
Differentiating the radial position r 50 cos 39 = 50 cos 7.5t mm with respect to time gives r = 375 sin 7.5t mm/s .0 2
r = —2812.5 cos 7.5t mm/s Then when t = 2 s, 9 = 5 rad = 286.48 r = —37.984 mm = —243.858 mm/s = 1376.93 mm/s2 E
9 = 2.5 rad/s 0 rad/s2 = E = —243.858 mm/s r6 = —94.960 mm/s v = —243.858 ér — 94.960 ée mm/s = 262 mm/s E 52.240 .0 2 2
r — r6 = 1614.33 mm/s .. .. 2
r6 + 2:6 = 1219.25 mm/s
5 = 1614.33 at  1219.25 ée mm/s 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 6(t) = 90 ~ 30 cos wt where w = 1.5 rad/s, t is in seconds, and 6 is in degrees.
For t = 5 s, Determine the radial distance r from
the pivot A to the pin 8. Determine the velocity components vr and V9 of the collar. c. Determine the acceleration components ar and as of the collar. d. 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 =
n/2 — ﬂ/6 cos 1.5t rad, the angular velocity and angular acceleration
of the arm are 6 = (N/4) sin 1.5t rad/s 9 = (3ﬁ/8) cos l.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 L sin 9 + r6 cos 9 = O .. . O O. O 2
r sin 9 + 2r9 cos 6 + r6 cos 9 — r6 sin 9 = O Then at t = 5 s = 79.600 6 = 0.73670 rad/s CD r = 2.0339 ft 5 2.03 ft . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
o u 2
r = —o.27491 ft/s r = 1.02553 ft/s
vr = r = —O.275 ft/s Ve = r6 = 1.498 ft/s . . . . . . . . .. Ans. 1.523 ft/s 5 79.600
to the radial if
direction or along
the horizontal rod to 2 2
= —0.0781 ft/s . . . . ... . . . . . . ............ Ans. l:
3; . ' 2
2r9 = 0.425 ft/s . . . . . . ... . . . . . . . . . . . . . . .. Ans. E 2 o
0.432 ft/s 5 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 0
where r is in millimeters. When 6 = 75 , a. Determine the velocity components VI and v9 of the pin P. b. Determine the acceleration components ar and a6 of the pin P. c. Verify that the velocity vector 3 is directed along the cam surface. 20+15Cm9 Solution since the arm rotates at a constant angular rate (é = 60 rev/min =
2n 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 '2 2
= —159 cos 9 mm/s 75°,
23.88229 mm
= —9l.03636 mm/s = —153.2665 mm/s2 = L = —9l.0 mm/s o = 150.1 mm/s = 175.5 mm/s F 16.240 2
= r — r6 = —lO96 mm/s2 r9 + 2L6 = 1144 mm/s2 .a 2 o
a = 1584 mm/s E 58.78 (Problem 13112 continues ...) 112 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 13112  cont.) The tangent to the surface is found from the slope tan ¢ = dy/dx r = 20 + 15 cos 9 (= 23.88229 mm) y = r sin 9 (= 23.06852 mm)
x = r cos 9 (= 6.18119 mm) dr/de = 1s sin 6 (= 14.48889 mm)
dy/de = dr/d9 sin 9 + r cos 9 (= 7.81400 mm) dx/d9 = dr/de cos 9 — r sin 9 (= —26.81852 mm)
= dx = dzgde = 7.81400 =
tan ¢ dx dx/de —26.81852 0'29137 ¢ = 16.24 which is the same direction as the velocity vector. “14° m: (6.240 f) ,0 [R \\'7a 113 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13119* A car drives over the
top of a hill that has a radius
of curvature of 110 ft. If the
normal component of acceleration
necessary to keep the car on the
road becomes greater than that
provided by gravity, the car will
become airborne. Determine the
maximum constant speed v at which
the car can go over the hill. Solution 5 9 = / (110)(32.2) ft/s = 40.6 mi/h 13—120* The car shown has a
speed of V = 100 km/h and its
speed is increasing at the rate
of ; = 5 m/s2 at the instant
shown. If the radius of
curvature at the bottom of the
hill is 80 m, determine the acceleration (magnitude and direction) of the car. Solution v = 100 km/h = 27.778 m/s
;  5 m/s2 — a
' t V2 27 7782 2
an=p—'=T= 9.645 m/S Therefore the total acceleration is 2 2
a = y 5 + 9.645 = 10.864 m/s2 0
at a 62.60 relative to the tangential (horizontal) direction. 2 0
10.86 m/s a 62.60 120 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13124* The bead shown is sliding around a circular ring 3 m in diameter. At the instant
0 shown (9 = 30 ), the speed of the bead is V = 4 m/s and the speed is increasing at a rate 0 2
of v = 5 m/s . Determine the acceleration (magnitude and direction) of the bead. Solution 10.667 m/s2 2 2 0
11.78 m/s & 34.89 123 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13126 The head shown moves along a circular path in a horizontal plane according to the 3 . .
relation 5 = t + 6t where s is measured in meters and t is the time in seconds. If the magnitude of the acceleration of the bead is 2
20 m/s when t = 2 5, determine the radius of the circle. Solution Differentiating the position of the bead along the circle s = t + 6t m Therefore . .  g n c  . n . . .  ~ a . ¢ n o v . . . u .  u g  n u . o o .. .:./ = 12 5 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13—129* A sphere slides along a rod that is bent in a vertical plane
into a shape that can be described by the equation x2 8y, where x and
y are both measured in feet. When
the sphere is at the point x 8 ft, y = 8 ft as shown, it is
moving along the rod (down and to the right) at a speed of v = 15 ft/s and is slowing down at a rate of 3 ft/sz. Determine the acceleration (magnitude and direction) of the sphere at this time. Solution
The radius of curvature is obtained from the derivatives of the shape
function
2
y = x /8
dy/dx x/4
2 2
d y/dx 1/4 2 2 L _ d dx _ 1(4
p ‘ 2 3/2 _ 2 3/2
[1+ ] [1+ W4) ] When x = —8 and y = 8, dy/dx = 2, 1/0 = 0.02236, 9 = 44.7214 ft, and 2
= v = 3 ft/s
v 2 2
an = —p—_ = W = “’31 ft/S t and the tangential direction makes an angle of —1
¢ tan —g§— tan —63.43o Therefore, a 128 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 14—16* Weights A and B are supported by cords
wrapped around a stepped drum as shown. At the
instant shown, weight A has a downward velocity
of 2 m/s and its speed is decreasing at a rate of 1.5 m/sz. For this instant, determine a. The acceleration of weight 3. b. The acceleration of point D on the rim of the wheel. Solution
Since the cords do not slip on the drums, the velocity and acceleration
of the weights are the same as the tangential components of the velocity and acceleration of points on the surface of the drums. Therefore, =0): = 13.333 rad/s C 10 rad/s2 D a: = 10(o.1) = 1 m/s2 ¢ Points on the surface of the drums move in circular motion; therefore, the acceleration of point D is E = «r a + w r E
D t n = (10)(o.15)(j) + (13.333)2(o.15)(—i) 26.7 i — 1.5 3 m/s2 2 o
26.7 m/s F 3.219 213 ENGINEERING MECHANICS  Dynamics 1421 It is desired to bring the two disks together without slipping. Initially, disk 8 is at rest and If disk A is . 2
given a constant angular deceleration of 5 rad/s disk A is rotating at 750 rev/min. and disk B is given a constant angular acceleration 2
of 8 rad/s , determine the time at which the two disks can be brought together without slipping and the angular velocity of each disk at that time. Solution Initially, wB = O and wA = 750 rev/min = 78.540 rad/s. given constant angular accelerations 2
uA —5 rad/s = constant 2
dB 8 rad/s = constant W.F. Riley & L.D. Both disks are and integrating the angular accelerations with respect to time gives the angular velocities of the disks 0A 78.540  5t rad/s w 8t ad
8 r /s The two disks will roll without slipping on each other when the velocities of the contact points are the same
=w=3w
V 2 A B
2(78.54O  St)
= 4.62 s
= 55.44 rad/s = 529 rev/min 36.96 rad/s 353 rev/min 218 Sturges ...
View
Full
Document
 Spring '08
 McVay

Click to edit the document details