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Unformatted text preview: ENGINEERING MECHANICS  Dynamics 181* A lO—lb uniform wheel l6—in. in diameter is at rest when it is W.F. Riley & L.D. Sturges placed in contact with a moving belt as shown. The kinetic coefficient of friction between the belt and the wheel is
Mk = 0.1 and the belt moves with a constant
speed of 30 ft/s. Determine the number of
revolutions that the wheel turns before it
begins to roll without slipping on the moving belt. Solution The mass center of the wheel is fixed +T 2F = ma : N'— W = 0 Y Y
N = W = 10 lb and while the wheel slips F = 0.1N = 1 1b Neither P, W, nor N'do work. The work done by the friction force F while slip occurs (and the wheel rotates through angle 6) is U = I F ds = J Fr d6 = (1)[—Eg—] I d9 = —I§— 6 lbft The initial kinetic energy of the wheel is zero. The wheel is in fixed—axis rotation and its final kinetic energy is given by T inZ
f ' 2 G
l 10 8 2 2
IG = §[3§TE][:3] = 0.06901 slug°ft
Therefore, the work—energy equation Ti + U = Tf gives
8 l 2
+ 6 = ——— . w
0 12 2 (0 06901) The wheel begins to roll without slipping when the velocity
of the contact point is the same as that of the moving belt 8 ——w 12 3o ft/s m 45 rad/s 6 = 104.81 rad = 16.68 rev ........... . . . . . .............. Ans . ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 185 A loolb rectangular plate of uniform thickness is supported by a pin and a cable as shown. If the cable at B breaks, determine for the ensuing motion a. The angular speed of the plate and the
magnitude of the pin reaction at A when the mass center of the plate is
directly below pin A. b. The angular speed of the plate and the magnitude of the pin reaction at A when O
the plate has rotated 90 (when pin B
is directly below pin A). ( c. The maximum angle Gmax through which the plate will rotate. Solution The plate rotates about a fixed axis
through A. The distance between A and
the mass center G is 11.18034 in., ¢ =
tan_1 5/10 = 26.5050, and the mass moment of inertia of the plate about A is 2 2
2 1 100 20 10
IA"IG+md'12[32.2][[12]+[12J] 2
O .
[A] = 3.59443 slug.ft2 32.2 12 The initial kinetic energy of the plate is zero
and the final kinetic energy of the plate is 2 2
T = L I m = %(3.59443)w A" a. As the plate rotates so that the
mass center is directly below A, the mass
center drops (moves in the direction of the
weight force) a distance of 11.18034  5 =
6.18034 in. Therefore, the work—energy ‘ + = T '
equation Ti U f gives
0 + lOO[—§Ll§9§ﬁ] = 1.797220.)2
12
w = 5.35 rad/s ......... . . . . . . . . . . . . . ... . . . . . . . . . . . . . .. Ans. (Problem 18—5 continues ...) ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 185  cont.) Then, the equations of motion for the plate 100 11.18034 2
+ = — = —— —— .
T 2F AD 100 [ 32.2 ][ 12 ](5 35322)
100 11.18034
9 = = —————— ——————————
+ 2F At [ 32.2 ][ 12 ]
+9 EMA = o = IAa
give
2
a = O rad/s An = 182.9180 lb
A = 182.9 lb . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 0
b. As the plate rotates 90 , the mass
center drops 10  5 = 5 in. and the work energy equation gives 5 2
+ — = . w
0 100[ 12 J 1 79722
w = 4.81 rad/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Now the equations of motion 100 11.18034 2
+ = — = ——— ————— .
2 BF AH 100 cos ¢ [ 32.2 ][ 12 ](4 81498)
. _ 100 11.18034
+2 2F — At + 100 Sln ¢ — [—EETE—J L————I§————)
5
+5 EMA — 12 (100) — 3.59443a
give
2
= 11.592 rad/s An = 156.525 1b At = —11.180 lb
= 156.9 lb . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. c. The maximum rotation of the plate
corresponds to a final kinetic energy of zero.
Therefore, the net work done by the weight
force must be zero  the weight must end up
at the same level as it started. Then 11.18034 sin(9 + ¢) = 11.18034 sin ¢
6 + ¢ = 1800  ¢
6 = 1800  2¢ = 126.870 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. '————————————————————_—_———
l ? ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges t———————————————————————————————————————————————————_—————________________ 1813 A 16—lb bowling ball is placed on a 28° inclined surface and
released from rest. Assume that the ball is a uniform sphere 8.6 in. in
diameter. If the static coefficient of friction between the ball and the
surface is 0.25, 3. Verify that the ball will begin to roll
without slipping. b. Determine the speed v and the angular
velocity w of the ball after it has rolled
20 ft down the incline. c. Compare the speed of part b with the speed
of a 16—lb particle that slides (without
friction) the same distance down the
incline. Solution a. From the equations of motion +5 2F = ma : N  16 cos 280 = 0 H II . ° _ 16 4.3
+F ZFt  mat. 16 Sin 28 F — [—§§T§—)[ 12 )a
4 3 2 16 4 3 2
c+ 2M6 = Ia“: F[ 12 J = 5 [ 32.2 ][ 12 ] a
Therefore
2
N = 14.12716 lb a = 30.13352 rad/s C
F = 2.14616 lb F/N = 0.15192 < us = 0.25 and the ball will roll without slipping. b. Neither the normal force N nor the friction force F do work. The
weight W = mg has a potential; the zero of gravitational potential energy
is set at the initial position. The kinetic energy is given by l 2 1 2 1 2 l 2 2 2 7 2
= ——— + ——— w = ——— + ——— ——— =
T 2 mVG 2 I6 2 va 2 [ 5 mr )(vg/r) lo va
and the workenergy equation Ti + vi + U = Tf + Vf gives
2 o
O + O + O = 7va/10 + mg(—20 sin 28 )
VG = 20.8 ft/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... Ans.
w = vG/r = 55.4 rad/s . . . . . ........... . . . . . . . . . . ... . . . . .. Ans.
c. VFor a particle sliding without friction, work—energy gives
2 o
O + O + O = mVP/2 + mg(20 sin 28 )
V = 24.6 ft/s .................... . . . . . . . . ............ Ans. P ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges ———_———_—_—_—_—_——— 18—21* Two 16—in. diameter disks and an 8—in. diameter cylinder are
fastened together to form a spool weighing 50 lb that has a radius of
gyration of 5 in. with respect to an axis through the mass center of the spool. The spool is initially at rest when
a constant force N of magnitude 50 lb is
applied to the spool through a cord wrapped
around the 8in. diameter cylinder. If the
spool rolls without slipping on the horizontal
surface, determine the velocity VG of the mass
center of the spool and the angular velocity w of the spool after it has rolled 10 ft to the right. Solution
Neither the normal force An, the frictional force Af, nor the weight
W do any work on the spool. As the
center of the spool moves 10 ft to
the right, the point where the 50 lb
force is applied moves 10(4/8) ft to the right and the 50 lb force does work on the spool 4
U = so 10 — = 250 lb°ft
[ [8]] Am The spool starts from rest (I: = O) and the kinetic energy at the final position is
l
= —~— +
Tf 2 G 2 G
l 50 8 2 + 1 50 5 2 wz
2 32.2 12 2 32.2 12 2
0.47986 0 lb’ft II and the work—energy equation Ti + U = Tf gives
2
0 + 250 = 0.47986 w
w = 22.8 rad/s ... . . . . . . . . . . .... . . . . . . . . . . . . . . . . . . . . . .. Ans.
V = (8/12)w = 15.22 ft/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. G ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 18—39* A l6—lb ladder 12 ft long slides in a
smooth corner as shown. The ladder is at rest
with 9 = 00 when the lower end is disturbed
slightly. Determine the angle 9, the angular
velocity w, and the velocity 38 of end B when end A loses contact with the vertical wall. Solution
Neither normal force A nor B /‘
does work. The weight W = mg has a a potential; the zero of gravitational
potential energy is at the level
of B. The moment of inertia of the rod is
1 16 2
Is ‘ 12 [32.2 ]‘12)
2
= 5.9627 slug°ft Using the relative velocity equation
to relate the velocities of A and 3
gives v j = VB 1 + 12w(cos 9 1 — sin 6 3) A
=—2m'6ft
VA 1 Sin /s E5
VB = 12w cos 9 ft/s
Then the velocity of the mass center G is G;
_'=~+_° = w 6"+w— 6"—'6"
VG VB VG/B 12 cos 1 6 ( cos 1 Sin J) 69
2 2 2 2 2
= + = 3 w
VG VGX VGy 6 (ft/s) 0‘
and the workenergy equation Ti + vi + U = Tf + V? gives f3 2 2
o + 16(6) + 0 %[%](36m ) + %(5.9627)w + 16(6 cos 9) 96 = 11.92550.)2 + 96 cos 9 2
w H 8.05(l  cos 9) (rad/s)2 Using the relative acceleration equation to
relate the accelerations of A and B gives A A A A 2 A A
= a 1 + 12a(cos 9 1 — sin 6 J) + 12w (sin 9 1 — cos 6 J) a
A J 8 (Problem 18—39 continues ...) ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges (Problem 1839  cont.) 2 2
a3 = 12a cos 9 — 12w sin 9 ft/s and then the acceleration of the mass center G is g a q 2 A
= = 9 — '
3G 38 + saw (120: cos 12w sm 6) 1
+ 6a(—cos 9 T  sin 6 j)
2 A A
+ 6w (sin 9 1  cos 6 J) 2 2
6a cos 6 — 60 sin 6 ft/s D)
II 2 2
6a sin 9 — 60 cos 9 ft/s m
I! Therefore the equations of motion are 16 2
_) Z = 3 = ———— _ '
+ I; maGX A [ 32.2 )(Gd cos 6 6w Sin 6)
16 2
+T 2 = : _ = ._.._____._. _ ' 6 _ w
Ey maGy B 16 [ 322 )( 6a Sin 6 cos 9)
C+ EMG = Igor: B(6 sin 6) — A(6 cos 6) = 5.96270: End A loses contact with the vertical wall when A = O at which time
2
a = w tan 9
= 8.05(1  cos 9) tan 9 (a) and
a = 1.06263 sin 9 2 .
1.0626(16  2.9814a sin 9  2.9814w cos 9) sin 9 2
17.002 sin 9 — 3.1680G sin 6  25.503(1  cos 9) sin 9 cos 9 (b)
Substituting Eq. a into Eq. b and solving for 9 gives
6 = 48.190 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
w = 1.638 rad/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
VB = 13.10 ft/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. (Solving with symbols instead of numbers would show that 9 = cos—1(2/3)
independent of W, C, or 9. Also, note that when 6 = 48.190 that end B is
still in contact with and sliding on the floor a = 3.00 rad/s2 B = 4.00 lb > O & L.D. ENGINEERING MECHANICS  Dynamics W.F. Riley 1847 The 5—lb uniform rod is 3 ft long.
rod move in frictionless guides and weigh 2 1b each.
the slider at A has a modulus of 5 = 15 lb/ft
and is stretched 2 ft when the rod is vertical. If the system is released from rest when the rod is vertical, determine
the angular velocity w of the rod and the velocity 68 of end B a. When the stretch in the spring is zero. b. When the bar is horizontal. Solution
Neither normal force A nor B does
work; the weights and the spring
force £5 all have potentials. The
zero of gravitational potential
energy is at the level of A. The moment of inertia of the rod is 1 5 2
Is ' 12 [ 32.2 ](3) 2
0.11646 slug'ft Using the relative velocity equation I <1 gives
VB J VA 1 3 (Sin 1 cos J) 00
VA = —3w sin 9 ft/s ﬁx
VB = —30 cos 9 ft/s to relate the velocities of A and B on Then the velocity of the mass center G is = —3w sin 9 i + 1.5w(sin 9 i  cos 6 3) <1 V=V+
G A 2 2 2 2 2
= + = . (1.)
VG VGX VGy 2 25 (ft/s) G/A (Problem 1847 continues Sturges The sliders at the ends of the A spring attached to ...) ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 18—47 — cont.) 0 Initially, 9 = O , VA = VB = w = 0, Ti = 0, and 2
vi 2(3) + 5(1.5) + o + —%— (15)(2) = 43.5 lb°ft The final kinetic and potential energy are l 5 2 l 2
Tf — 2 [ 32.2 )(2.250 ) + 2 (0.11646)w
1 2 2 _ 2 l 2 2 2
+ 2 [ 32.2 ](9w Sin 9) + 2 [ 32.2 ](9w cos 9) 2
0.51242w lb'ft Vf = 2(3 sin 9) + 5(1.5 sin 9) + 2(0) + % (15)(2  3 cos 9)2 13.5 sin 9 + 7.5(2  3 cos 9)2 lb°ft Then the workenergy equation Ti + Vi + U = Tf + V? gives 2 2
o + 43.5 + o = 0.51242w + 13.5 sin e + 7.5(2  3 cos 6) a. The stretch in the spring is zero when 2  3 cos 6 = 0 6 = 48.190
m = 8.08 rad/s D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
v3 = —3(8.08) cos 48.190 = —16.16 ft/s = 16.16 ft/s ¢ . . . . . . . . . . . . . . . . . . . . . . . .. Ans. b. The bar is horizontal when 9 = O , w = 8.38 rad/s D . . . . . . . . . . . . . . . . . . . . . . . . . . ..' . . . . . . . . . . .. Ans.
0
VB = 3(8.38) cos 0 = 25.1 ft/s = 25.1 ft/s ¢ . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ...
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