CHAPTER 12
PROBLEM 12.1
The value of g at any latitude f may be obtained from the formula
g = 9.78(1 + 0.0053 sin 2f ) m/s2
which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly
spherical. Determine to four significant figures (a) the weight in N, (b) the mass in kg, at the latitudes of 0°, 45°,
and 60°, of a silver bar, the mass of which has been officially designated as 5 kg.
Solution
g = 9.78(1 + 0.0053 sin 2f ) m/s2
(a)
(b)
Weight:
f = 0∞ :
g = 9.78 m/s2
f = 45∞ :
g = 9.8059 m/s2
f = 60∞ :
g = 9.8189 m/s2
W = mg
f = 0∞ :
W = (5 kg)(9.78 m/s2 ) = 48.9 N b
f = 45∞ :
W = (5 kg)(9.8059 m/s2 ) = 49.0 N
b
f = 60∞ :
W = (5 kg)(9.8189 m/s2 ) = 49.1 N b
m = 5.000 kg b
Mass: At all latitudes:
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3
PROBLEM 12.2
The acceleration due to gravity on the moon is 1.62 m/s2 . Determine (a) the weight in newtons, (b) the mass in
kilograms, on the moon, of a gold bar, the mass of which has been officially designated as 2 kg.
Solution
(a)
Weight:
(b)
Mass: Same as on earth:
W = mg = (2 kg)(1.62 m/s2 )
W = 3.24 N b
m = 2.00 kg b
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4
PROBLEM 12.3
A 200-kg satellite is in a circular orbit 1500 km above the surface of Venus. The acceleration due to the
gravitational attraction of Venus at this altitude is 5.52 m/s2 . Determine the magnitude of the linear momentum
of the satellite knowing that its orbital speed is 23.4 ¥ 103 km/h.
Solution
First note
v = 23.4 ¥ 103 km/h = 6500 m/s
Now
L = mv = 200 kg ¥ 6500 m/s
L = 1.300 ¥ 106 kg ◊ m/s b
or
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5
PROBLEM 12.4
A spring scale A and a lever scale B having equal lever arms
are fastened to the roof of an elevator, and identical packages
are attached to the scales as shown. Knowing that when the
elevator moves downward with an acceleration of 1.2 m/s2 the
spring scale indicates a load of 70 N, determine (a) the weight
of the packages, (b) the load indicated by the spring scale and
the mass needed to balance the lever scale when the elevator
moves upward with an acceleration of 1.2 m/s2.
Solution
Assume g = 9.81 m/s2
m=
(a)
+
SF = ma : W - Fs =
W
g
W
a
g
Ê
aˆ
W Á1 - ˜ = Fs
Ë
g¯
W=
or
Fs
a
1g
=
70
1.2
19.81
W = 79.8 N b
m=
(b)
SF = ma : Fs - W =
W 79.756 N
=
= 8.1300 kg
g 9.81 m/s2
W
a
g
Ê
aˆ
Fs = W Á1 + ˜
Ë
g¯
1.2 ˆ
Ê
= 79.756 Á1 +
Ë 9.81˜¯
Fs = 89.5 N b
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6
Problem 12.4 (continued)
For the balance system B,
SM 0 = 0 : bFw - bFp = 0
Fw = Fp
But
Ê
Fw = Ww Á1 +
Ë
and
Ê
aˆ
Fp = W p Á 1 + ˜
Ë
g¯
so that
Ww = W p
and
mw =
Wp
g
=
aˆ
g ˜¯
79.756
9.81
mw = 8.13 kg b
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7
PROBLEM 12.5
A hockey player hits a puck so that it comes to rest in 9 s after sliding 30 m on the ice. Determine (a) the initial
velocity of the puck, (b) the coefficient of friction between the puck and the ice.
Solution
(a)
Assume uniformly decelerated motion.
Then
v = v0 + at
At t = 9 s:
0 = v0 + a(9)
or
a=-
v 2 = v02 + 2a( x - 0)
Also
At t = 9 s:
0 = v02 + 2a(30)
Substituting for a
Ê v ˆ
0 = v02 + 2 Á - 0 ˜ (30) = 0
Ë 9¯
v0 = 6.6667 m/s
or
a=-
and
(b)
v0
9
or v0 = 6.67 m/s b
6.6667
= -0.74074 m/s2
9
We have
+
or
N = W = mg
F = m k N = m k mg
Sliding:
+
or
SFy = 0 : N - W = 0
SFx = ma : -F = ma
mk = -
or - m k mg = ma
a
-0.74074 m/s2
=g
9.81 m/s2
or m k = 0.0755 b
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8
PROBLEM 12.6
Determine the maximum theoretical speed that an automobile starting from rest can reach after traveling
400 m. Assume that the coefficient of static friction is 0.80 between the tires and the pavement and that (a) the
automobile has front-wheel drive and the front wheels support 62 percent of the automobile’s weight, (b) the
automobile has rear-wheel drive and the rear wheels support 43 percent of the automobile’s weight.
Solution
(a)
For maximum acceleration
FF = Fmax = m s N F = 0.8(0.62 W )
= 0.496 W = 0.496 mg
Now
+
SFx = ma: FF = ma
0.496 mg = ma
or
a = 0.496(9.81 m/s2 ) = 4.86576 m/s2
Then
Since a is constant, we have
v 2 = 0 + 2 a( x - 0 )
When
2
x = 400 m: vmax
= 2( 4.86576 m/s2 )( 400 m)
vmax = 62.391 m/s
or
vmax = 225 km/h b
or
(b)
For maximum acceleration
FR = Fmax = m s N R = 0.8(0.43 W )
= 0.344 W = 0.344 mg
Now
or
Then
+
SFx = ma: FR = ma
0.344 mg = ma
a = 0.344(9.81 m/s2 ) = 3.37464 m/s2
Since a is constant, we have
v 2 = 0 + 2 a( x - 0 )
When
or
2
x = 400 m: vmax
= 2(3.37464 m/s2 )( 400 m)
vmax = 51.959 m/s
vmax = 187.1 km/h b
or
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9
PROBLEM 12.7
In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 0.9 m/s2 while still on a level
section of the highway. Knowing that the speed of the bus is 100 km/h as it begins to climb the grade and that
the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus
up the grade when its speed has decreased to 80 km/h.
Solution
First consider when the bus is on the level section of the highway.
alevel = 0.9 m/s2
+
We have
SFx = ma: P =
W
alevel
g
Now consider when the bus is on the upgrade.
We have
+
SFx = ma: P - W sin 7∞ =
Substituting for P
W
a¢
g
W
W
alevel - W sin 7∞ = a ¢
g
g
a ¢ = alevel - g sin 7∞
or
= (0.9 - 9.81sin 7∞) m/s2
= - 0.29554 m/s2
For the uniformly decelerated motion
v 2 = ( v0 )2upgrade + 2a ¢( xupgrade - 0)
Noting that 100 km/h =
250
Ê 5 ˆ
m/s, then when v = 80 km/h Á =
v , we have
Ë 0.8 0 ˜¯
9
2
2
250
ˆ
Ê 250
Ê
ˆ
m/s˜ = Á
m/s˜ + 2( -0.29554 m/s2 ) xupgrade
ÁË 0.8 ¥
¯
Ë 9
¯
9
or
xupgrade = 469.949 m
xupgrade = 0.470 km b
or
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10
PROBLEM 12.8
If an automobile’s braking distance from 100 km/h is 50 m on level pavement, determine the automobile’s
braking distance from 100 km/h when it is (a) going up a 5° incline, (b) going down a 3-percent incline.
Assume the braking force is independent of grade.
Solution
Assume uniformly decelerated motion in all cases.
For braking on the level surface,
v0 = 100 km/h =
250
m/s, v f = 0
9
x f - x0 = 50 m
v 2f = v02 + 2a( x f - x0 )
a=
v 2f - v02
2( x f - x0 )
Ê 250 ˆ
0-Á
Ë 9 ˜¯
=
2(50)
2
= -7.7160 m/s2
Braking force.
Fb = ma
W
a
g
7.7160
=W
9.81
= -0.78654 W
=
(a)
Going up a 5° incline.
SF = ma
W
a
g
F + W sin 5∞
g
a=- b
W
= -(0.78654 + sin 5∞)(9.81)
- Fb - W sin 5∞ =
= -8.56211 m/s2
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11
Problem 12.8 (continued)
x f - x0 =
v 2f - v02
2a
2
Ê 250 ˆ
0-Á
Ë 9 ˜¯
=
(2)( -8.5621)
(b)
x f - x0 = 45.1 m b
Going down a 3 percent incline.
3
b = 1.71835∞
100
W
- Fb + W sin b = a
g
a = -(0.78654 - sin b )(9.81)
= -7.4130 m/s
tan b =
2
Ê 250 ˆ
0-Á
Ë 9 ˜¯
x f - x0 =
2( -7.4130)
x f - x0 = 52.0 m b
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12
PROBLEM 12.9
A 20-kg package is at rest on an incline when a force P is applied to it.
Determine the magnitude of P if 10 s is required for the package to travel
5 m up the incline. The static and kinetic coefficients of friction between
the package and the incline are both equal to 0.3.
Solution
Kinematics: Uniformly accelerated motion. ( x0 = 0, v0 = 0)
1
x = x0 + v0 t + at 2 ,
2
a=
+
+
or
2x
t
2
=
(2)(5)
(10)2
= 0.100 m/s2
SFy = 0 : N - P sin 50∞ - mg cos 20∞ = 0
N = P sin 50∞ + mg cos 20∞
+
or
SFx = ma : P cos 50∞ - mg sin 20∞ - m N = ma
P cos 50∞ - mg sin 20∞ - m ( P sin 50∞ + mg cos 20∞) = ma
P=
ma + mg (sin 20∞ + m cos 20∞)
cos 50∞ - m sin 50∞
For motion impending, set a = 0 and m = m s = 0.4
(20)(0) + (20)(9.81)(sin 20∞ + 0.4 cos 20∞)
cos 50∞ - 0.4 sin 50∞
= 419 N
P=
b
For motion with a = 0.100 m/s2 , use m = m k = 0.3.
P=
(20)(0.100) + (20)(9.81)(sin 20∞ + 0.3 cos 20∞)
cos 50∞ - 0.3 sin 50∞
P = 301 N b
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13
PROBLEM 12.10
The acceleration of a package sliding at Point A is 3 m/s2.
Assuming that the coefficient of kinetic friction is the same
for each section, determine the acceleration of the package at
Point B.
Solution
For any angle q.
Use x and y coordinates as shown.
ay = 0
+
SFy = ma y : N - mg cos q = 0
N = mg cos q
+
SFx = max : mg sin q - m k N = max
ax = g (sin q - m k cos q )
At Point A.
q = 30∞, ax = m/s2
g sin 30∞ - ax
g cos 30∞
9.81 sin 30∞ - 3
=
9.81 cos 30∞
= 0.22423
mk =
At Point B.
q = 15∞, m k = 0.22423
ax = 9.81(sin 15∞ - 0.22423 cos 15∞)
a = 0.414 m/s2
= 0.414 m/s
15∞ b
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14
PROBLEM 12.11
The two blocks shown are originally at rest. Neglecting the masses of
the pulleys and the effect of friction in the pulleys and between block A
and the horizontal surface, determine (a) the acceleration of each block,
(b) the tension in the cable.
Solution
From the diagram
x A + 3 y B = constant
Then
v A + 3v B = 0
and
aA + 3aB = 0
aA = -3aB or
(a)
+
A:
SFx = mA a A :
(1)
-T = mA aB
T = 3mA aB
Using Eq. (1)
B:
+
SFy = mB aB : WB - 3T = mB aB
Substituting for T
mB g - 3(3mA aB ) = mB aB
A:
aB =
or
B:
(b)
g
1+ 9
mA
mB
=
9.81 m/s2
= 0.83136 m/s2
30 kg
1+ 9
25 kg
Then
a A = 2.49 m/s2 Æ b
and
a B = 0.831 m/s2Ø b
We have
T = 3 ¥ 30 kg ¥ 0.83136 m/s2
T = 74.8 N b
or
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15
PROBLEM 12.12
The two blocks shown are originally at rest. Neglecting the masses
of the pulleys and the effect of friction in the pulleys and assuming that
the coefficients of friction between block A and the horizontal surface
are m s = 0.25 and m k = 0.20, determine (a) the acceleration of each
block, (b) the tension in the cable.
Solution
From the diagram
x A + 3 y B = constant
Then
v A + 3v B = 0
and
aA + 3aB = 0
aA = -3aB or
(1)
First determine if the blocks will move with aA = aB = 0. We have
+
1
SFy = 0 : WB - 3T = 0 or T = mB g
3
B:
+
SFx = 0:
FA =
Then
+
A:
A:
Also,
FA - T = 0
1
¥ 25 kg ¥ 9.81 m/s2 = 81.75 N
3
SFy = 0 : WA - N A = 0 or
N A = mA g
( FA ) max = ( m s ) A N A = ( m s ) A mA g
B:
= 0.25 ¥ 30 kg ¥ 9.81 m/s2
= 73.575 N
FA ( FA ) max, which implies that the blocks will move.
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16
Problem 12.12 (continued)
(a)
A:
+
+
or
SFx = mA a A : FA - T = mA a A
T = 0.20 mA g + 3mA aB
Using Eq. (1)
or
N A = mA g
FA = ( m k ) A N A = 0.20 mA g
Sliding:
B:
SFy = 0 : WA - N A = 0 or
+
SFy = mB aB : WB - 3T = mB aB
mB g - 3(0.20 mA g + 3mA aB ) = mB aB
Ê
m ˆ
g Á1 - 0.6 A ˜
mB ¯
Ë
aB =
m
1+ 9 A
mB
=
Ê
30 kg ˆ
(9.81 m/s2 ) Á1 - 0.6
Ë
25 kg ˜¯
30 kg
1 + 9 25 kg
= 0.23278 m/s2
a A = 0.698 m/s2 Æ b
Then
a B = 0.233 m/s2 Ø b
and
(b)
We have
T = (30 kg)(0.20 ¥ 9.81 + 3 ¥ 0.23278) m/s2
T = 79.8 N b
or
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17
PROBLEM 12.13
The coefficients of friction between the load and the flat-bed
trailer shown are m s = 0.40 and m k = 0.30. Knowing that the
speed of the rig is 70 km/h, determine the shortest distance in
which the rig can be brought to a stop if the load is not to shift.
Solution
Load: We assume that sliding of load relative to trailer is impending:
F = Fm
= ms N
Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration a max .
+
SFy = 0 : N - W = 0 N = W
Fm = m s N = 0.40 W
+
SFx = ma : Fm = mamax
0.40 W =
W
amax
g
amax = 3.92 m/s2
amax = 0.4 ¥ 9.81 m/s2 = 3.924 m/s2 a max = 3.92 m/s2 Æ
Uniformly accelerated motion.
v 2 = v02 + 2ax with v = 0
v0 = 70 km/h =
175
m/s
9
a = - amax = -3.924 m/s2
2
Ê 175 ˆ
+ 2( -3.924) x 0=Á
Ë 9 ˜¯
x = 48.2 m b
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18
PROBLEM 12.14
A tractor-trailer is traveling at 90 km/h when the driver applies
his brakes. Knowing that the braking forces of the tractor and
the trailer are 18 kN and 70 kN, respectively, determine (a) the
distance traveled by the tractor-trailer before it comes to a stop,
(b) the horizontal component of the force in the hitch between
the tractor and the trailer while they are slowing down.
Solution
(a)
+
SFx = ma: - ( Fbr )trac - ( Fbr ) trl =
Wtotal
a
g
a=-
or
9.81 m/s2
(18, 000 + 70, 000) N
(87, 000 + 75, 000) N
= -5.3289 m/s2
For uniformly decelerated motion
v 2 = v02 + 2a( x - x0 ) v0 = 90 km/h = 25 m/s
When v = 0:
0 = (25 m/s)2 + 2( -5.3289 m/s2 )( Dx )
Dx = 58.6 m b
or
(b)
+
SFx = mtrl a : - ( Fbr ) trl + Phitch =
Then
Wtrl
a
g
Phitch = 70, 000 N +
87, 000 N
¥ ( -5.3289 m/s2 )
2
9.81 m/s
= 22, 740 N
Phitch = 22, 700 N (tension) b
or
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19
PROBLEM 12.15
Block A has a mass of 40 kg, and block B has a mass of 8 kg.
The coefficients of friction between all surfaces of contact are
m s = 0.20 and m k = 0.15. If P = 0, determine (a) the acceleration
of block B, (b) the tension in the cord.
Solution
From the diagram
2x A + x B /A = constant
Then
2 v A + v B /A = 0
and
2 a A + a B /A = 0
Now
a B = a A + a B /A
Then
aB = aA + ( -2aA )
or
aB = - aA
(1)
First we determine if the blocks will move for the given value of q. Thus, we seek the value of q for which the
blocks are in impending motion, with the impending motion of A down the incline.
B:
SFy = 0 : N AB - WB cosq = 0
+
or
N AB = mB g cosq
Now
FAB = m s N AB
B:
= 0.2mB g cos q
+
SFx = 0 : - T + FAB + WB sinq = 0
T = mB g (0.2 cos q + sin q )
or
A:
+
SFy = 0 : N A - N AB - W A cosq = 0
or
N A = ( mA + mB ) g cosq
Now
FA = m s N A
A:
= 0.2( mA + mB ) g cos q
+
SFx = 0 : - T - FA - FAB + W A sinq = 0
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20
Problem 12.15 (continued)
T = mA g sin q - 0.2( mA + mB ) g cos q - 0.2mB g cos q
or
= g[mA sin q - 0.2( mA + 2mB ) cos q ]
Equating the two expressions for T
mB g (0.2 cos q + sin q ) = g[mA sin q - 0.2( mA + 2mB ) cos q ]
8(0.2 + tan q ) = [40 tan q - 0.2( 40 + 2 ¥ 8)]
or
tan q = 0.4
or
or q = 21.8∞. For impending motion. Since q 25∞, the blocks will move. Now consider the motion of the
blocks.
(a)
+
SFy = 0 : N AB - WB cos 25∞ = 0
or
N AB = mB g cos 25∞
Sliding:
FAB = m k N AB = 0.15mB g cos 25∞
+
B:
SFx = mB aB : - T + FAB + WB sin 25∞ = mB aB
T = mB [ g (0.15 cos 25∞ + sin 25∞) - aB ]
or
A:
= 8[9.81(0.15 cos 25∞ + sin 25∞) - aB ]
= 8(5.47952 - aB )
+
( N)
SFy = 0 : N A - N AB - W A cos 25∞ = 0
N A = ( mA + mB ) g cos 25∞
or
FA = m k N A = 0.15( mA + mB ) g cos 25∞
Sliding:
+
SFx = mA a A : - T - FA - FAB + W A sin 25∞ = mA a A
Substituting and using Eq. (1)
T = mA g sin 25∞ - 0.15( mA + mB ) g cos 25∞
- 0.15mB g cos 25∞ - mA ( - aB )
= g[mA sin 25∞ - 0.15( mA + 2mB ) cos 25∞] + mA aB
= 9.81[40 sin 25∞ - 0.15( 40 + 2 ¥ 8) cos 25∞] + 40aB
= 91.15202 + 40aB
( N)
Equating the two expressions for T
8(5.47952 - aB ) = 91.15202 + 40aB
or
aB = -0.98575 m/s2
a B = 0.986 m/s2
(b)
We have
25∞ b
T = 8 [5.47952 - ( -0.98575)]
T = 51.7 N b
or
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21
PROBLEM 12.16
Block A has a mass of 40 kg, and block B has a mass of 8 kg.
The coefficients of friction between all surfaces of contact are
m s = 0.20 and m k = 0.15. If P = 40 N Æ, determine (a) the
acceleration of block B, (b) the tension in the cord.
Solution
From the diagram
2x A + x B /A = constant
Then
2 v A + v B /A = 0
and
2 a A + a B /A = 0
Now
a B = a A + a B /A
Then
aB = aA + ( -2aA )
or
aB = - aA
(1)
First we determine if the blocks will move for the given value of P. Thus, we seek the value of P for which the
blocks are in impending motion, with the impending motion of a down the incline.
B:
+
SFy = 0 : N AB - WB cos 25∞ = 0
or
N AB = mB g cos 25∞
Now
FAB = m s N AB
B:
= 0.2 mB g cos 25∞
+
SFx = 0 : - T + FAB + WB sin 25∞ = 0
A:
T = 0.2 mB g cos 25∞ + mB g sin 25∞
or
= (8 kg)(9.81 m/s2 )(0.2 cos 25∞ + sin 25∞)
= 47.39249 N
A:
or
+
SFy = 0 : N A - N AB - W A cos 25∞ + P sin 25∞ = 0
N A = ( mA + mB ) g cos 25∞ - P sin 25∞
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22
Problem 12.16 (continued)
Now
FA = m s N A
or
FA = 0.2[( mA + mB ) g cos 25∞ - P sin 25∞]
+
SFx = 0 : - T - FA - FAB + W A sin 25∞ + P cos 25∞ = 0
-T - 0.2[( mA + mB ) g cos 25∞ - P sin 25∞] - 0.2mB g cos 25∞ + mA g sin 25∞ + P cos 25∞ = 0
or
or
P (0.2 sin 25∞ + cos 25∞) = T + 0.2[( mA + 2mB ) g cos 25∞] - mA g sin 25∞
Then
P(0.2 sin 25∞ + cos 25∞) = 47.39249 N + 9.81 m/s2 {0.2[( 40 + 2 ¥ 8) cos 25∞ - 40 sin 25∞] kg}
P = -19.04 N for impending motion.
or
Since P 40 N, the blocks will move. Now consider the motion of the blocks.
(a)
+
SFy = 0 : N AB - WB cos 25∞ = 0
or
N AB = mB g cos 25∞
Sliding:
FAB = m k N AB
B:
= 0.15 mB g cos 25∞
+
SFx = mB aB : - T + FAB + WB sin 25∞ = mB aB
T = mB [ g (0.15 cos 25∞ + sin 25∞) - aB ]
or
= 8[9.81(0.15 cos 25∞ + sin 25∞) - aB ]
= 8(5.47952 - aB )
( N)
A:
+
SFy = 0 : N A - N AB - W A cos 25∞ + P sin 25∞ = 0
or
N A = ( mA + mB ) g cos 25∞ - P sin 25∞
Sliding:
FA = m k N A
= 0.15[( mA + mB ) g cos 25∞ - P sin 25∞]
+
SFx = mA a A : - T - FA - FAB + W A sin 25∞ + P cos 25∞ = mA aA
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23
Problem 12.16 (continued)
Substituting and using Eq. (1)
T = mA g sin 25∞ - 0.15[( mA + mB ) g cos 25∞ - P sin 25∞]
- 0.15 mB g cos 25∞ + P cos 25∞ - mA ( - aB )
= g[mA sin 25∞ - 0.15( mA + 2mB ) cos 25∞]
+ P (0.15 sin 25∞ + cos 25∞) + mA aB
= 9.81[40 sin 25∞ - 0.15( 40 + 2 ¥ 8) cos 25∞]
+ 40(0.15 sin 25∞ + cos 25∞) + 40aB
= 129.94004 aB
( N)
Equating the two expressions for T
8(5.47952 - aB ) = 129.94004 + 40aB
or
aB = -1.79383 m/s2
a B = 1.794 m/s2
(b)
We have
25° b
T = 8 [5.47952 - ( -1.79383)]
T = 58.2 N b
or
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24
PROBLEM 12.17
Boxes A and B are at rest on a conveyor belt that is initially at
rest. The belt is suddenly started in an upward direction so that
slipping occurs between the belt and the boxes. Knowing that
the coefficients of kinetic friction between the belt and the
boxes are ( m k ) A = 0.30 and ( m k ) B = 0.32, determine the initial
acceleration of each box.
Solution
Assume that a B a A so that the normal force NAB between the boxes is zero.
+
A:
SFy = 0 : NA - WA cos 15∞ = 0
or
NA = WA cos 15∞
Slipping:
FA = ( m k ) A NA
A:
= 0.3WA cos 15∞
+
or
SFx = mA a A : FA - WA sin 15∞ = mA a A
0.3WA cos 15∞ - WA sin 15∞ =
WA
aA
g
aA = (9.81 m/s2 )(0.3 cos 15∞ - sin 15∞)
or
= 0.3037 m/s2
B:
+
SFy = 0 : N B - WB cos 15∞ = 0
or
N B = WB cos 15∞
Slipping:
FB = ( m k ) B N B
B:
= 0.32WB cos 15∞
+
or
or
SFx = mB aB : FB - WB sin 15∞ = mB aB
0.32WB cos 15∞ - WB sin 15∞ =
WB
aB
g
aB = (9.81 m/s2 )(0.32 cos 15∞ - sin 15∞) = 0.49322 m/s2
aB aA fi assumption is correct
a A = 0.304 m/s2
15° b
a B = 0.493 m/s2
15° b
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25
Problem 12.17 (continued)
Note: If it is assumed that the boxes remain in contact ( NAB π 0), then
a A = aB
and find ( SFx = ma)
A:
0.3WA cos 15∞ - WA sin 15∞ - N AB =
WA
a
g
B:
0.32WB cos 15∞ - WB sin 15∞ + N AB =
WB
a
g
Solving yields a = 0.3878 m/s2 and NAB = - 4.2168 N, which contradicts the assumption.
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26
PROBLEM 12.18
Knowing that the system shown starts from rest,
find the velocity at t = 1.2 s of (a) collar A, (b)
collar B. Neglect the masses of the pulleys and
the effect of friction.
Solution
Referring to solution of Problem 11.51 we note that
1
aB = - aA
2
where minus sign indicates that a A and a B have opposite sense.
(1)
Block B.
+
SF = ma : 25 - 2T = 7.5aA +
SF = ma : 2T - T = 10 aA
(2)
Block A.
T = 10 aA
(a)
Substituting for T from (3) into (2):
25 - 2(10 aA ) = 7.5aA
25 = 27.5aA
v A = ( v A )0 + a At = 0 + 0.909(1.2), (b)
(3)
a A = 0.909 m/s2 Æ
v A = 1.091 m/s Æ b
a B = 0.455 m/s2 ¨
Substituting a = 0.909 into Eq. (1):
v B = ( v B )0 + aB t = 0 + 0.455(1.2) v B = 0.545 m/s ¨ b
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27
PROBLEM 12.19
Each of the systems shown is initially at rest.
Neglecting axle friction and the masses of the
pulleys, determine for each system (a) the
acceleration of block A, (b) the velocity of block A
after it has moved through 3 m, (c) the time required
for block A to reach a velocity of 6 m/s.
Solution
Let y be positive downward for both blocks.
Constraint of cable: y A + y B = constant
a A + aB = 0
or
aB = - a A
WA
aA
g
or
T = WA -
For blocks A and B, + SF = ma :
Block A:
WA - T =
Block B:
P + WB - T =
WB
W
aB = - B a A
g
g
P + WB - W A +
WA
W
aA = - B aA
g
g
Solving for aA,
aA =
WA
aA
g
W A - WB - P
g
W A + WB
(1)
v 2A - ( v A )20 = 2a A [ y A - ( y A )0 ] with ( v A )0 = 0
v A = 2aA [ y A - ( y A )0 ]
v A - ( v A ) 0 = a At
t=
(2)
with (v A )0 = 0
vA
aA
(3)
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28
Problem 12.19 (continued)
(a)
Acceleration of block A.
System (1):
By formula (1),
System (2):
By formula (1),
System (3):
By formula (1),
(b)
(c)
W A = 100 g, WB = 50 g, P = 0
( aA )1 =
100 - 50
(9.81)
100 + 50
(a A )1 = 3.27 m/s2Ø b
W A = 100 g, WB = 0, P = 50 g N
( a A )2 =
100 - 50
(9.81) = 4.905 m/s2
100
(a A )2 = 4.91 m/s2Ø b
W A = 1100 g, WB = 1050 g, P = 0
( a A )3 =
1100 - 1050
(9.81)
1100 + 1050
(a A )3 = 0.228 m/s2 Ø b
v A at y A - ( y A )0 = 3 m. Use formula (2).
System (1):
( v A )1 = (2)(3.27)(3) ( v A )1 = 4.43 m/s Ø b
System (2):
( v A )2 = (2)( 4.905)(3) (v A )2 = 5.42 m/s Ø b
System (3):
( v A )3 = (2)(0.228)(3) ( v A )3 = 1.170 m/s Ø b
Time at v A = 6 m/s. Use formula (3).
System (1):
t1 =
6
3.27
t1 = 1.835 s b
System (2):
t2 =
6
4.905
t2 = 1.223 s b
System (3):
t3 =
6
0.228
t3 = 26.3 s b
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29
PROBLEM 12.20
A man standing in an elevator that is moving with a constant
acceleration holds a 3-kg block B between two other blocks in such a
way that the motion of B relative to A and C is impending. Knowing
that the coefficients of friction between all surfaces are m s = 0.30
and m k = 0.25, determine (a) the acceleration of the elevator if it is
moving upward and each of the forces exerted by the man on blocks
A and C has a horizontal component equal to twice the weight of
B, (b) the horizontal components of the forces exerted by the man
on blocks A and C if the acceleration of the elevator is 2.0 m/s2
downward.
Solution
First we observe that because B is not moving relative to A and to C that a B = a EL .
(a)
F = ms N
We have
= 0.30(2WB )
= 0.6WB = 0.6mB g
For a EL to be ≠, the net vertical force must be ≠, which requires that the frictional forces be acting as
shown. It then follows that the impending motion of B relative to A and C is downward. Then
+
SFy = mB aEL : 2 F - WB = mB aEL
2(0.6mB g ) - mB g = mB aEL
or
aEL = 0.2 ¥ 9.81 m/s2
or
a EL = 1.962 m/s2 ≠ b
or
(b)
F = ms N
We have
= 0.30 N
Now we observe that because the direction of the impending motion is unknown, the directions of the
frictional forces is also unknown (although Fnet must be downward).
+
SFy = mB aEL : ± 2 F - WB = - mB |aEL |
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30
Problem 12.20 (continued)
or
±2 F = mB ( g - |aEL |)
= 3 kg ¥ (9.81 - 2) m/s2
Since the magnitude of F must be positive, it then follows that F ≠, and that the impending motion of B
relative to A and C is downward. Finally
2(0.30 N ) = 3 kg ¥ (9.81 - 2) m/s2
N AB = N BC = 39.1 N b
or
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31
PROBLEM 12.21
A package is at rest on a conveyor belt, which is initially at rest. The belt
is started and moves to the right for 1.3 s with a constant acceleration of
2 m/s2. The belt then moves with a constant deceleration a 2 and comes to
a stop after a total displacement of 2.2 m. Knowing that the coefficients
of friction between the package and the belt are m s = 0.35 and m k = 0.25,
determine (a) the deceleration a 2 of the belt, (b) the displacement of the
package relative to the belt as the belt comes to a stop.
Solution
(a)
Kinematics of the belt. vo = 0
1.
Acceleration phase: with a1 = 2 m/s2Æ
v1 = vo + a1t1 = 0 + (2)(1.3) = 2.6 m/s
1
1
x1 = xo + vot1 + a1t12 = 0 + 0 + (2)(1.3)2 = 1.69 m
2
2
2.
Deceleration phase: v2 = 0 since the belt stops.
v22 - v12 = 2a2 ( x2 - x1 )
a2 =
t2 - t1 =
(b)
v22 - v12
0 - (2.6)2
=
= -6.63
2( x2 - x1 ) 2(2.2 - 1.69)
a 2 = 6.63 m/s2 ¨ b
v2 - v1 0 - 2.6
= 0.3923 s
=
a2
-6.63
Motion of the package
1.
Acceleration phase. Assume no slip. ( a p )1 = 2 m/s2Æ
SFy = 0 : N - W = 0 or N = W = mg
+
SFx = ma : F f = m( a p )1
The required friction force is Ff .
The available friction force is m s N = 0.35W = 0.35 mg
Ff
m
= ( a p )1 ms N
= m s g = (0.35)(9.81) = 3.43 m/s2
m
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32
Problem 12.21 (continued)
Since 2.0 m/s2 3.43 m/s2, the package does not slip.
( v p )1 = v1 = 2.6 m/s and (x p )1 = 1.69 m.
2.
Deceleration phase. Assume no slip. ( a p )2 = -11.52 m/s2
+
SFx = ma : - F f = m( a p )2
Ff
m
= ( a p )2 = -6.63 m/s2
m s N m s mg
=
= m s g = 3.43 m/s2 6.63 m/s2
m
m
Since the available friction force m s N is less than the required friction force Ff for no slip, the
package does slip.
( a p )2 6.63 m/s2 ,
+
F f = mk N
SFx = m( a p )2 : - m k N = m( a p )2
mk N
= - mk g
m
= -(0.25)(9.81)
( a p )2 = -
= -2.4525 m/s2
( v p )2 = ( v p )1 + ( a p )2 (t2 - t1 )
= 2.6 + ( -2.4525)(0.3923)
= 1.638 m/s2
1
( x p )2 = ( x p )1 + ( v p )1 (t2 - t1 )2 + ( a p )2 (t2 - t1 )2
2
1
= 1.69 + (2.6)(0.3923) + ( -2.4525)(0.3923)2
2
= 2.521 m
Position of package relative to the belt
( x p )2 - x2 = 2.521 - 2.2 = 0.321
x p/belt = 0.321 m Æb
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33
PROBLEM 12.22
To transport a series of bundles of shingles A
to a roof, a contractor uses a motor-driven lift
consisting of a horizontal platform BC which
rides on rails attached to the sides of a ladder.
The lift starts from rest and initially moves with
a constant acceleration a1 as shown. The lift then
decelerates at a constant rate a2 and comes to rest
at D, near the top of the ladder. Knowing that the
coefficient of static friction between a bundle
of shingles and the horizontal platform is 0.30,
determine the largest allowable acceleration a1
and the largest allowable deceleration a2 if the
bundle is not to slide on the platform.
Solution
Acceleration a1 : Impending slip. F1 = m s N1 = 0.30 N1
SFy = mA a y :
N1 - WA = mA a1 sin 65∞
N1 = W A + mA a1 sin 65∞
= mA ( g + a1 sin 65∞)
+
SFx = mA ax : F1 = mA a1 cos 65∞
F1 = m s N
mA a1 cos 65∞ = 0.30mA ( g + a1 sin 65∞)
or
0.30 g
cos 65∞ - 0.30 sin 65∞
= (1.990)(9.81)
a1 =
= 19.53 m/s2
a1 = 19.53 m/s2
65∞ b
Deceleration a 2 : Impending slip. F2 = m s N 2 = 0.30 N 2
SFy = ma y : N1 - WA = - mA a2 sin 65∞
N1 = W A - mA a2 sin 65∞
+
SFx = max :
F2 = mA a2 cos 65∞
F2 = m s N 2
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34
Problem 12.22 (continued)
or
mA a2 cos 65∞ = 0.30 mA ( g - a2 cos 65∞)
0.30 g
cos 65∞ + 0.30 sin 65∞
= (0.432)(9.81)
a2 =
= 4.24 m/s2
a 2 = 4.24 m/s2
65∞ b
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35
PROBLEM 12.23
To unload a bound stack of plywood from a truck, the driver first tilts
the bed of the truck and then accelerates from rest. Knowing that the
coefficients of friction between the bottom sheet of plywood and the bed
are m s = 0.40 and m k = 0.30, determine (a) the smallest acceleration of the
truck which will cause the stack of plywood to slide, (b) the acceleration
of the truck which causes corner A of the stack to reach the end of the
bed in 0.9 s.
Solution
Let a P be the acceleration of the plywood, aT be the acceleration of the truck, and a P /T be the acceleration of
the plywood relative to the truck.
(a)
Find the value of aT so that the relative motion of the plywood with respect to the truck is impending.
aP = aT and F1 = m s N1 = 0.40 N1
+
SFy = mP a y : N1 - WP cos 20∞ = - mP aT sin 20∞
N1 = mP ( g cos 20∞ - aT sin 20∞)
+
SFx = max : F1 - WP sin 20∞ = mP aT cos 20∞
F1 = mP ( g sin 20∞ + aT cos 20∞)
mP ( g sin 20∞ + aT cos 20∞) = 0.40 mP ( g cos 20∞ - aT sin 20∞)
(0.40 cos 20∞ - sin 20∞)
g
cos 20∞ + 0.40 sin 20∞
= (0.03145)(9.81)
aT =
= 0..309
aT = 0.309 m/s2 Æ b
(b)
1
1
xP /T = ( xP /T )o + ( vP /T )t + aP /T t 2 = 0 + 0 + aP /T t 2
2
2
aP /T =
2 xP /T
t
2
=
(2)(2)
= 4.94 m/s2
2
(0.9)
a P /T = 4.94 m/s2
20∞
a P = aT + a P /T = ( aT Æ) + ( 4.94 m/s2
+
20∞)
Fy = mP a y : N 2 - WP cos 20∞ = - mP aT sin 20∞
N 2 = mP ( g cos 20∞ - aT sin 20∞)
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36
Problem 12.23 (continued)
+
SFx = Smax : F2 - WP sin 20∞ = mP aT cos 20∞ - mP aP /T
F2 = mP ( g sin 20∞ + aT cos 20∞ - aP /T )
For sliding with friction
F2 = m k N 2 = 0.30 N 2
mP ( g sin 20∞ + aT cos 20∞ - aP /T ) = 0.30mP ( g cos 20∞ + aT sin 20∞)
aT =
(0.30 cos 20∞ - sin 20∞) g + aP /T
cos 20∞ + 0.30 sin 20∞
= ( -0.05767)(9.81) + (0.9594)( 4.94)
= 4.17 aT = 4.17 m/s2 Æ b
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37
PROBLEM 12.24
The propellers of a ship of weight W can produce a propulsive force F0; they produce a force of the same
magnitude but of opposite direction when the engines are reversed. Knowing that the ship was proceeding
forward at its maximum speed v0 when the engines were put into reverse, determine the distance the ship travels
before coming to a stop. Assume that the frictional resistance of the water varies directly with the square of the
velocity.
Solution
F0 = kv02 = 0 k =
At maximum speed a = 0.
F0
v02
When the propellers are reversed, F0 is reversed.
+
SFx = ma : - F0 - kv 2 = ma
- F0 - F0
v2
= ma
v02
dx =
x
Ú0
a-
F0
mv02
mv02 vdv
vdv
=
a
F0 v02 + v 2
dx = -
x==-
(
mv02
F0
0
2
0
+ v2
)
)
vdv
+ v2
0
Úv
(v
v02
mv02 1
ln v02 + v 2
F0 2
(
)
0
v0
mv 2
mv02 È 2
ln v0 - ln 2v02 ˘ = 0 ln 2
˚ 2 F0
2 F0 Î
( )
x = 0.347
m0 v02
b
F0
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38
PROBLEM 12.25
A constant force P is applied to a piston and rod of total mass m to
make them move in a cylinder filled with oil. As the piston moves,
the oil is forced through orifices in the piston and exerts on the piston
a force of magnitude kv in a direction opposite to the motion of the
piston. Knowing that the piston starts from rest at t = 0 and x = 0, show
that the equation relating x, v, and t, where x is the distance traveled
by the piston and v is the speed of the piston, is linear in each of the
variables.
Solution
+
SF = ma : P - kv = ma
dv
P - kv
=a=
dt
m
t
v m dv
Ú0 dt = Ú0 P - kv
m
v
= - ln ( P - kv ) 0
k
m
= - [ln ( P - kv ) - ln P ]
k
m P - kv
P - kv
kt
or
=t = - ln
ln
k
P
m
m
P
P - kv
or
v = (1 - e - kt/m )
= e - kt/m
m
k
Pt
x = Ú v dt =
0
k
t
t
0
PÊ k
ˆ
- Á - e - kt/m ˜
¯
Ë
k
m
t
0
Pt P - kt/m
Pt P
=
+ (e
- 1) =
- (1 - e - kt/m )
k m
k m
x=
Pt kv
- , which is linear.
k
m
b
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39
PROBLEM 12.26
A spring AB of constant k is attached to a support at A and to
a collar of mass m. The unstretched length of the spring is l.
Knowing that the collar is released from rest at x = x0 and
neglecting friction between the collar and the horizontal rod,
determine the magnitude of the velocity of the collar as it passes
through Point C.
Solution
Choose the origin at Point C and let x be positive to the right. Then x is a position coordinate of the slider B and
x0 is its initial value. Let L be the stretched length of the spring. Then, from the right triangle
L = l2 + x 2
The elongation of the spring is e = L - l, and the magnitude of the force exerted by the spring is
Fs = ke = k ( l2 + x 2 - l)
cosq =
By geometry,
x
l + x2
2
SFx = max : - Fs cosq = ma
+
- k ( l 2 + x 2 - l)
a=-
x
l2 + x 2
= ma
kÊ
lx ˆ
Áx- 2
˜
mË
l + x2 ¯
0
v
Ú0 v dv = Úx0 a dx
0
v
k 0Ê
k Ê1 2
1 2
lx ˆ
2
2ˆ
=
+
v = - Ú Áxdx
x
x
l
l
˜¯
Á
˜
m x0 Ë
mË2
2 0
l2 + x 2 ¯
x0
1
kÊ
1 2
ˆ
v = - Á 0 - l2 - x02 + l l2 + x02 ˜
¯
2
mË
2
(
)
k
2l2 + x02 - 2l l2 + x02
m
k
= ÍÈ l2 + x02 - 2l l2 + x02 + l2 ˙˘
˚
mÎ
v2 =
(
)
answer: v =
k
m
(
)
l2 + x02 - l b
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40
PROBLEM 12.27
Determine the maximum theoretical speed that a 1350 kg automobile starting from rest can reach after traveling
a 400 m if air resistance is considered. Assume that the coefficient of static friction between the tires and the
pavement is 0.70, that the automobile has front-wheel drive, that the front wheels support 62 percent of the
automobile’s weight, and that the aerodynamic drag D has a magnitude D = 0.667 v2, where D and v are
expressed in N and m/s, respectively.
Solution
F = Fmax for v = vmax
F = m s N F = 0.70(0.62 mg)
= 0.434 mg
+
SFx = ma : F - D = ma
a=
or
Now
At x = 0, v = 0:
or
v
(0.434 mg - 0.667 v 2 )
m
(0.434 mg - 0.667 v 2 )
dv
=a=
dx
m
v
1 x
vdv
dx = Ú
Ú
0
0
m
0.434 mg - 0.667 v 2
v
1
-1
ln ÈÎ0.434 mg - 0.667 v 2 ˘˚
x=
0
m
2(0.667)
Ê 0.434 mgg - 0.667 v 2 ˆ
-1
1
ln Á
x=
˜
0.434 mg
m
1.334 Ë
¯
or
0.434 mg - 0.667 v 2
Ê -1.334 x ˆ
= exp Á
˜
Ë
m ¯
0.434 mg
Substituting m = 1350 kg, g = 9.81 m/s2, x = 400 m fi vmax = 53.042 m/s = 190.951 km/h
vmax = 191.0 km/h b
or
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41
PROBLEM 12.28
The coefficients of friction between blocks A and C and the
horizontal surfaces are m s = 0.24 and m k = 0.20. Knowing that
mA = 5 kg, mB = 10 kg, and mC = 10 kg, determine (a) the tension
in the cord, (b) the acceleration of each block.
Solution
We first check that static equilibrium is not maintained:
( FA )m + ( FC ) m = m s ( mA + mC ) g
= 0.24(5 + 10) g
= 3.6 g
Since WB = mB g = 10 g 3.6 g , equilibrium is not maintained.
SFy : N A = mA g
Block A:
FA = m k N A = 0.2mA g
+
SFl = mA aA : T - 0.2mA g = mA aA (1)
SFy : N C = mC g
Block C:
FC = m k N C = 0.2mC g
+
SFx = mC aC : T - 0.2mC g = mC aC Block B:
+
(2)
SFy = mB aB
mB g - 2T = mB aB (3)
1
aB = ( a A + aC )
2
(a) Tension in cord. Given data:
mA = 5 kg
From kinematics:
(4)
mB = mC = 10 kg
Eq. (1): T - 0.2(5) g = 5a A
aA = 0.2T - 0.2 g (5)
Eq. (2): T - 0.2(10) g = 10aC
aC = 0.1T - 0.2 g (6)
Eq. (3): 10 g - 2T = 10aB
aB = g - 0.2T (7)
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42
Problem 12.28 (continued)
Substitute into (4):
1
g - 0.2T = (0.2T - 0.2 g + 0.1T - 0.2 g )
2
24
24
1.2 g = 0.35T
T=
g = (9.81 m/ss2 )
7
7
(b)
T = 33.6 N b
Substitute for T into (5), (7), and (6):
Ê 24 ˆ
aA = 0.2 Á g ˜ - 0.2 g = 0.4857(9.81 m/s2 )
Ë 7 ¯
Ê 24 ˆ
aB = g - 0.2 Á g ˜ = 0.3143(9.81 m/s2 ) Ë 7 ¯
Ê 24 ˆ
aC = 0.1Á g ˜ - 0.2 g = 0.14286(9.81 m/s2 ) Ë 7 ¯
a A = 4.76 m/s2 Æ b
a B = 3.08 m/s2 Ø b
aC = 1.401 m/s2 ¨ b
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43
PROBLEM 12.29
Solve Problem 12.28, assuming that mA = 5 kg, mB = 10 kg, and
mC = 20 kg.
PROBLEM 12.28 The coefficients of friction between blocks A
and C and the horizontal surfaces are m s = 0.24 and m k = 0.20.
Knowing that mA = 5 kg, mB = 10 kg, and mC = 10 kg, determine
(a) the tension in the cord, (b) the acceleration of each block.
Solution
We first check that static equilibrium is not maintained:
( FA )m + ( FC ) m = m s ( mA + mC ) g = 0.24(5 + 20) g = 6 g
WB = mB g = 10 g 6 g , equilibrium is not maintained.
Since
We shall assume that all 3 blocks move and use Eqs. (1), (2), (3), (4) derived in solution of Problem 12.28.
(a)
Tension in cord.
mA = 5 kg, mB = 10 kg, mC = 20 kg
Given data:
Eq. (1): T - 0.2(5) g = 5a A
aA = 0.2T - 0.2 g (5¢)
Eq. (2): T - 0.2(20) g = 20aC
aC = 0.05T - 0.2 g
(6¢)
Eq. (3): 10 g - 2T = 10 aB
aB = g - 0.2T (7¢)
Substituting into (4):
1
g - 0.2T = (0.2T - 0.2 g + 0.05T - 0.2 g )
2
48
48
1.2 g = 0.325T
T=
g = (9.81 m/s2 )
13
13
Substituting into (6¢):
T = 36.2 N v
Ê 48 ˆ
aC = 0.05 Á g ˜ - 0.2 g = -0.0154 g (impossible)
Ë 13 ¯
This means that our assumption that block C moves was wrong.
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44
Problem 12.29 (continued)
aC = 0 v
Assuming now that C does not move, we have
Substituting aC = 0 and from (5¢) and (6¢) into Eq. (4):
1
g - 0.2T = (0.2T - 0.2 g )
2
11
11
1.1g = 0.3T
T = g = (9.81 m/s2 ) = 35.97 N
3
3
T = 36.0 N v
Substituting for T into Eqs. (5¢) and (7¢):
Ê 11 ˆ
aA = 0.2 Á g ˜ - 0.2 g = 0.5333(9.81 m/s2 )
Ë3 ¯
a A = 5.23 m/s2 Æ b
Ê 11 ˆ
aB = g - 0.2 Á g ˜ = 0.2667(9.81 m/s2 )
Ë3 ¯
a B = 2.62 m/s2 Ø b
g is less than ( FC ) m = 0.24(20 g ) = 4.8 g.
We also check that T = 11
3
aC = 0 b
The value assumed for aC is thus correct:
T = 36.0 N b
as well as that obtained for T:
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45
PROBLEM 12.30
Blocks A and B weigh 10 kg each, block C weighs 7 kg, and block D weighs
8 kg. Knowing that a downward force of magnitude 120 N is applied to
block D, determine (a) the acceleration of each block, (b) the tension in cord
ABC. Neglect the weights of the pulleys and the effect of friction.
Solution
Note: As shown, the system is in equilibrium.
From the diagram:
Cord 1:
2 y A + 2 y B + yC = constant
Then
2v A + 2v B + vC = 0
and
2aA + 2aB + aC = 0
( yD - y A ) + ( yD - y B ) = constant
Cord 2:
Then
2v D - v A - v B = 0
and
2aD - aA - aB = 0 A:
(a)
+
B:
(2)
SFy = mA aA : mA g - 2T1 + T2 = mA a A
10 g - 2T1 + T2 = 10 aA
or
+
(1)
(3)
SFy = mB aB : mB g - 2T1 + T2 = mB aB
10 g - 2T1 + T2 = 10 aB
or
(4)
Note: Eqs. (3) and (4) fi a A = a B
Then
Eq. (1) fi aC = -4 aA
Eq. (2) fi aD = aA
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46
Problem 12.30 (continued)
C:
+
SFy = mC aC : mC g - T1 = mC aC
T1 = 7( g - aC )
or
T1 = 7 ( g + 4 aA )
+
D:
(5)
SFy = mD aD : mD g - 2T2 + ( FD )ext = mD aD
1
T2 = [8( g - aD ) + 120]
2
T2 = 4 g - 4 aA + 60
or
Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3)
10 g - 14( g + 4 aA ) + ( 4 g - 4aA + 60) = 10 aA
fi - 56aA - 4aA + 60 = 10a A
aA =
or
6
m/s2
7
a A = a B = a D = 0.857 m/s2Ø b
24
m/s2 7
Substituting into Eq. (5)
aC = -
and
(b)
24 ˆ
Ê
T1 = 7 Á 9.81 + ˜ Ë
7¯
or aC = 3.43 m/s2 ≠ b
or T1 = 92.7 N b
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47
PROBLEM 12.31
Blocks A and B weigh 10 kg each, block C weighs 7 kg, and block D weighs
8 kg. Knowing that a downward force of magnitude 120 N is applied to block
B and that the system starts from rest, determine at t = 3 s the velocity (a) of
D relative to A, (b) of C relative to D. Neglect the weights of the pulleys and
the effect of friction.
Solution
Note: As shown, the system is in equilibrium.
From the diagram:
Cord 1:
2 y A + 2 y B + yC = constant
Then
2v A + 2v B + vC = 0
and
2aA + 2aB + aC = 0
( yD - y A ) + ( yD - y B ) = constant
Cord 2:
2v D - v A - v B = 0
Then
2aD - aA - aB = 0
and
A:
+
(a)
+
(2)
SFy = mA a A : mA g - 2T1 + T2 = mA aA
10 g - 2T1 + T2 = 10 a A
or
(3)
SFy = mB aB : mB g - 2T1 + T2 + ( FB )ext = mB aB
10 g - 2T1 + T2 + 50 = 10 aB or
(4)
(3) - ( 4) fi -50 = 10( a A - aB )
Forming
B:
(1)
aB = a A + 5
or
-5 = aA - aB
aB = a A + 5
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48
Problem 12.31 (continued)
Then
C:
Eq. (1):
2aA + 2( a A + 5) + aC = 0
aC = -4 aA - 10
or
Eq. (2):
2aD - a A - ( a A + 5) = 0
aD = aA +
or
+
SFy = mC aC : mC g - T1 = mC aC
T1 = 7 ( g - aC ) = 7 ( g + 4a A + 10)
or
D:
+
5
2
(5)
SFy = mD aD : mD g - 2T2 = mD aD
T2 =
or
1
-5 ˆ
Ê
¥ 8( g - aD ) = 4 Á g - a A ˜ Ë
2
2¯
(6)
Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3)
-5 ˆ
Ê
10 g - 14( g + 4a A + 10) + 4 Á g - a A ˜ = 10a A
Ë
2¯
fi 10 g - 14 g - 56a A - 140 + 4 g - 4aA - 10 = 10aA
aA = -
or
-15
m/s2
7
-10
Ê -15 ˆ
- 10 =
aC = -4 ¥ Á
m/s2
Ë 7 ˜¯
7
-15 5 5
aD =
+ = m/s2
7
2 14
Then
Note: We have uniformly accelerated motion, so that
v = 0 + at
(a)
We have
v D /A = v D - v A
or
È 5 Ê -15 ˆ ˘ m
v D /A = aD t - aA t = Í - Á
˜˙ 2 ¥ 3s
Î14 Ë 7 ¯ ˚ s
v D /A = 7.50 m/s Ø b
or
(b)
And
v C /D = v C - v D
or
-75
Ê -10 5 ˆ
vC /D = aC t - aD t = Á
- ˜ m/s2 ¥ 3s =
m/s
Ë 7
14 ¯
14
v C /D = 5.36 m/s ≠ b
or
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49
PROBLEM 12.32
The 15-kg block B is supported by the 25-kg block A and is attached
to a cord to which a 225-N horizontal force is applied as shown.
Neglecting friction, determine (a) the acceleration of block A, (b)
the acceleration of block B relative to A.
Solution
(a)
First we note a B = a A + a B /A , where a B /A is directed along the inclined surface of A.
+
B:
SFx = mB ax : P - WB sin 25∞ = mB aA cos 25∞ + mB aB /A
225 - 15 g sin 25∞ = 15( a A cos 25∞ + aB /A )
or
15 - g sin 25∞ = aA cos 25∞ + aB /A
+
or
or
A:
or
+
B:
(1)
SFy = mB a y : N AB - WB cos 25∞ = - mB aA sin 25∞
N AB = 15( g cos 25∞ - a A sin 25∞)
A:
SFx ¢ = mA aA : P - P cos 25∞ + N AB sin 25∞ = mA aA
N AB = [25aA - 225(1 - cos 25∞)] / sin 25∞
Equating the two expressions for N AB
15( g cos 25∞ - a A sin 25∞) =
or
25aA - 225(1 - cos 25∞)
sin 25∞
3(9.81) cos 25∞ sin 25∞ + 45(1 - cos 25∞)
5 + 3 sin 2 25∞
2
= 2.7979 m/s
aA =
a A = 2.80 m/s2 ¨ b
(b)
From Eq. (1)
aB /A = 15 - (9.81)sin 25∞ - 2.7979 cos 25∞
a B /A = 8.32 m/s2
or
25∞ b
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50
PROBLEM 12.33
Block B of mass 10 kg rests as shown on the upper surface of a 22-kg
wedge A. Knowing that the system is released from rest and neglecting
friction, determine (a) the acceleration of B, (b) the velocity of B relative to
A at t = 0.5 s.
Solution
+
(a)
A:
SFx = mA a A : W A sin 30∞ + N AB cos 40∞ = mA a A
1 ˆ
Ê
22 Á aA - g ˜
Ë
2 ¯
or
NAB =
cos 40∞
Now we note: a B = a A + a B /A , where a B /A is directed along the top surface of A.
B:
+
SFy¢ = mB a y¢ : NAB - WB cos 20∞ = - mB aA sin 50∞
or
NAB = 10 ( g cos 20∞ - a A sin 50∞)
Equating the two expressions for NAB
1 ˆ
Ê
22 Á aA - g ˜
Ë
2 ¯
= 10( g cos 20∞ - aA sin 50∞)
cos 40∞
aA =
+
(9.81)(1.1 + cos 20∞ cos 40∞)
= 6.4061 m/s2
2.2 + cos 40∞ sin 50∞
or
SFx ¢ = mB ax ¢ : WB sin 20∞ = mB aB /A - mB a A cos 50∞
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51
Problem 12.33 (continued)
or
aB /A = g sin 20∞ + aA cos 50∞
= (9.81sin 20∞ + 6.4061cos 50∞) m/s2
= 7.4730 m /s2
Finally
a B = a A + a B/ A
We have
aB2 = 6.40612 + 7.47302 - 2(6.4061 ¥ 7.4730) cos 50∞
or
aB = 5.9447 m/s2
and
or
7.4730 5.9447
=
sin a
sin 50∞
a = 74.4∞
a B = 5.94 m/s2
(b)
75.6∞ b
Note: We have uniformly accelerated motion, so that
v = 0 + at
Now
v B /A = v B - v A = a B t - a A t = a B /A t
At t = 0.5 s :
v B /A = 7.4730 m/s2 ¥ 0.5 s
v B /A = 3.74 m/s
or
20∞ b
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52
PROBLEM 12.34
A 20-kg sliding panel is supported by rollers at B and C. A 12.5-kg counterweight A is attached to a cable as
shown and, in cases a and c, is initially in contact with a vertical edge of the panel. Neglecting friction, determine
in each case shown the acceleration of the panel and the tension in the cord immediately after the system is
released from rest.
Solution
(a)
F = Force exerted by counterweight
Panel:
+
SFx = ma :
T - F = 20a
(1)
Counterweight A: Its acceleration has two components
a A = a P + a A /P = a Æ + a Ø
+
+
SFx = max : F = 12.5a (2)
SFy = ma y : 12.5 g - T = 12.5a (3)
Adding (1), (2), and (3):
T - F + F + 12.5 g - T = (20 + 12.5 + 12.5)a
a=
12.5 g 12.5 ¥ 9.81
=
= 2.725 m/s2 45
45
a = 2.73 m/s2 b
Substituting for a into (3):
12.5 ¥ 9.81 - T = 12.5 ¥ (2.725) T = 12.5 ¥ (9.81 - 2.725)
T = 88.6 N b
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53
Problem 12.34 (continued)
(b)
Panel:
+
SFx = ma :
T = 20a
(1)
Counterweight A:
+
SFy = ma :
Adding (1) and (2):
12.5 g - T = 12.5a (2)
T + 12.5 g - T = 32.5a
a=
5
g
13
a = 3.77 m/s2 ¨ b
Substituting for a into (1):
Ê5 ˆ
T = 20 Á g ˜ Ë 13 ¯
(c)
T = 75.5 N b
Since panel is accelerated to the left, there is no force exerted by panel on counterweight and vice versa.
Panel:
+
SFx = ma :
T = 20a
(1)
+
Counterweight A: Same free body as in Part (b):
SFy = ma :
12.5 g - T = 12.5a (2)
Since Eqs. (1) and (2) are the same as in (b), we get the same answers:
a = 3.77m/s2 ¨; T = 75.5 N b
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54
PROBLEM 12.35
A 250-kg crate B is suspended from a cable attached to a 20-kg trolley A which
rides on an inclined I-beam as shown. Knowing that at the instant shown the
trolley has an acceleration of 0.4 m/s2 up and to the right, determine (a) the
acceleration of B relative to A, (b) the tension in cable CD.
Solution
(a)
First we note: a B = a A + a B /A , where a B /A is directed perpendicular to cable AB
SFx = mB ax : 0 = - mB aB / A + mB aA cos 25∞
B:
+
aB /A = (0.4 m/s2 ) cos 25∞
or
a B /A = 0.363 m/s2 ¨ b
or
For crate B
+
(b)
SFy = mB a y : TAB - mB g = mB a A sin 25∞
A:
(Not that ( aB ) y = aA sin 25o )
TAB = 250( g + aA sin 25∞)
or
= 2494.76 N
For trolley A
+
or
SFx = mAaA : TCD - TAB sin 25∞ - mA g sin 25∞ = mAa A
TCD = (2494.76 N )sin 25∞ + 20(9.81 ¥ sin 25∞ + 0.4)
TCD = 1145 N b
or
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55
PROBLEM 12.36
During a hammer thrower’s practice swings, the 7.1-kg head A of
the hammer revolves at a constant speed v in a horizontal circle as
shown. If r = 0.93 m and q = 60∞, determine (a) the tension in wire
BC, (b) the speed of the hammer’s head.
Solution
(a)
+
First we note
aA = an =
v 2A
r
SFy = 0 : TBC sin 60∞ - WA = 0
or
7.1 kg ¥ 9.81 m/s2
sin 60∞
= 80.426 N
TBC =
TBC = 80.4 N b
+
(b)
or
SFx = mA a A : TBC cos 60∞ = mA
v 2A =
v 2A
r
(80.426 N ) cos 60∞ ¥ 0.93 m
7.1 kg
v A = 2.30 m/s b
or
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56
PROBLEM 12.37
A 450-g tetherball A is moving along a horizontal circular path at
a constant speed of 4 m/s. Determine (a) the angle q that the cord
forms with pole BC, (b) the tension in the cord.
Solution
aA = an =
First we note
r = l AB sin q
+
where
(a)
v 2A
r
SFy = 0 : TAB cosq - W A = 0
TAB =
or
+
mA g
cosq
SFx = mA a A : TAB sinq = mA
v 2A
r
Substituting for TAB and r
mA g
v 2A
sin q = mA
cos q
l AB sin q
1 - cos2 q =
or
sin 2 q = 1 - cos2 q
( 4 m/s)2
1.8 m ¥ 9.81 m/s2
cos q
cos2 q + 0.906105 cos q - 1 = 0
cos q = 0.64479
Solving
q = 49.9∞ b
or
(b)
From above
or
TAB =
mA g 0.450 kg ¥ 9.81 m/s2
=
cos q
0.64479
TAB = 6.85 N b
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57
PROBLEM 12.38
A single wire ACB of length 2 m in. passes through a ring at C that is attached
to a sphere which revolves at a constant speed v in the horizontal circle shown.
Knowing that q1 = 60∞ and q 2 = 30∞ and that the tension is the same in both
portions of the wire, determine the speed v.
Solution
aC = aN =
First we note
r = LAC sin 30∞ = LBC sin 60∞
where
Now
or
LAC + LBC = LABC
1 ˆ
Ê 1
= 2m
rÁ
+
Ë sin 30∞ sin 60∞ ˜¯
r = 0.633975 m
+
or
+
or
SFy = 0 : TCA cos 30∞ + TCB cos 60∞ - mC g = 0
T=
or
or
vC2
r
mC g
= 0.73205 mC g
cos 30∞ + cos 60∞
SFx = mC aC : TCA sin 30∞ + TCB sin 60∞ = mC
0.73205 mC g (sin 30∞ + sin 60∞) = mC
vC2
r
vC2
r
vC2 = 0.73205 (9.81 m/s2 ) (sin 30∞ + sin 60∞)(0.633975 m)
vC = 2.49 m/s b
or
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58
PROBLEM 12.39
A single wire ACB passes through a ring at C that is attached to a 6 kg sphere
which revolves at a constant speed v in the horizontal circle shown. Knowing
that q1 = 50o and d = 750 mm and that the tension in both portions of the wire is
40 N, determine (a) the angle q 2 , (b) the speed v.
(a)
+
Solution
SFy = 0 : T cos q1 + T cos q 2 - mg = 0
cos q 2 =
=
mg
- cos q1
T
6 ¥ 9.81
- cos 50∞ = 0.82871
40
q 2 = 34.033∞
(b)
q 2 = 34.0∞ b
q3 = q1 - q 2 = 50∞ - 34.033∞ = 15.967∞
l1
sin q 2
=
d
sin q3
or
l1 =
d sin q 2
sin q3
Radius of horizontal circle
r = l1 sin q1 =
d sin q 2 sin q1
sin q3
(0.75)(sin 34.033∞)(sin 50∞)
sin(15.967∞)
= 1.16891 m
mv 2
SFx = man : T sin q1 + T sin q 2 =
r
r
sin
q
+
sin
q
T
(
)
1
2
v2 =
m
(1.16891)( 40)(sin 50∞ + sin 34.033∞)
=
6
2 2
= 10.3309 m /s
=
+
v = 3.21 m/s b
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59
PROBLEM 12.40
Two wires AC and BC are tied at C to a 7-kg sphere which revolves at a constant
speed v in the horizontal circle shown. Knowing that q1 = 55∞ and q 2 = 30∞ and
that d = 1.4 m, determine the range of values of v for which both wires remain
taut.
Solution
aC = an =
First we note
vC2
r
r = l tan 55∞ and
where
r = ( d + l ) tan 30∞
Then
r ˆ
Ê
r = Ád +
˜ tan 30∞
Ë
tan 55∞ ¯
or
r=
1.4 m
1
1
tan 30∞ tan 55∞
= 1.35680 m
+
+
SFx = mC aC : TCA sin 30∞ + TCB sin 55∞ = mC
(1)
SFy = 0 : TCA cos 30∞ + TCB cos 55∞ - WC = 0
or
TCA cos 30∞ + TCB cos 55∞ = mC g Case 1: TCA = 0:
Eq. (2) fi TCB =
Substituting into Eq. (1)
vC2
r
(2)
mC g
cos 55∞
mC g
v2
sin 55∞ = mC C
cos 55∞
r
or
( vC2 )TCA = 0 = (1.35680 m)(9.81 m) tan 55∞
or
( vC )TCA= 0 = 4.36 m/s
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60
Problem 12.40 (continued)
Now we form
(cos 30∞)(1) - (sin 30∞)(2)
TCB sin 55∞ cos 30∞ - TCB cos 55∞ sin 30∞ = mC
vC2
cos 30∞ - mC g sin 30∞
r
TCB sin 25∞ = mC
vC2
cos 30∞ - mC g sin 30∞
r
or
(vC)max occurs when TCB = (TCB ) max , which occurs when TCA = 0.
(vC)max = 4.36 m/s and wire AC will be taut if vC 4.36 m/s.
mC g
cos 30∞
Case 2: TCB = 0:
Eq. (2) fi TCA =
Substituting into Eq. (1)
mC g
v2
sin 30∞ = mC C
cos 30∞
r
(vC2 )T g =0 = (1.35680 m)(9.81 m/s2 ) tan 30∞
or
C
( vC )TC g = 0 = 2.77 m/s
or
Now we form
(cos 55∞)(1) - (sin 55∞)(2)
TCA sin 30∞ cos 55∞ - TCA cos 30∞ sin 55∞ = mC
vC2
cos 55∞ - mC g sin 55∞
r
-TCA sin 25∞ = mC
vC2
cos 55∞ - mC g sin 55∞
r
or
(vC)min occurs when TCA = (TCA ) max , which occurs when TCB = 0.
(vC)min = 2.77 m/s and wire BC will be taut if vC > 2.77 m/s.
Both wires are taut when
2.77 m/s v 4.36 m/s b
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61
PROBLEM 12.41
A 100-g sphere D is at rest relative to drum ABC, which rotates
at a constant rate. Neglecting friction, determine the range of
the allowable values of the velocity v of the sphere if neither of the
normal forces exerted by the sphere on the inclined surfaces of the
drum is to exceed 1.1 N.
Solution
aD = an =
First we note
vD2
r
r = 0.2 m
where
SFx = mD aD : N1 cos 60∞ + N 2 cos 20∞ = mD
+
+
vD2
r
SFy = 0 : N1 sin 60∞ + N 2 sin 20∞ - w D = 0
N1 sin 60∞ + N 2 sin 20∞ = mD g
or
(1)
(2)
Case 1: N1 is maximum.
N1 = 1.1 N
Let
Eq. (2)
(1.1 N ) sin 60∞ + N 2 sin 20∞ = (0.1 kg) (9.81 m/s2 )
N 2 = 0.082954 N
or
( N 2 )( N1 )max 1.1 N.
O.K.
0.2 m
(1.1cos 60∞ + 0.082954 cos 20∞) N
0.1 kg
Eq. (1)
( vD2 )( N1 )max =
or
( vD )( N1 )max = 1.121 m/s
Now we form
(sin 20∞)(1) - (cos 20∞)(2)
N1 cos 60∞ sin 20∞ - N1 sin 60∞ cos 20∞ = mD
vD2
sin 20∞ - mD g cos 20∞
r
- N1 sin 40∞ = mD
vD2
sin 20∞ - mD g cos 20∞
r
or
( vD ) min occurs when N1 = ( N1 ) max
( vD ) min = 1.121 m/s
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62
Problem 12.41 (continued)
Case 2: N 2 is maximum.
N 2 = 1.1 N
Let
N1 sin 60∞ + (1.1 N )sin 20∞ = (0.1 kg)(9.81 m/s2 )
Eq. (2)
N1 = 0.69834 N
or
( N1 )( N 2 )max 1.1 N.
O.K.
0.2 m
(0.69834 cos 60∞ + 1.1cos 20∞) N
0.1 kg
Eq. (1)
( vD2 )( N 2 )max =
or
( vD )( N 2 )max = 1.663 m/s
Now we form
(sin 60∞)(1) - (cos 60∞)(2)
N 2 cos 20∞ sin 60∞ - N 2 sin 20∞ cos 60∞ = mD
vD2
sin 60∞ - mD g cos 60∞
r
N 2 cos 40∞ = mD
vD2
sin 60∞ - mD g cos 60∞
r
or
( vD ) maxoccurs when N 2 = ( N 2 ) max
( vD ) max = 1.663 m/s
For N1 , N 2 1.1 N
1.121 m/s vD 1.663 m/s b
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63
PROBLEM 12.42*
As part of an outdoor display, a 6 kg model C of the earth is attached to wires AC and
BC and revolves at a constant speed v in the horizontal circle shown. Determine the
range of the allowable values of v if both wires are to remain taut and if the tension in
either of the wires is not to exceed 125 N.
Solution
aC = an =
First note
vC2
r
r = 0.9 m
where
+
+
SFx = mC aC : TCA sin 40∞ + TCB sin15∞ = mC
vC2
r
SFy = 0 : TCA cos 40∞ - TCB cos15∞ - mC g = 0
(1)
(2)
Note that Eq. (2) implies that
(a)
when
TCB = (TCB ) max,
TCA = (TCA ) max
(b)
when
TCB = (TCB ) min ,
TCA = (TCA ) min
Case 1: TCA is maximum.
Let
Eq. (2)
or
TCA = 125 N
(125 N ) cos 40∞ - TCB cos15∞ - (6 ¥ 9.81) = 0
TCB = 68.4732 N
(TCB )(TCA )max < 125 N
O.K.
[(TCB ) max = 68.4732 N ]
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64
Problem 12.42* (continued)
Eq. (1)
( vC2 )(TCA )max =
0.9
(125 sin 40∞ + 68.4732 sin 15∞) N
6
( vC )(TCA )max = 3.8354 m/s
or
Now we form
(cos15∞) (1) + (sin 15∞) (2)
TCA sin 40∞ cos15∞ + TCA cos 40∞ sin 15∞ = mC
or
TCA sin 55∞ = mC
vC2
cos15∞ + mC g sin 15∞
r
vC2
cos15∞ + mC g sin 15∞
r
(3)
( vc ) max occurs when TCA = (TCA ) max
( vC ) max = 3.84 m/s
Case 2: TCA is minimum.
Because (TCA ) min occurs when TCB = (TCB ) min,
let TCB = 0 (note that wire BC will not be taut).
Eq. (2)
or
TCA cos 40∞ - (6 g ) = 0
TCA = 76.8363 N 125 N O.K.
Note: Eq. (3) implies that when TCA = (TCA ) min , vC = ( vC ) min . Then
0.9
(76.8363)sin 40∞
6
Eq. (1)
( vC2 ) min =
or
( vC ) min = 2.7218 m/s
0 TCA , TCB 125 N when
2.72 m/s vC 3.84 m/s b
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65
PROBLEM 12.43*
The 600 g flyballs of a centrifugal governor revolve at a constant speed v
in the horizontal circle of 150 mm radius shown. Neglecting the weights
of links AB, BC, AD, and DE and requiring that the links support only
tensile forces, determine the range of the allowable values of v so that the
magnitudes of the forces in the links do not exceed 80 N.
Solution
v2
r
First note
a = an =
where
r = 0.15 m
+
+
SFx = ma : TDA sin 20∞ + TDE sin 30∞ = m
v2
r
SFy = 0 : TDA cos 20∞ - TDE cos 30∞ - mg = 0
FBD of flyball
(1)
(2)
Note that Eq. (2) implies that
(a)
when
TDE = (TDE ) max,
TDA = (TDA ) max
(b)
when
TDE = (TDE ) min,
TDA = (TDA ) min
Case 1: TDA is maximum.
Let
Eq. (2)
TDA = 80 N
(80 N ) cos 20∞ - TDE cos 30∞ - (0.6 ¥ 9.81) = 0
or
TDE = 80.01 N unacceptable ( 80 N )
Now let
TDE = 80 N
Eq. (2)
or
TDA cos 20∞ - (80 N ) cos 30∞ - (0.6 ¥ 9.81) = 0
TDA = 79.9921 N O.K. ( 80 N)
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66
Problem 12.43* (continued)
(TDA ) max = 79.9921 N
(TDE ) max = 80 N
( v 2 )(TDA )max =
Eq. (1)
0.15 m
(79.9921sin 20∞ + 80 sin 30∞) N
0.6 kg
v(TDA )max = 4.10362 m/s
or
(cos 30∞)(1) + (sin 30∞)(2)
Now form
TDA sin 20∞ cos 30∞ + TDA cos 20∞ sin 30∞ = m
TDA sin 50∞ = m
or
v2
cos 30∞ + mg sin 30∞
r
v2
cos 30∞ + mg sin 30∞
r
(3)
vmax occurs when TDA = (TDA ) max
vmax = 4.10362 m/s
Case 2: TDA is minimum.
Because (TDA ) min occurs when TDE = (TDE ) min ,
let TDE = 0.
Eq. (2)
TDA cos 20∞ - (0.6 ¥ 9.81) = 0
or
TDA = 6.26375 N 80 N
O.K.
Note: Eq. (3) implies that when TDA = (TDA ) min , v = vmin . Then
Eq. (1)
( v 2 ) min =
or
(0.15 m)
(6.26375 N )sin 20∞
(0.6 kg)
vmin = 0.53558 m/s
0 TAB , TBC , TAD , TDE 80 N
when
0.536 m/s v 4.10 m/s b
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67
PROBLEM 12.44
A child having a mass of 22 kg sits on a swing and is held in the
position shown by a second child. Neglecting the mass of the
swing, determine the tension in rope AB (a) while the second
child holds the swing with his arms outstretched horizontally,
(b) immediately after the swing is released.
Solution
Note: The factors of “ 12 ” are included in the following free-body diagrams because there are two ropes and only one
is considered.
(a)
For the swing at rest
1
SFy = 0 : TBA cos 35∞ - W = 0
2
TBA =
or
22 kg ¥ 9.81 m/s2
2 cos 35∞
TBA = 131.7 N b
or
(b)
an =
At t = 0, v = 0, so that
+
or
v2
=0
r
1
SFn = 0 : TBA - W cos 35∞ = 0
2
1
TBA = (22 kg)(9.81 m/s2 ) cos 35∞
2
TBA = 88.4 N b
or
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68
PROBLEM 12.45
A 60-kg wrecking ball B is attached to a 15-m-long steel cable AB
and swings in the vertical arc shown. Determine the tension in the
cable (a) at the top C of the swing, (b) at the bottom D of the swing,
where the speed of B is 4.2 m/s.
Solution
(a)
At C, the top of the swing, v B = 0; thus
an =
+
v B2
=0
LAB
SFn = 0 : TBA - WB cos 20∞ = 0
TBA = (60 kg)(9.81 m/s2 ) ¥ cos 20∞
or
TBA = 553 N b
or
+
(b)
SFn = man : TBA - WB = mB
or
( v B )2D
LAB
È
( 4.2 m/s)2 ˘
TBA = (60 kg) Í9.81 m/s2 +
˙
15 m ˚
Î
TBA = 659 N b
or
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69
PROBLEM 12.46
During a high-speed chase, a 1000 kg sports car traveling at a
speed of 160 km/h just loses contact with the road as it reaches
the crest A of a hill. (a) Determine the radius of curvature r of
the vertical profile of the road at A. (b) Using the value of r
found in part a, determine the force exerted on a 80 kg driver
by the seat of his 1200 kg car as the car, traveling at a constant
speed of 75 km/h, passes through A.
Solution
(a)
Note:
+
160 km/h =
400
m/s
9
S Fn = man : Wcar =
Wcar v 2A
g r
r=
or
Ê 400
ˆ
m/s˜
ÁË
¯
9
2
9.81 m/s2
= 201.357 m
r = 201 m b
or
(b)
Note: v is constant fi at = 0; 75 km/h =
125
m/s
6
+
SFn = man : W - N =
W v 2A
g r
2
or
N =W -
mv A2
S
Ê 125 ˆ
80 ¥ Á
Ë 6 ˜¯
= 80 ¥ 9.81 = 612.359 N
201.357
N = 612 N ≠ b
or
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70
PROBLEM 12.47
The portion of a toboggan run shown is contained in a vertical plane.
Sections AB and CD have radii of curvature as indicated, and section
BC is straight and forms an angle of 20° with the horizontal. Knowing
that the coefficient of kinetic friction between a sled and the run is 0.10
and that the speed of the sled is 7.5 m/s at B, determine the tangential
component of the acceleration of the sled (a) just before it reaches B,
(b) just after it passes C.
Solution
Note: Just before B, r B = 18 m
+
(a)
S Fn = man : N - W cos 20∞ =
W v B2
g rB
or
Ê
v2 ˆ
N = W Á cos 20∞ + B ˜
g rB ¯
Ë
Sliding:
F = mk N
Ê
v2 ˆ
= m kW Á cos 20∞ + B ˜
g rB ¯
Ë
+
SFt = mat : W sin 20∞ - F =
W
at
g
or
at = g (sin 20∞ - mt cos 20∞) - m k
v B2
rB
Then
at = 9.81(sin 20∞ - 0.1cos 20∞) -
0.1(7.5 m/s)2
18
at = 2.12 m/s2
or
20° b
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71
Problem 12.47 (continued)
(b)
It is first necessary to determine vC .
+
For section BC
SFy = 0 : N BC - W cos 20∞ = 0
or
N BC = W cos 20∞
Sliding:
FBC = m k N BC = m kW cos 20∞
+
SFx = maBC : W sin 20∞ - FBC =
W
aBC
g
aBC = g (sin 20∞ - m k cos 20∞)
or
= (9.81 m/s2 )(sin 20∞ - 0.1cos 20∞)
= 2.43338 m/s2
For this uniformly accelerated motion we have
vC2 = v B2 + 2aBC D x BC
= (7.5 m/s)2 + 2(2.43338 m/s2 )(12 m)
vC = 10.7075 m/s
or
Now, just after C, rC = 42 m
+
SFn = man : W cos 20∞ - N =
W vC2
g rC
or
Ê
v2 ˆ
N = W Á cos 20∞ - C ˜
g rC ¯
Ë
Sliding:
F = mk N
Ê
v2 ˆ
= m kW Á cos 20∞ - C ˜
g rC ¯
Ë
+
or
Note:
Then
SFt = mat : W sin 20∞ - F =
W
at
g
at = g (sin 20∞ - m k cos 20∞) + m k
vC2
rC
g (sin 20∞ - m k cos 20∞) = aBC
at = 2.43338 m/s2 + 0.1
(10.7075 m/s)2
= 2.7068 m/s2
42 m
at = 2.71 m/s2
or
20° b
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72
PROBLEM 12.48
A series of small packages, each with a mass of 0.5 kg are discharged from a
conveyor belt as shown. Knowing that the coefficient of static friction between
each package and the conveyor belt is 0.40, determine (a) the force exerted by
the belt on a package just after it has passed Point A, (b) the angle q defining
Point B where the packages first slip relative to the belt.
Solution
Assume package does not slip.
at = 0, F f m s N
On the curved portion of the belt
an =
v 2 (1 m/s)2
=
= 4 m/s2
r 0.250 m
+
For any angle q
SFy = ma y : N - mg cosq = - man = -
mv 2
r
+
N = mg cosq -
mv 2
r
SFx = max : -F f + mg sinq = mat = 0
F f = mg sinq (a)
At Point A,
At Point B,
(2)
q = 0∞
N = (0.5)(9.81)(1.000) - (0.5)( 4)
(b)
(1)
N = 2.905 N b
Ff = ms N
mg sin q = m s ( mg cos q - man )
Ê
a ˆ
4 ˘
È
sin q = m s Á cos q - n ˜ = 0.40 Ícos q Ë
g¯
9
.
81˙˚
Î
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73
Problem 12.48 (continued)
Squaring and using trigonometic identities,
1 - cos2 q = 0.16 cos2 q - 0.130479 cos q + 0.026601
1.16 cos2 q - 0.130479 cos q - 0.97340 = 0
cos q = 0.97402
q = 13.09∞ b
Check that package does not separate from the belt.
N=
Ff
ms
=
mg sin q
ms
N 0.
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74
PROBLEM 12.49
A 54-kg pilot files a jet trainer in a half vertical loop of 1200-m radius
so that the speed of the trainer decreases at a constant rate. Knowing
that the pilot’s apparent weights at Points A and C are 1680 N and
350 N, respectively, determine the force exerted on her by the seat of
the trainer when the trainer is at Point B.
Solution
+
First we note that the pilot’s apparent weight is equal to the vertical force that she exerts on the seat of the jet
trainer.
v2
At A: SFn = man : N A - W = m A
r
Ê 1680 N
ˆ
2
v A = (1200 m) Á
- 9.81 m/s2 ˜
or
Ë 54 kg
¯
= 25, 561.3 m2 /s2
+
At C:
SFn = man : N C + W = m
or
vC2
r
Ê 350 N
ˆ
vC2 = (1200 m) Á
+ 9.81 m/s2 ˜
Ë 54 kg
¯
= 19, 549.8 m2 /s2
Since at = constant, we have from A to C
vC2 = v 2A + 2at Ds AC
or
or
19, 549.8 m2 /s2 = 25, 561.3 m2 /s2 + 2at (p ¥ 1200 m)
at = -0.79730 m/s2
Then from A to B
v B2 = v 2A + 2at Ds AB
Êp
ˆ
= 25,561.3 m2/s2 + 2( -0.79730 m/s2 ) Á ¥ 1200 m˜
¯
Ë2
= 22,555 m2/s2
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75
Problem 12.49 (continued)
+
At B:
SFn = man : N B = m
v B2
r
22,555 m2 /s2
1200 m
or
N B = 54 kg
or
N B = 1014.98 N ¨
+
SFt = mat : W + PB = m | at |
or
PB = (54 kg)(0.79730 - 9.81) m/s2
or
PB = 486.69 N ≠
Finally,
( Fpilot ) B = N B2 + PB2 = (1014.98)2 + ( 486.69)2
= 1126 N
( Fpilot ) B = 1126 N
or
25.6° b
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76
PROBLEM 12.50
A 250-g block B fits inside a small cavity cut in arm OA, which rotates
in the vertical plane at a constant rate such that v = 3 m/s. Knowing
that the spring exerts on block B a force of magnitude P = 1.5 N and
neglecting the effect of friction, determine the range of values of q for
which block B is in contact with the face of the cavity closest to the
axis of rotation O.
Solution
+
SFn = man : P + mg sinq - Q = m
v2
r
To have contact with the specified surface, we need Q 0,
or
Q = P + mg sinq -
sinq Data:
mv 2
0
r
1 Ê v2 P ˆ
g ÁË r m ˜¯
(1)
m = 0.250 kg, v = 3 m/s, P = 1.5 N, r = 0.9 m
Substituting into (1):
sin q 1 È (3)2 1.5 ˘
Í
˙
9.81 Î 0.9 0.25 ˚
sin q 0.40775
24.1∞ q 155.9∞ b
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77
PROBLEM 12.51
A curve in a speed track has a radius of 300 m and a rated speed of 190 km/h. (See
sample Problem 12.6 for the definition of rated speed). Knowing that a racing
car starts skidding on the curve when traveling at a speed of 290 km/h, determine
(a) the banking angle q, (b) the coefficient of static friction between the tires and
the track under the prevailing conditions, (c) the minimum speed at which the
same car could negotiate that curve.
Solution
W = mg
Weight
Acceleration
a=
v2
r
SFx = max : F + W sin q = ma cos q
F=
mv 2
cos q - mg sin q r
(1)
SFy = ma y : N - W cos q = ma sin q
N=
(a)
mv 2
sin q + mg cos q r
Banking angle. Rated speed v = 190 km / h =
0=
(2)
475
m / s. F = 0 at rated speed.
9
mv 2
cos q - mg sin q
r
2
Ê 475 ˆ
ÁË
˜
v2
9 ¯
tan q =
=
r g (300)(9.81)
q = 43.425∞
(b)
Slipping outward.
q = 43.5∞ b
Ê 725 ˆ
v = 290 km / h = Á
m/s
Ë 9 ˜¯
F = mN
m=
F v 2 cos q - r g sin q
=
N v 2 sin q + r g cos q
2
m=
Ê 725 ˆ
ÁË
˜ cos ( 43.425∞) - (300)(9.81)sin 43.425∞
9 ¯
2
Ê 725 ˆ
ÁË
˜ sin ( 43.425∞) + (300)(9.81) cos 43.425∞
9 ¯
= 0.40768
m = 0.408 b
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78
Problem 12.51 (continued)
(c)
Minimum speed.
F = - mN
-m =
v 2 cos q - r g sin q
v 2 sin q + r g cos q
r g (sin q - m cos q )
v2 =
cos q + m siin q
=
(300)(9.81)(sin 43.425∞ - 0.40768 cos 43.425∞)
(cos 43.425∞ + 0.40768 sin 43.425∞)
= 1144 m2 /s2
v = 33.8258 m/s
v = 121.8 km/h b
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79
PROBLEM 12.52
A car is traveling on a banked road at a constant speed v. Determine the range of values of v for which the car
does not skid. Express your answer in terms of the radius r of the curve, the banking angle q, and the angle of
static friction f s between the tires and the pavement.
Solution
Case 1: v = vmax
Note: R = F + N
+
SFn = man :
+
SFy = 0:
or
(1)
Forming ( 2)
R sin (q + f s )
=
R cos (q + f s )
m
2
vmax
r
R cos (q + f s ) - W = 0
R sin (q + f s ) = m
(1)
R cos (q + f s ) = mg (2)
2
vmax
r
mg
vmax = gr tan (q + f s )
or
Case 2: v = vmin
Note: R = F + N
+
+
2
vmin
r
R cos (q - f s ) - W = 0
SFn = man :
S Fy = 0:
or
( 3)
Forming ( 4 )
or
R sin (q - f s )
=
R cos (q - f s )
R sin (q - f s ) = m
(3)
R cos (q - f s ) = mg (4)
2
vmin
r
mg
m
vmin = gr tan (q - f s )
For the car not to skid
gr tan (q - f s ) v gr tan (q + f s ) b
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80
PROBLEM 12.53
Tilting trains, such as the American Flyer, which will run from
Washington to New York and Boston, are designed to travel safely
at high speeds on curved sections of track, which were built for
slower, conventional trains. As it enters a curve, each car is tilted
by hydraulic actuators mounted on its trucks. The tilting feature of
the cars also increases passenger comfort by eliminating or greatly
reducing the side force Fs (parallel to the floor of the car) to which
passengers feel subjected. For a train traveling at 160 km/h on a
curved section of track banked through an angle q = 6∞ and with a
rated speed of 100 km/h, determine (a) the magnitude of the side
force felt by a passenger of weight W in a standard car with no
tilt (f = 0), (b) the required angle of tilt f if the passenger is to feel
no side force. (See Sample Problem 12.6 for the definition of rated
speed.)
Solution
vR = 100 km/h =
Rated speed:
250
400
m/s, 160 km/h =
m/s
9
9
From Sample Problem 12.6,
vR2 = g r tan q
2
Ê 250 ˆ
ÁË
˜
2
vR
9 ¯
r=
=
= 748.352 m
g tan q 9.81 tan 6∞
or
Let the x-axis be parallel to the floor of the car.
+
SFx = max : Fs + W sin (q + f ) = man cos (q + f )
=
(a)
f = 0.
mv 2
cos (q + f )
r
È v2
˘
Fs = W Í cos (q + f ) - sin (q + f ) ˙
Î gr
˚
2
˘
È
Ê 400 ˆ
˙
Í
ÁË
˜¯
9
=W Í
cos 6∞ - sin 6∞˙
˙
Í (9.81)(748.352)
˙
Í
˚
Î
= 0.16306W
Fs = 0.1631W b
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81
Problem 12.53 (continued)
(b)
For Fs = 0,
v2
cos (q + f ) - sin (q + f ) = 0
gr
2
Ê 400 ˆ
ÁË
˜
v2
9 ¯
tan (q + f ) =
=
= 0.26907
g r (9.81)(748.352)
q + f = 15.060∞∞
f = 15.06∞ - 6∞
f = 9.06∞ b
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82
PROBLEM 12.54
Tests carried out with the tilting trains described in Problem 12.53
revealed that passengers feel queasy when they see through the car
windows that the train is rounding a curve at high speed, yet do not
feel any side force. Designers, therefore, prefer to reduce, but not
eliminate that force. For the train of Problem 12.53, determine the
required angle of tilt f if passengers are to feel side forces equal to
10% of their weights.
PROBLEM 12.53 Tilting trains, such as the American Flyer, which
will run from Washington to New York and Boston, are designed to
travel safely at high speeds on curved sections of track, which were
built for slower, conventional trains. As it enters a curve, each car
is tilted by hydraulic actuators mounted on its trucks. The tilting
feature of the cars also increases passenger comfort by eliminating
or greatly reducing the side force Fs (parallel to the floor of the car)
to which passengers feel subjected. For a train traveling at 160 km/h
on a curved section of track banked through an angle q = 6∞ and
with a rated speed of 100 km/h, determine (a) the magnitude of the
side force felt by a passenger of weight W in a standard car with no
tilt (f = 0), (b) the required angle of tilt f if the passenger is to feel
no side force. (See Sample Problem 12.6 for the definition of rated
speed.)
Solution
vR = 100 km/h =
Rated speed:
250
400
m/s, 160 km/h =
m/s
9
9
From Sample Problem 12.6,
vR2 = g r tan q
2
Ê 250 ˆ
ÁË
˜
vR2
9 ¯
r=
=
= 748.352 m
g tan q 9.81 tan 6∞
or
Let the x-axis be parallel to the floor of the car.
+
SFx = max : Fs + W sin (q + f ) = man cos (q + f )
=
Solving for Fs,
mv 2
cos (q + f )
r
È v2
˘
Fs = W Í cos (q + f ) - sin (q + f ) ˙
Î gr
˚
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83
Problem 12.54 (continued)
2
Ê 400 ˆ
ÁË
˜
v
9 ¯
=
= 0.26907 and
g r (9.81)(748.352)
2
Now
So that
Let
Then
Fs = 0.10W
0.10W = W [0.26907 cos (q + f ) - sin (q + f )]
u = sin (q + f )
cos (q + f ) = 1 - u 2
0.10 = 0.26907 1 - u 2 - u or 0.26907 1 - u 2 = 0.10 + u
Squaring both sides,
or
0.072399(1 - u 2 ) = 0.01 + 0.2u + u 2
1.072399u 2 + 0.2u - 0.062399 = 0
The positive root of the quadratic equation is u = 0.16537
Then,
q + f = sin -1 u = 9.5187∞
f = 9.5187∞ - q f = 3.52∞ b
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84
PROBLEM 12.55
A small, 300-g collar D can slide on portion AB of a rod which is bent as shown. Knowing
that a = 40∞ and that the rod rotates about the vertical AC at a constant rate of 5 rad/s,
determine the value of r for which the collar will not slide on the rod if the effect of friction
between the rod and the collar is neglected.
Solution
vD = rq& ABC
First note
+
N=
or
+
or
or
SFy = 0 : N sin 40∞ - W = 0
mg
sin 40∞
SFn = man : N cos 40∞ = m
vD2
r
( rq& ABC )2
mg
cos 40∞ = m
sin 40∞
r
r=
g
2
q ABC
1
tan 40∞
9.81 m/s2
1
2
(5 rad/s) tan 40∞
= 0.468 m
=
r = 468 mm b
or
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85
PROBLEM 12.56
A small, 200-g collar D can slide on portion AB of a rod which is bent as shown. Knowing
that the rod rotates about the vertical AC at a constant rate and that a = 30∞ and r = 600 mm,
determine the range of values of the speed v for which the collar will not slide on the rod if
the coefficient of static friction between the rod and the collar is 0.30.
Solution
Case 1: v = vmin , impending motion downward
+
SFx = max : N - W sin 30∞ = m
v2
cos 30∞
r
Ê
ˆ
v2
N = m Á g sin 30∞ + cos 30∞˜
r
Ë
¯
or
+
SFy = ma y : F - W cos 30∞ = - m
v2
sin 30∞
r
or
Ê
ˆ
v2
F = m Á g cos 30∞ - sin 30∞˜
r
Ë
¯
Now
F = ms N
Then
or
Ê
ˆ
Ê
ˆ
v2
v2
m Á g cos 30∞ - sin 30∞˜ = m s ¥ m Á g sin 30∞ + cos 30∞˜
r
r
Ë
¯
Ë
¯
v 2 = gr
1 - m s tan 30∞
m s + tan 30∞
= (9.81 m/s2 )(0.6 m)
or
1 - 0.3 tan 30∞
0.3 + tan 30∞
vmin = 2.36 m/s
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86
Problem 12.56 (continued)
Case 2: v = vmax , impending motion upward
+
SFx = max : N - W sin 30∞ = m
v2
cos 30∞
r
Ê
ˆ
v2
N = m Á g sin 30∞ + cos 30∞˜
r
Ë
¯
or
+
SFy = ma y : F + W cos 30∞ = m
v2
sin 30∞
r
or
Ê
ˆ
v2
F = m Á - g cos 30∞ + sin 30∞˜
r
Ë
¯
Now
F = ms N
Then
Ê
ˆ
Ê
ˆ
v2
v2
m Á - g cos 30∞ + sin 30∞˜ = m s ¥ m Á g sin 30∞ + cos 30∞˜
r
r
Ë
¯
Ë
¯
or
v 2 = gr
1 + m s tan 30∞
tan 30∞ - m s
= (9.81 m/s2 )(0.6 m)
or
1 + 0.3 tan 30∞
tan 30∞ - 0.3
vmax = 4.99 m/s
For the collar not to slide
2.36 m/s v 4.99 m/s b
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87
PROBLEM 12.57
A small, 250 g collar D can slide on portion AB of a rod which is bent as shown. Knowing
that r = 200 mm and that the rod rotates about the vertical AC at a constant rate of 10 rad/s,
determine the smallest allowable value of the coefficient of static friction between the collar
and the rod if the collar is not to slide when (a) a = 15∞, (b) a = 45∞. Indicate for each case
the direction of the impending motion.
Solution
First we note that v = rq& ABC = (0.2 m)(10 rad/s) = 2 m/s, and that requiring m s = ( m s ) min implies that sliding of
collar D is impending.
m s = tan f s
Also,
Now we consider the two possible cases of impending motion.
Case 1: Impending motion downward.
+
SFx = max : N - W sin a =
W v2
cos a
g r
Ê
ˆ
v2
N = W Á sin a - cos a ˜
gr
Ë
¯
or
+
SFy = ma y : F - W cos a = -
W v2
sin a
g r
or
Ê
ˆ
v2
F = W Á cos a - sin a ˜
gr
Ë
¯
Now
F = ms N
Then
or
Ê
ˆ
Ê
ˆ
v2
v2
W Á cos a - sin a ˜ = m s ¥ W Á sin a + cos a ˜
gr
gr
Ë
¯
Ë
¯
v 2 1 - m s tan a 1 - tan f s tan a
=
=
tan a + tan f s
gr tan a + m s
=
1
tan (a + f s )
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88
Problem 12.57 (continued)
Case 2: Impending motion upward.
+
SFx = max : N - W sin a =
W v2
cos a
g r
Ê
ˆ
v2
N = W Á sin a + cos a ˜
gr
Ë
¯
or
+
SFy = ma y : F + W cos a =
W v2
sin a
g r
or
Ê
ˆ
v2
F = W Á - cos a + sin a ˜
gr
Ë
¯
Now
F = ms N
Then
Ê
ˆ
Ê
ˆ
v2
v2
W Á - cos a + sin a ˜ = m s ¥ W Á sin a + cos a ˜
gr
gr
Ë
¯
Ë
¯
v 2 1 + m s tan a 1 + tan a tan f s
=
=
tan a - tan f s
gr tan a - m s
or
=
1
tan (a - f s )
gr (9.81 m/s2 )(0.2 m)
=
= 0.4905
(2 m/s)2
v2
Now
tan (a ± f s ) = 0.4905
Then
a ± f s = 26.128∞, f s 0
or
and where the “+” corresponds to impending motion downward and the “–” to impending motion upward.
(a)
a = 15∞ :
We have
f s 0 fi “+” so that
15∞ ± f s = 26.128∞
f s = 11.128∞
Then
( m s ) min = tan 11.128∞
or
( m s ) min = 0.1967, motion impending downward
b
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89
Problem 12.57 (continued)
(b)
a = 45∞ :
We have
f s 0 fi “–”
Then
or
45∞ ± f s = 26.128∞
so that
f s = 18.872∞
( m s ) min = tan 18.872∞
m s = 0.342 motion impending upward
b
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90
PROBLEM 12.58
A semicircular slot of 250-mm radius is cut in a flat plate which rotates about
the vertical AD at a constant rate of 14 rad/s. A small, 400-g block E is designed
to slide in the slot as the plate rotates. Knowing that the coefficients of friction
are m s = 0.35 and m k = 0.25, determine whether the block will slide in the slot
if it is released in the position corresponding to (a) q = 80∞, (b) q = 40∞. Also
determine the magnitude and the direction of the friction force exerted on the
block immediately after it is released.
Solution
r = (650 - 250 sin q ) mm
First note
=(0.65 - 0.25 sin q ) m
vE = rf& ABCD
an =
Then
vE2
= r(f& ABCD )2
r
= (0.65 - 0.25 sin q ) m (14 rad/s)2
= (127.4 - 49 sin q ) m/s2
Assume that the block is at rest with respect to the plate.
+
vE2
sin q
r
Ê
ˆ
v2
N = W Á - cos q + E sin q ˜
gr
Ë
¯
or
+
or
SFx = max : N + W cos q = m
SFy = ma y : -F + W sin q = - m
vE2
cos q
r
Ê
ˆ
v2
F = W Á sin q + E cos q ˜
gr
Ë
¯
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91
Problem 12.58 (continued)
(a)
We have
q = 80∞
Then
È
˘
1
N = (0.4 ¥ 9.81) Í- cos 80∞ +
¥ (127.4 - 49 sin 80∞) ¥ sin 80∞˙
2
9.81 m/s
Î
˚
= 30.4954 N
È
˘
1
F = (0.4 ¥ 9.81) Ísin 80∞ +
¥ (127.4 - 49 sin 80∞) ¥ cos 80∞˙
2
9.81m/s
Î
˚
= 9.3617 N
Now
Fmax = m s N = 0.35(30.4954) = 10.6734 N
The block does not slide in the slot, and
F = 9.3617 N
(b)
We have
80° b
q = 40∞
Then
1
È
˘
N = (0.4 ¥ 9.81) Í- cos 40∞ +
¥ (127.4 - 49 sin 40∞) ¥ sin 40∞˙
9.81
Î
˚
= 21.652 N
1
È
˘
F = (0.4 ¥ 9.81) Ísin 40∞ +
¥ (127.4 - 49 sin 40∞) ¥ cos 40∞˙
9.81
Î
˚
= 31.9088 N
Now
Fmax = m s N , from which it follows that
F Fmax
Block E will slide in the slot
and
a E = a n + a E /plate
= a n + (a E /plate )t + (a E /plate ) n
+
At t = 0, the block is at rest relative to the plate, thus (a E /plate ) n = 0 at t = 0, so that a E /plate must be directed
tangentially to the slot.
v2
SFx = max : N + W cos 40∞ = m E sin 40∞
r
or
Ê
ˆ
v2
N = W Á - cos 40∞ + E sin 40∞˜
gr
Ë
¯
(as above)
= 21.652 N
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92
Problem 12.58 (continued)
Sliding:
F = mk N
= 0.25(21.652 N)
= 5.413 N
Noting that F and a E /plane must be directed as shown (if their directions are reversed, then SFx is
while ma xis ), we have
the block slides downward in the slot and
F = 5.413 N
40° b
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93
PROBLEM 12.59
Three seconds after a polisher is started from rest, small tufts of
fleece from along the circumference of the 225-mm-diameter
polishing pad are observed to fly free of the pad. If the polisher
is started so that the fleece along the circumference undergoes a
constant tangential acceleration of 4 m/s2 , determine (a) the speed
v of a tuft as it leaves the pad, (b) the magnitude of the force required
to free a tuft if the average mass of a tuft is 1.6 mg.
Solution
(a)
at = constant fi uniformly acceleration motion
Then
v = 0 + at t
At t = 3 s:
v = (4 m/s2 )(3 s)
v = 12.00 m/s b
or
(b)
SFt = mat : Ft = mat
or
Ft = (1.6 ¥ 10 -6 kg)( 4 m/s2 )
= 6.4 ¥ 10 -6 N
SFn = man : Fn = m
At t = 3 s:
Fn = (1.6 ¥ 10 -6 kg)
v2
r
(12 m/s)2
Ê 0.225 ˆ
m˜
ÁË
¯
2
= 2.048 ¥ 10 -3 N
Finally,
Ftuft = Ft2 + Fn2
= (6.4 ¥ 10 -6 N )2 + (2.048 ¥ 10 -3 N)2
Ftuft = 2.05 ¥ 10 -3 N b
or
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94
PROBLEM 12.60
A turntable A is built into a stage for use in a theatrical production.
It is observed during a rehearsal that a trunk B starts to slide on
the turntable 10 s after the turntable begins to rotate. Knowing that
the trunk undergoes a constant tangential acceleration of 0.24 m/s2 ,
determine the coefficient of static friction between the trunk and
the turntable.
Solution
First we note that ( aB )t = constant implies uniformly accelerated motion.
v B = 0 + ( a B )t t
v B = (0.24 m/s2 )(10 s) = 2.4 m/s
At t = 10 s:
In the plane of the turntable
SF = mB a B : F = mB (a B )t + mB (a B ) n
F = mB ( aB )t2 + ( aB )2n
Then
( aB )t2
= mB
+
Ê v2 ˆ
+Á B˜
Ë r¯
SFy = 0 : N - W = 0
or
N = mB g
At t = 10 s:
F = m s N = m s mB g
Then
2
m s mB g = mB
( aB )t2
Ê v2 ˆ
+Á B˜
Ë r¯
2
1/ 2
or
2
Ï
È (2.4 m/s)2 ˘ ¸Ô
Ô
2 2
0
24
ms =
+
(
.
m/s
)
Ì
Í
˙ ˝
9.81 m/s2 Ô
Î 2.5 m ˚ Ô˛
Ó
1
m s = 0.236 b
or
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95
PROBLEM 12.61
The parallel-link mechanism ABCD is
used to transport a component I between
manufacturing processes at stations E, F,
and G by picking it up at a station when
q = 0 and depositing it at the next station
when q = 180º. Knowing that member
BC remains horizontal throughout its
motion and that links AB and CD rotate at
a constant rate in a vertical plane in such
a way that vB = 0.7 m/s, determine (a) the
minimum value of the coefficient of static
friction between the component and BC
if the component is not to slide on BC
while being transferred, (b) the values of
q for which sliding is impending.
SOLUTION
SFx = max : F =
+
W v B2
cos q
g r
+
SFy = ma y : N - W = -
or
Now
W v B2
sin q
g r
Ê
ˆ
v2
N = W Á1 - B sin q ˜
gr
Ë
¯
Ê
ˆ
v2
Fmax = m s N = m s W Á1 - B sin q ˜
gr
Ë
¯
and for the component not to slide
F £ Fmax
or
or
Ê
ˆ
v2
W v B2
cos q m s W Á1 - B sin q ˜
g r
Ë gr
¯
ms cos q
gr
- sin q
v B2
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96
Problem 12.61 (continued)
We must determine the values of q which maximize the above expression. Thus
Ê gr
ˆ
Ê cos q ˆ - sin q Á 2 - sin q ˜ - (cos q ) ( - cos q )
v
Ë
¯
d Á
B
˜=
gr
=0
2
dq Á 2 - sin q ˜
g
r
Ê
ˆ
Ë vB
¯
ÁË v 2 - sin q ˜¯
B
or
sin q =
Now
sin q =
for
m s = ( m s ) min
(0.7 m/s)2
(9.81 m/s2 ) (0.25 m)
q = 11.525º
or
(a)
v B2
gr
and
= 0.199796
q = 168.475º
From above,
v2
cos q
where sin q = B
gr
gr
- sin q
2
vB
cos q sin q
cos q
=
= tan q
( m s ) min =
2
1
1
sin
q
- sin q
sin q
= tan 11.525∞
( m s ) min =
(ms)min = 0.204 b
or
(b)
We have impending motion
to the left for
q = 11.53° b
to the right for
q = 168.5° b
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97
PROBLEM 12.62
Knowing that the coefficients of friction between the component I and member BC of the mechanism of
Problem 12.61 are ms = 0.35 and mk = 0.25, determine (a) the maximum allowable constant speed vB if the
component is not to slide on BC while being transferred, (b) the values of q for which sliding is impending.
SOLUTION
SFx = max : F =
+
W v B2
cos q
g r
+
SFy = ma y : N - W = -
W v B2
sin q
g r
Ê
ˆ
v2
N = W Á1 - B sin q ˜
Ë gr
¯
or
Fmax = m s N
Now
Ê
ˆ
v2
= m s W Á1 - B sin q ˜
Ë gr
¯
and for the component not to slide
F Fmax
or
or
Ê
ˆ
v2
W v B2
cos q m s W Á1 - B sin q ˜
g r
gr
Ë
¯
v B2 m s
( )
To ensure that this inequality is satisfied, v B2
gr
cos q + m s sin q
max
(1)
must be less than or equal to the minimum value of
m s g r / (cos q + m s sin q ), which occurs when (cos q + m s sin q ) is maximum. Thus
d
(cos q + m s sin q ) = - sin q + m s cos q = 0
dq
or
tan q = m s
m s = 0.35
or
q = 19.2900∞
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98
Problem 12.62 (continued)
(a)
The maximum allowed value of v B is then
(v )
2
B
max
= ms
gr
cos q + m s sin q
= gr
tan q
= gr sin q
cos q + (tan q ) sin q
where tan q = m s
= (9.81)(0.25m ) sin 19.29∞
( v B ) max = 0.900 m/s b
or
(b)
First note that for 90∞ q 180∞, Eq. (1) becomes
v B2 m s
gr
cos a + m s sin a
where a = 180∞ - q . It then follows that the second value of q for which motion is impending is
q = 180∞ - 19.2900∞
= 160.7100∞
we have impending motion
to the left for
q = 19.29∞ b
to the right for
q = 160.7∞ b
Alternative solution.
SF = ma : W + R = ma n
Then
For impending motion, f = f s . Also, as shown above, the values of q for which motion is impending
(
minimize the value of vB, and thus the value of an is an =
that an is minimum when ma n and R are perpendicular.
v B2
r
). From the above diagram, it can be concluded
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99
Problem 12.62 (continued)
Therefore,
from the diagram
q = f s = tan -1 m s
and
or
or
(as above)
man = W sin f s
m
v B2
= mg sin q
r
v B2 = g r sin q
(as above)
For 90∞ q 180∞, we have
from the diagram
a = 180∞ - q
(as above)
a = fs
and
or
man = W sin f s
v B2 = g r sin q
(as above)
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100
PROBLEM 12.63
In the cathode-ray tube shown, electrons emitted by the cathode
and attracted by the anode pass through a small hole in the anode
and then travel in a straight line with a speed v0 until they strike the
screen at A. However, if a difference of potential V is established
between the two parallel plates, the electrons will be subjected to
a force F perpendicular to the plates while they travel between the
plates and will strike the screen at Point B, which is at a distance d
from A. The magnitude of the force F is F = eV /d , where -e is the
charge of an electron and d is the distance between the plates. Derive
an expression for the deflection d in terms of V, v0 , the charge -e
and the mass m of an electron, and the dimensions d, l, and L.
SOLUTION
Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = l
at the right edge of the plate. At the screen,
x=
l
+L
2
Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0.
Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when
it impacts on the screen. For uniform horizontal motion,
x = v0 t ,
so that
t1 =
and
t2 =
l
v0
l
2v0
+
L
.
v0
Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection
produced by the electric force. While the electron is between plates (0 t t1 ), the vertical force on the
electron is Fy = eV /d . After it passes the plates (t1 t t2 ), it is zero.
For 0 t t1 ,
SFy = ma y : a y =
Fy
m
=
eV
md
v y = ( v y )0 + a y t = 0 +
eVt
md
eVt 2
1
y = y0 + ( v y )0 t + a y t 2 = 0 + 0 +
2
2md
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101
Problem 12.63 (continued)
At t = t1 ,
( v y )1 =
eVt1
md
and y1 =
eVt12
2md
For t1 t t2 , a y = 0
y = y1 + ( v y )1 (t - t1 )
At t = t2 ,
y2 = d = y1 + ( v y )1 (t2 - t1 )
d=
=
eVt12 eVt1
eVt
1
+
(t2 - t1 ) = 1 ÊÁË t2 - t1 ˆ˜¯
2md md
2
md
eV l Ê l
L 1 lˆ
+ Á
mdv0 Ë 2v0 v0 2 v0 ˜¯
or
d=
eV lL
mdv02
b
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102
PROBLEM 12.64
In Problem 12.63, determine the smallest allowable value of the ratio
d/l in terms of e, m, v0, and V if at x = l, the minimum permissible
distance between the path of the electrons and the positive plate is
0.05d .
PROBLEM 12.63 In the cathode-ray tube shown, electrons emitted
by the cathode and attracted by the anode pass through a small hole
in the anode and then travel in a straight line with a speed v0 until
they strike the screen at A. However, if a difference of potential V
is established between the two parallel plates, the electrons will be
subjected to a force F perpendicular to the plates while they travel
between the plates and will strike the screen at Point B, which is
at a distance d from A. The magnitude of the force F is F = eV /d ,
where –e is the charge of an electron and d is the distance between
the plates. Neglecting the effects of gravity, derive an expression for
the deflection d in terms of V, v0, the charge –e and the mass m of
an electron, and the dimensions d, l, and L.
SOLUTION
Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = l
at the right edge of the plate. At the screen,
x=
l
+L
2
Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0.
Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when
it impacts on the screen. For uniform horizontal motion,
x = v0 t ,
so that
t1 =
and
t2 =
l
v0
l
2v0
+
L
.
v0
Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection
produced by the electric force. While the electron is between the plates (0 t t1 ), the vertical force on the
electron is Fy = eV/d . After it passes the plates (t1 t t2 ), it is zero.
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103
Problem 12.64 (continued)
Fy
=
eV
md
v y = ( v y )0 + a y t = 0 +
eVt
md
SFy = ma y : a y =
For 0 t t1 ,
m
1
eVt 2
y = y0 + ( v y )0 t + a y t 2 = 0 + 0 +
2
2md
At t = t1 ,
But
so that
l
v0
, y=
y
eV l2
2mdv02
d
- 0.05d = 0.450 d
2
eV l2
0.450 d
2mdv02
d2
1
eV
eV
= 1.111 2 2
2
0.450 2mv0
l
mv0
d
eV
1.054
b
l
mv02
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104
PROBLEM 12.65
The current model of a cathode-ray tube is to be modified so that the
length of the tube and the spacing between the plates are reduced
by 40 percent and 20 percent, respectively. If the size of the screen
is to remain the same, determine the new length l ¢ of the plates,
assuming that all of the other characteristics of the tube are to remain
unchanged. (See Problem 12.63 for a description of a cathode-ray
tube.)
PROBLEM 12.63 In the cathode-ray tube shown, electrons emitted
by the cathode and attracted by the anode pass through a small hole
in the anode and then travel in a straight line with a speed v0 until
they strike the screen at A. However, if a difference of potential V
is established between the two parallel plates, the electrons will be
subjected to a force F perpendicular to the plates while they travel
between the plates and will strike the screen at Point B, which is
at a distance d from A. The magnitude of the force F is F = eV /d ,
where –e is the charge of an electron and d is the distance between
the plates. Neglecting the effects of gravity, derive an expression for
the deflection d in terms of V, v0, the charge –e and the mass m of an
electron, and the dimensions d, l, and L.
SOLUTION
Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = l
at the right edge of the plate. At the screen,
x=
l
+L
2
Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0.
Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when
it impacts on the screen. For uniform horizontal motion,
x = v0 t ,
so that
t1 =
and
t2 =
l
v0
l
2v0
+
L
.
v0
Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection
produced by the electric force. While the electron is between the plates (0 t t1 ), the vertical force on the
electron is Fy = eV /d . After it passes the plates (t1 t t2 ), it is zero.
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105
Problem 12.65 (continued)
For 0 t t1 ,
Fy
eV
md
eVt
v y = ( v y )0 + a y t = 0 +
md
eVt 2
1
y = y0 + ( v y )0 t + a y t 2 = 0 + 0 +
2
2md
SFy = ma y : a y =
m
=
Let d, l, L, and d be dimensions taken from the current model and d ¢, l ¢, L¢, and d¢ be those of the modified
model.
d=
d¢ =
eV lL
mdv02
eV l ¢L ¢
md ¢v02
Since e, V, m, and v0 are constants
d ¢ Ê l¢ ˆ Ê L¢ ˆ Ê d ˆ
=Á ˜Á ˜Á ˜
d Ë l ¯ Ë L ¯ Ë d¢¯
But
Then
d ¢ = d , L ¢ = L - 0.40 L = 0.6 L,
d ¢ = d - 0.20 L = 0.8 d
d ¢ Ê l ¢ ˆ Ê 0.6 L ˆ Ê d ˆ
=Á ˜Á
˜
˜Á
d Ë l ¯ Ë L ¯ Ë 0.8d ¯
L = 0.75
l¢
l
l ¢ = 1.333 l b
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106
PROBLEM 12.66
Rod OA rotates about O in a horizontal plane. The motion of the 300 g
collar B is defined by the relations r = 300 + 100 cos (0.5pt) and
q = p(t2 - 3t), where r is expressed in millimeters, t in seconds, and q
in radians. Determine the radial and transverse components of the force
exerted on the collar when ( a) t = 0, (b) t = 0.5 s.
SOLUTION
Polar coordinates and their derivatives.
(a)
For t = 0.
r = 0.3 + 0.1 cos (0.5p t ) m
q = p (t 2 - 3t ) rad
r& = -0.05p sin (0.5p t ) m/s
q& = p (2t - 3) rad/s
&&
r = -.025p 2 cos (0.5p t ) m/s2
q&& = 2p rad/s2
r = 0.4 m
q =0
r& = 0
q& = -3p rad/s
&&
r = -0.025p 2 m/s2
q&& = 2p rad/s2
Components of acceleration.
ar = &&
r - rq& 2
= -0.025p 2 - 0.4( -3p )2
= -35.777 m/s2
aq = rq&& + 2r&q&
= (0.4)(2p ) + 2( -3p )(0)
= 2.513 m/s2
Components of force.
Fr = mar = (0.3)( -35.777) Fr = -10.73 N b
Fq = maq = (0.3)(2.513) (b)
For t = 0.5 s.
Fq = 0.754 N b
r = 0.37071 m
q = -3.927 rad
r& = -0.11107 m/s
q& = -2p rad/s
&&
r = -0.17447 m/s2
q&& = 2p rad/s2
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107
Problem 12.66 (continued)
Components of acceleration.
ar = &&
r - rq& 2
= -14.809 m/s2
aq = rq&& + 2r&q&
= 3.725 m/s2
Components of force.
Fr = mar = (0.3)( -14.809) Fr = -4.44 N b
Fq = maq = (0.3)(3.725) Fq = 1.118 N b
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108
PROBLEM 12.67
For the motion defined in Problem 12.66, determine the radial and
transverse components of the force exerted on the collar when t = 1.5 s.
PROBLEM 12.66 Rod OA rotates about O in a horizontal plane. The
motion of the 300 g collar B is defined by the relations r = 300 + 100 cos
(0.5pt) and q = p(t2 - 3t), where r is expressed in millimeters, t in seconds,
and q in radians. Determine the radial and transverse components of the
force exerted on the collar when ( a) t = 0, (b) t = 0.5 s.
SOLUTION
Polar coordinates and their derivatives.
For t = 1.5 s.
r = 0.3 + 0.1 cos (0.5p t ) m
q = p (t 2 - 3t ) rad
r& = -0.05p sin (0.5p t ) m/s
q& = p (2t - 3) rad/s
&&
r = -.025p 2 cos (0.5p t ) m/s2
q&& = 2p rad/s2
r = 0.22929 m
q = -7.068 rad
r& = -0.11107 m/s
q& = 0
&&
r = 0.17447 m/s2
q&& = 2p rad/s2
Components of acceleration.
ar = &&
r - rq& 2 = 0.17447 m/s2
a = rq&& + 2r&q& = 1.441 m/s2
q
Components of force.
Fr = mar = (0.3)(0.17447)
Fr = 0.0523 N b
Fq = maq = (0.3)(1.44107)
Fq = 0.432 N b
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109
PROBLEM 12.68
Rod OA oscillates about O in a horizontal plane. The motion of the 2.5 kg
collar B is defined by the relations r = 250/(t + 4) and q = (2/p) sin pt,
where r is expressed in mm, t in seconds, and q in radians. Determine the
radial and transverse components of the force exerted on the collar when
(a) t = 1 s, (b) t = 6 s.
SOLUTION
r=
Then
r& = -
and
&&
r=
(a)
At t = 1 s:
Now
ˆ
Ê2
q = Á sin p t ˜ rad
¯
Ëp
250
0.25
mm =
m
t+4
t+4
We have
0.25
m/s
(t + 4) 2
q& = (2 cos p t ) rad/s
0.5
m/s2
( t + 4 )3
q&& = -(2p sin p t ) rad/s2
r = 0.05 m
r& = -0.01 m/s
q = -2 rad/s
&&
r = 0.004 m/s2
q&& = 0
ar = &&
r - rq& 2
= (0.004 m/s2 ) - (0.05 m)( -2 rad/s)2
= -0.196 m/s2
and
aq = rq&& + 2r&q&
= 0 + 2( -0.01)( -2)
= 0.04 m/s2
Finally
Fr = mB ar
= 2.5 kg( -0.196 m/s2 )
Fr = -0.49 N b
or
Fq = mB aq
= (2.5 kg)(0.04 m/s2 )
Fq = 0.1 N b
or
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110
Problem 12.68 (continued)
(b)
At t = 6 s:
r = 0.025 m
r& = -0.0025 m/s
q& = 2 rad/s
&&
r = 5 ¥ 10 -4 m/s2
q&& = 0
Now
ar = &&
r - rq& 2 = 5 ¥ 10 -4 (0.025m )(2 rad/s2 ) = -0.0995 m/s2
and
aq = rq&& + 2r&q& = 0 + 2( -0.0025 m/s)(2 rad/s) = -0.01 m/s2
Finally
Fr = mB ar
(
= (2.5 kg ) -0.0995 m/s2
)
Fr = -0.249 N b
or
Fq = mB aq
(
= (2.5 kg ) -0.01 m/s2
or
)
Fq = -0.0250 N b
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111
PROBLEM 12.69
A collar B of mass m slides on the frictionless arm AA¢. The arm
is attached to drum D and rotates about O in a horizontal plane at
the rate q& = ct, where c is a constant. As the arm-drum assembly
rotates, a mechanism within the drum winds in the cord so that the
collar moves toward O with a constant speed k. Knowing that at
t = 0, r = r0, express as a function of m, c, k, r0, and t, (a) the tension
T in the cord, (b) the magnitude of the horizontal force Q exerted on
B by arm AA¢.
SOLUTION
Kinematics
dr
= r& = - k
dt
We have
At t = 0, r = r0 :
r
t
Ú r dr = Ú 0 -kdt
0
or
r = r0 - kt
Also,
&&
r=0
q& = ct
q&& = c
ar = &&
r - rq& 2
Now
= 0 - ( r0 - kt )(ct )2
= -c 2 ( r0 - kt )t 2
aq = rq&& + 2r&q&
and
= ( r0 - kt )(c) + 2( - k )(ct )
= c( r0 - 3kt )
(a)
+
Kinetics
SFr = mar :
-T = m[-c 2 ( r0 - kt )t 2 ]
T = mc 2( r0 - kt )t 2 b
or
+
(b)
SFq = maq :
Q = m[c( r0 - 3kt )]
Q = mc( r0 - 3kt ) b
or
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112
PROBLEM 12.70
The 3-kg collar B slides on the frictionless arm AA¢. The arm is
attached to drum D and rotates about O in a horizontal plane at the
rate q& = 0.75t , where q& and t are expressed in rad/s and seconds,
respectively. As the arm-drum assembly rotates, a mechanism within
the drum releases cord so that the collar moves outward from O with
a constant speed of 0.5 m/s. Knowing that at t = 0, r = 0, determine
the time at which the tension in the cord is equal to the magnitude of
the horizontal force exerted on B by arm AA¢.
SOLUTION
Kinematics
dr
= r& = 0.5 m/s
dt
We have
r
t
Ú 0 dr = Ú 0 0.5 dt
At t = 0, r = 0 :
or
r = (0.5t ) m
Also,
&&
r=0
q& = (0.75t ) rad/s
q&& = 0.75 rad/s2
Now
ar = &&
r - rq& 2 = 0 - [(0.5t ) m][(0.75t ) rad/s]2 = -(0.28125t 3 ) m/s2
and
aq = rq&& + 2r&q&
= [(0.5t ) m][0.75 rad/s2 ] + 2(0.5 m/s)[(0.75t ) rad/s]
= (1.125t ) m/s2
+
Kinetics
S Fr = mar : - T = (3 kg)( -0.28125t 3 ) m/s2
T = (0.84375t 3 ) N
or
SFq = mB aq : Q = (3 kg)(1.125t ) m/s2
+
or
Q = (3.375t ) N
Now require that
T =Q
or
or
(0.84375t 3 ) N = (3.375t ) N
t 2 = 2.00s
t 2 = 2.00s b
or
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113
PROBLEM 12.71
The 100-g pin B slides along the slot in the rotating arm OC
and along the slot DE which is cut in a fixed horizontal plate.
Neglecting friction and knowing that rod OC rotates at the constant
rate q&0 = 12 rad/s, determine for any given value of q (a) the radial
and transverse components of the resultant force F exerted on pin
B, (b) the forces P and Q exerted on pin B by rod OC and the wall
of slot DE, respectively.
SOLUTION
Kinematics
From the drawing of the system, we have
0.2
r=
m
cosq
Then
sin q & ˆ
Ê
q ˜ m/s
r& = Á 0.2
Ë
cos2 q ¯
q& = 12 rad/s
q&& = 0
and
cos q (cos2q ) - sin q ( -2 cos q sin q ) & 2
q
cos 4 q
Ê 1 + sin 2q 2 ˆ
= Á 0.2
q& ˜ m/s2
cos3 q
Ë
¯
&&
r = 0.2
Substituting for q&
r& = 0.2
&&
r = 0.2
Now
Ê
sin q ˆ
(12) = Á 2.4
˜ m/s
cos q
cos2 q ¯
Ë
sin q
2
Ê
1 + sin 2 q ˆ
2
2
(
12
)
=
28
.
8
Á
˜ m/s
cos3 q ¯
cos3 q
Ë
1 + sin 2 q
ar = &&
r - rq& 2
Ê
1 + sin 2 q ˆ Ê 0.2 ˆ
2
= Á 28.8
˜ (12)
˜ -Á
cos3 q ¯ Ë cos q ¯
Ë
Ê
sin 2 q ˆ
2
= Á 57.6
˜ m/s
cos3 q ¯
Ë
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114
Problem 12.71 (continued)
aq = rq&& + 2r&q&
and
Ê
sin q ˆ
= 0 + 2 Á 2.4
˜ (12)
cos2 q ¯
Ë
Ê
sin q ˆ
= Á 57.6
m/s2
2 ˜
cos
q
Ë
¯
Kinetics
(a)
ÈÊ
˘
sin 2 q ˆ
Fr = mB ar = (0.1 kg) ÍÁ 57.6
m/s2 ˙
3 ˜
cos q ¯
ÍÎË
˙˚
We have
Fr = (5.76 N ) tan 2 q sec q b
or
ÈÊ
˘
sin q ˆ
Fq = mB aq = (0.1 kg) ÍÁ 57.6
m/s2 ˙
2 ˜
cos q ¯
ÍÎË
˚˙
and
Fq = (5.76 N ) tan q sec q b
or
(b)
Now
+
SFy : Fq cos q + Fr sin q = P cos q
P = 5.76 tan q sec q + (5.76 tan 2 q sec q ) tan q
or
P = (5.76 N ) tan q sec2 q
or
+
or
qb
SFr : Fr = Q cos q
Q = (5.76 tan 2 q sec q )
1
cos q
Q = (5.76 N ) tan 2 q sec2 q Æ b
or
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115
PROBLEM 12.72*
Slider C has a weight of 250 g and may move in a slot cut in arm AB,
which rotates at the constant rate q&0 = 10 rad/s in a horizontal plane.
The slider is attached to a spring of constant k = 50 N/m, which is
unstretched when r = 0. Knowing that the slider is released from rest
with no radial velocity in the position r = 500 mm and neglecting
friction, determine for the position r = 300mm (a) the radial and
transverse components of the velocity of the slider, (b) the radial
and transverse components of its acceleration, (c) the horizontal
force exerted on the slider by arm AB.
SOLUTION
Let l0 be the radial coordinate when the spring is unstretched. Force exerted by the spring.
Fr = - k ( r - l0 )
SFr = mar : - k ( r - l0 ) = m( &&
r - rq& 2 )
kl
kˆ
Ê
&&
r = Á q& 2 - ˜ r + 0
¯
Ë
m
m
But
(1)
d
dr& dr
dr&
( r&) =
= r&
dt
dr dt
dr
kl ˘
kˆ
ÈÊ
& & = &&
rdr
rdr = ÍÁ q& 2 - ˜ r + 0 ˙ dr
¯
Ë
m˚
m
Î
&&
r=
Integrate using the condition r& = r&0 when r = r0 .
1 2 r& È 1 Ê & 2 k ˆ 2 kl0 ˘ r
r
r&
= Áq - ˜ r +
m ˙˚ r0
2 r&0 ÍÎ 2 Ë
m¯
(
)
kl
1 2 1 2 1 Ê &2 k ˆ 2
r& - r&0 = Á q - ˜ r - r02 + 0 ( r - r0 )
m
2
2
2Ë
m¯
2kl
kˆ
Ê
r&2 = r&02 + Á q& 2 - ˜ r 2 - r02 + 0 ( r - r0 )
¯
Ë
m
m
(
Data:
)
m = 0.25 kg
q& = 10 rad/s, k = 50 N/m, l0 = 0
r&0 = ( vr )0 = 0, r0 = 500 mm = 0.5 m, r = 300 mm = 0.3 m
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116
Problem 12.72* (continued)
(a)
Components of velocity when r = 0.3 m
50 ˆ
Ê
2
2
r&2 = 0 + Á102 ˜ (0.3 - 0.5 ) + 0
Ë
0.25 ¯
= 16 m2 /s2
vr = r& = ±4.00 m/s
Since r is decreasing, vr is negative
r& = -4 m/s
vr = -4.00 m/s b
vq = rq& = (0.3)(10)
(b)
vq = 3.00 m/s b
Components of acceleration.
Fr = - kr + kl0 = -(50)(0.3) + 0 = -15 N
F
15
ar = r = 0.25
m
aq = rq&& + 2r&q& = 0 + (2)( -4)(10)
aq = -80.0 m/s2 b
(c)
ar = -60.0 m/s2 b
Transverse component of force.
Fq = maq = (0.25)( -80)
Fq = -20.0 N b
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117
PROBLEM 12.73*
Solve Problem 12.72, assuming that the spring is unstretched when
slider C is located 50 mm to the left of the midpoint O of arm
AB (r = 50 mm).
PROBLEM 12.72 Slider C has a weight of 250 g and may move in
a slot cut in arm AB, which rotates at the constant rate q&0 = 10 rad/s
in a horizontal plane. The slider is attached to a spring of constant
k = 50 N/m, which is unstretched when r = 0. Knowing that the
slider is released with no radial velocity in the position r = 500
mm and neglecting friction, determine for the position r = 300 mm
(a) the radial and transverse components of the velocity of the slider,
(b) the radial and transverse components of its acceleration, (c) the
horizontal force exerted on the slider by arm AB.
SOLUTION
Let l0 be the radial coordinate when the spring is unstretched. Force exerted by the spring.
Fr = - k ( r - l0 )
SFr = mar : - k ( r - l0 ) = m( &&
r - rq& 2 )
kl
kˆ
Ê
&&
r = Á q& 2 - ˜ r + 0
Ë
m
m¯
But
(1)
d
dr& dr
dr&
( r&) =
= r&
dt
dr dt
dr
kl ˘
kˆ
ÈÊ
& & = &&
rdr
rdr = ÍÁ q& 2 - ˜ r + 0 ˙ dr
¯
Ë
m
m˚
Î
&&
r=
Integrate using the condition r& = r&0 when r = r0 .
1 2 r& È 1 Ê & 2 k ˆ 2 kl0
r&
= Áq - ˜ r +
m
2 r&0 ÍÎ 2 Ë
m¯
(
˘ r
r˙
˚ r0
)
kl
1 2 1 2 1 Ê &2 k ˆ 2
r& - r&0 = Á q - ˜ r - r02 + 0 ( r - r0 )
2
2
2Ë
m¯
m
2kl
kˆ
Ê
r&2 = r&02 + Á q& 2 - ˜ r 2 - r02 + 0 ( r - r0 )
Ë
m
m¯
(
)
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118
Problem 12.73* (continued)
m = 0.25kg
Data:
q& = 10 rad/s, k = 50 N/m, l0 = 0 - 50 mm = -50 mm = -0.05 m
r&0 = ( vr )0 = 0, r0 = 0.5 m, r = 0.3 m
(a)
Components of velocity when r = 0.3 m
(2)(50) ( -0.05)
50 ˆ
Ê
(0.32 - 0.52 ) +
(0.3 - 0.5)
r&2 = 0 + Á102 ˜
Ë
0.25 ¯
0.25
= 20 m2 /s2
vr = r& = ±4.47214 m/s
Since r is decreasing, vr is negative.
(b)
r& = -4.47214 m/s
vr = -4.47 m/s b
vq = rq& = (0.3)(10)
vq = 3.00 m/s b
Components of acceleration.
Fr = - kr + kl0 = -(50)(0.3) + (50) (-0.05) = -17.5 N
ar =
Fr
17.5
=
0.25
m
ar = -70.0 m/s2 b
aq = rq&& + 2r&q& = 0 + (2)( -4.47214)(10)
aq = -89.4 m/s2 b
(c)
Transverse component of force.
Fq = maq = (0.25)( -89.4428) Fq = -22.4 N b
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119
PROBLEM 12.74
A particle of mass m is projected from Point A with an initial velocity v0
perpendicular to line OA and moves under a central force F along a semicircular
path of diameter OA. Observing that r = r0 cosq and using Eq. (12.27), show
that the speed of the particle is v = v0 /cos2 q .
SOLUTION
Since the particle moves under a central force, h = constant.
Using Eq. (12.27),
h = r 2q& = h0 = r0 v0
or
rv
rv
v0
q& = 0 20 = 2 0 0 2 =
r
r0 cos q r0 cos2 q
Radial component of velocity.
vr = r& =
d
( r0 cos q ) = -( r0 sin q )q&
dt
Transverse component of velocity.
vq = rq& = ( r0 cos q )q&
Speed.
v = vr 2 + vq 2 = r0q& =
r0 v0
2
r0 cos q
v=
v0
cos2 q
b
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120
PROBLEM 12.75
For the particle of Problem 12.74, determine the tangential component Ft of
the central force F along the tangent to the path of the particle for (a) q = 0,
(b) q = 45∞.
SOLUTION
Since the particle moves under a central force, h = constant
Using Eq. (12.27),
h = r 2q& = h0 = r0 v0
rv
rv
v0
q& = 0 20 = 2 0 0 2 =
r
r0 cos q r0 cos2 q
Radial component of velocity.
vr = r& =
d
( r0 cos q ) = -( r0 sin q )q&
dt
Transverse component of velocity.
vq = rq& = ( r0 cos q )q&
Speed.
v = vr 2 + vq 2 = r0q& =
r0 v0
2
r0 cos q
=
v0
cos2 q
Tangential component of acceleration.
at =
v0
dv
( -2)( - sin q )q& 2v0 sin q
= v0
=
◊
3
3
dt
cos q
cos q r0 cos2 q
=
Tangential component of force.
Ft = mat : Ft =
(a)
q = 0,
(b)
q = 45∞,
2v0 2 sin q
r0 cos5 q
2mv0 2 sin q
r0 cos5 q
Ft = 0
Ft =
Ft = 0 b
2mv0 sin 45∞
cos5 45∞
Ft =
8mv0 2
b
r0
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121
PROBLEM 12.76
A particle of mass m is projected from point A with an initial velocity v0
perpendicular to line OA and moves under a central force F directed away
from the center of force O. Knowing that the particle follows a path defined
by the equation r = r0 / cos 2q and using Eq. (12.27), express the radial and
transverse components of the velocity v of the particle as functions of q.
SOLUTION
Since the particle moves under a central force, h = constant.
Using Eq. (12.27),
h = r 2q& = h0 = r0 v0
or
rv
r v cos 2q v0
q& = 0 20 = 0 0 2
= cos 2q
r0
r
r0
Radial component of velocity.
vr = r& =
= r0
dr & d Ê r0 ˆ &
sin 2q &
q=
q = r0
q
˜
Á
dq
dq Ë cos 2q ¯
(cos 2q )3/ 2
sin 2q v0
cos 2q (cos 2q )3/ 2 r
vr = v0
sin 2q
b
cos 2q
Transverse component of velocity.
vq =
h r0 v0
=
cos 2q r
r0
vq = v0 cos 2q b
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122
PROBLEM 12.77
For the particle of Problem 12.76, show (a) that the velocity of the particle
and the central force F are proportional to the distance r from the particle
to the center of force O, (b) that the radius of curvature of the path is
proportional to r3.
PROBLEM 12.76 A particle of mass m is projected from Point A with an
initial velocity v0 perpendicular to line OA and moves under a central force F
directed away from the center of force O. Knowing that the particle follows
a path defined by the equation r = r0 / cos 2q and using Eq. (12.27), express
the radial and transverse components of the velocity v of the particle as
functions of q.
SOLUTION
Since the particle moves under a central force, h = constant.
Using Eq. (12.27),
h = r 2q& = h0 = r0 v0
or
rv
r v cos 2q v0
q& = 0 2n = 0 0 2
= cos 2q
r0
r
r0
Differentiating the expression for r with respect to time,
r& =
sin 2q
sin 2q &
sin 2q v0
dr & d Ê r0 ˆ &
q=
q = r0
q = r0
cos 2q = v0
3/ 2
3/ 2
r
dq
dq ÁË cos 2q ˜¯
cos 2q
(cos 2q )
(cos 2q )
0
Differentiating again,
&&
r=
(a)
2 cos2 2q + sin 2 2q & v0 2 2 cos2 2q + sin 2 2q
sin 2q ˆ &
dr& & d Ê
q=
q
=
v
v
q=
0
0
˜
r0
dq
dq ÁË
cos 2q ¯
cos 2q
(cos 2q )3/ 2
vr = r& = v0
v r
sin 2q
= 0 sin 2q
r0
cos 2q
v r
vq = rq& = 0 cos 2q
r0
v = ( vr )2 + ( vq )2 =
v0 r
sin 2 2q + cos2 2q r0
v=
v0 r
b
r0
v 2 2 cos2 2q + sin 2 2q
r0
v0 2
ar = &&
r - rq& 2 = 0
cos2 2q
r0
cos 2q
cos 2q r0 2
=
v0 2 cos2 2q + sin 2 2q
v0
v 2r
=
= 02
r0
cos 2q
r0 cos 2q
r0
Fr = mar =
mv02 r
r02
Fr =
:
mv02 r
r02
b
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123
Problem 12.77 (continued)
Since the particle moves under a central force, aq = 0.
Magnitude of acceleration.
a = ar 2 + aq 2 =
v0 2 r
r0 2
Tangential component of acceleration.
at =
v0 2 r
dv d Ê v0 r ˆ v0
&
= Á
=
r
=
sin 2q
dt dt Ë r0 ˜¯ r0
r0 2
Normal component of acceleration.
at = a2 - at 2 =
Êr ˆ
cos 2q = Á 0 ˜
Ë r¯
But
an =
Hence,
(b)
But an =
v2
r
or
r=
v0 2 r
r0 2
1 - sin 2 2q =
v0 2 r cos 2q
r0 2
2
v0 2
r
v 2 v0 2 r 2 r
= 2 ◊ 2
an
r0
v0
r=
r3
b
r02
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124
PROBLEM 12.78
The radius of the orbit of a moon of a given planet is equal to twice the radius of that planet. Denoting
by r the mean density of the planet, show that the time required by the moon to complete one full revolution
about the planet is (24 p /G r )1/2 , where G is the constant of gravitation.
SOLUTION
For gravitational force and a circular orbit,
Fr =
GMm
r2
=
mv 2
r
or
v=
GM
r
Let t be the periodic time to complete one orbit.
vt = 2p r
Solving for t,
But
Then
t=
t
GM
= 2p r
r
hence,
GM = 2
or
2p r 3/2
GM
4
M = p R3 r ,
3
3p Ê r ˆ
t=
Á ˜
Gr Ë R ¯
p
G r R3/2
3
3/2
Using r = 2R as a given leads to
t = 23/2
3p
24p
=
Gr
Gr
t = (24p /G r )1/2 b
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125
PROBLEM 12.79
Show that the radius r of the orbit of a moon of a given planet can be determined from the radius R of the planet,
the acceleration of gravity at the surface of the planet, and the time t required by the moon to complete one full
revolution about the planet. Determine the acceleration of gravity at the surface of the planet Jupiter knowing
that R = 71,492 km and that t = 3.551 days and r = 670.9 ¥ 103 km for its moon Europa.
SOLUTION
Mm
r2
We have
F =G
and
F = Fn = man = m
Then
or
Now
so that
G
Mm
r2
=m
v2 =
[Eq. (12.28)]
v2
r
v2
r
GM
r
GM = gR2
v2 =
[Eq. (12.30)]
gR2
r
or
v=R
g
r
2p r
2p r
=
v
g
R
r
For one orbit,
t=
or
Ê gt 2 R2 ˆ
r=Á
˜
Ë 4p 2 ¯
Solving for g,
g = 4p 2
1/3
Q.E.D.
b
r3
t 2 R2
and noting that t = 3.551 days = 306,806 s, then
gJupiter = 4p 2
= 4p 2
3
rEur
2
t Eur
RJup
(670.9 ¥ 106 m)3
(306, 806 s)2 (71.4492 ¥ 106 m)2
gJupiter = 24.8 m/s2 b
or
Note:
gJupiter ª 2.53gEarth
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126
PROBLEM 12.80
Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete
one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to the
ground. Determine (a) the altitude of these satellites above the surface of the earth, (b) the velocity with which
they describe their orbit. Give the answers in SI units.
SOLUTION
For gravitational force and a circular orbit,
GMm mv 2
Fr = 2 =
r
r
Let t be the period time to complete one orbit.
or
v 2t 2 =
v=
GM t 2
= 4p 2 r 2
r
But
vt = 2p r
Then
r3 =
Data:
t = 23.934 h = 86.1624 ¥ 103 s
(a)
g = 9.81 m/s2 , R = 6.37 ¥ 106 m
GM t 2
4p 2
or
or
GM
r
Ê GM t 2 ˆ
r=Á
˜
Ë 4p 2 ¯
1/ 3
GM = gR2 = (9.81)(6.37 ¥ 106 )2 = 398.06 ¥ 1012 m3 /s2
1/ 3
È (398.06 ¥ 1012 )(86.1624 ¥ 103 )2 ˘
r=Í
˙
4p 2
Î
˚
= 42.145 ¥ 106 m
altitude h = r - R = 35.775 ¥ 106 m
h = 35, 800 km b
(b)
v=
398.06 ¥ 1012
GM
= 3.07 ¥ 103 m/s
=
6
r
42.145 ¥ 10
v = 3.07 km/s b
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127
PROBLEM 12.81
Determine the mass of the earth, knowing that the mean radius of the moon’s orbit about the earth is 384,000 km
and that the moon requires 27.32 days to complete one full revolution about the earth.
SOLUTION
Mm
[Eq. (12.28)]
r2
v2
F = Fn = man = m
r
F =G
We have
and
Then
or
Now
G
v2
Mm
=
m
r
r2
r
M = v2
G
v=
2p r
t
2
so that
M=
2
1 Ê 2p ˆ 3
r Ê 2p r ˆ
˜¯ = ÁË ˜¯ r
ÁË
G t
G t
Noting that
t = 27.32 days = 2.3604 ¥ 106 s
and
r = 384, 000 km = 3.84 ¥ 108 m
we have
M=
2
1
66.73 ¥ 10 -12
m3
kg.s2
Ê
ˆ
2p
(3.84 ¥ 108 )3
Á
6 ˜
Ë 2.3604 ¥ 10 s ¯
M = 6.01 ¥ 1024 kg b
or
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128
PROBLEM 12.82
A spacecraft is placed into a polar orbit about the planet Mars at an altitude of 380 km. Knowing that the mean
density of Mars is 3.94 Mg/m3 and that the radius of Mars is 3397 km, determine (a) the time t required for
the spacecraft to complete one full revolution about Mars, (b) the velocity with which the spacecraft describes
its orbit.
SOLUTION
(a)
From the solution to Problem 12.78, we have
t=
where
3p Ê r ˆ
Á ˜
Gr Ë R ¯
3/ 2
r = R + h = (3397 + 380) km
= 3777 km
1/ 2
Then
È
˘ Ê 3777 km ˆ
3p
t =Í
˜
-12 3
2
3
3 ˙ Á
Î (66.73 ¥ 10 m /kg ◊ s )(3.94 ¥ 10 kg/m ) ˚ Ë 3397 km ¯
= 7019.5 s
t = 1 h 57 min b
or
(b)
We have
3/ 2
2p r
t
2p (3777 ¥ 103 m)
=
7019.5 s
v=
v = 3380 m/s b
or
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129
PROBLEM 12.83
A satellite is placed into a circular orbit about the planet Saturn at an altitude of 3400 km. The satellite describes
its orbit with a velocity of 87, 500 km/h. Knowing that the radius of the orbit about Saturn and the periodic time
of Atlas, one of Saturn’s moons, are 136.9 ¥ 103 and 0.6017, respectively, determine (a) the radius of Saturn,
(b) the mass of Saturn. (The periodic time of a satellite is the time it requires to complete one full revolution
about the planet.)
SOLUTION
where
2p rA
tA
rA = 136.9 ¥ 103 km = 136.9 ¥ 106 m
and
t A = 0.6017 days = 51, 987 s
Velocity of Atlas.
vA =
vA =
Gravitational force.
F=
(2p )(136.9 ¥ 106 )
= 16.54583 ¥ 103 m/s
51, 987
GMm
r2
=
mv 2
r
from which
GM = rv 2 = constant
For the satellite,
rs vs2 = rA v 2A
rs =
rA v 2A
vs 2
vs = 87, 500 km/h = 24.30556 ¥ 103 m/s
where
rs =
(136.9 ¥ 106 )(16.54583 ¥ 103 )2
3 2
(24.30556 ¥ 10 )
rs = 63, 400 km
(a)
= 63.4409 ¥ 106 m
Radius of Saturn.
R = rs - (altitude) = 63,400 - 3400 (b)
R = 60, 000 km b
Mass of Saturn.
M=
rA v 2A (136.9 ¥ 106 )(16.54583 ¥ 103 )2
=
G
66.73 ¥ 10 -12
M = 5.62 ¥ 1024 kg b
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130
PROBLEM 12.84
The periodic times (see Problem 12.83) of the planet Uranus’ moons Juliet and Titania have been observed to
be 0.4931 days and 8.706 days, respectively. Knowing that the radius of Juliet’s orbit is 64,360 km, determine
(a) the mass of Uranus, (b) the radius of Titania’s orbit.
SOLUTION
2p rJ
tJ
Velocity of Juliet.
vJ =
where
rJ = 64, 360 km = 64.36 ¥ 106 m
and
t J = 0.4931 days = 42, 604 s
(2p )(64.36 ¥ 106 )
= 9.4917 ¥ 103 m/s
42, 604
GMm mv 2
F= 2 =
r
r
vJ =
Gravitational force.
from which
(a)
Mass of Uranus.
GM = rv 2 = constant
rJ v J2
G
(64.36 ¥ 106 )(9.4917 ¥ 103 )2
M=
66.73 ¥ 10 -12
= 86.893 ¥ 1024 kg
M=
M = 86.9 ¥ 1024 kg b
(b)
Radius of Titania’s orbit.
4p 2 rT3
GM = rT vT2 =
rT3
=
Êt
rJ3 Á T
ËtJ
tT 2
=
4p 2 rJ3
tJ2
2
2
ˆ
6 3 Ê 8.706 ˆ
24
3
˜¯ = (64.36 ¥ 10 ) ÁË 0.4931˜¯ = 81.103 ¥ 10 m
rT = 436.39 ¥ 106 m
rT = 436, 000 km b
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131
PROBLEM 12.85
A 600-kg spacecraft first is placed into a circular orbit about the earth at an altitude of 4500 km and then is
transferred to a circular orbit about the moon. Knowing that the mass of the moon is 0.01230 times the mass
of the earth and that the radius of the moon is 1740 km, determine (a) the gravitational force exerted on the
spacecraft as it was orbiting the earth, (b) the required radius of the orbit of the spacecraft about the moon if
the periodic times (see Problem 12.83) of the two orbits are to be equal, (c) the acceleration of gravity at the
surface of the moon.
SOLUTION
First note that
RE = 6370 km
Then
rE = RE + hE = (6370 + 4500) km
= 10870 km
(a)
We have
and
(b)
F =G
MM
[Eq. (12.28)]
r2
GM = gR2
[Eq. (12.29)]
Ê Rˆ
= mg Á ˜
2
Ë r¯
r
m
2
Then
F = gR2
For the earth orbit,
Ê 6370 km ˆ
F = 600 ¥ 9.81 ¥ Á
Ë 10870 km ˜¯
or
F = 2021.343 N
2
F = 2020 N b
From the solution to Problem 12.81, we have
2
M=
Then
t=
1 Ê 2p ˆ 3
Á ˜ r
GË t ¯
2p r 3/ 2
GM
2p rE3/ 2
Now
t E = t M fi
or
ÊM ˆ
rM = Á M ˜
Ë ME ¯
or
GM E
=
2p rM3/ 2
GM M
(1)
1/ 3
rE = (0.01230)1/ 3 (10870 km)
= 2509.176 km
rM = 2510 km b
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132
Problem 12.85 (continued)
(c)
We have
GM = gR2
[Eq.(12.29)]
Substituting into Eq. (1)
2p rE3/ 2
RE g E
=
2p rM3/ 2
RM g M
2
or
gM
3
2
Ê R ˆ Êr ˆ
Ê R ˆ ÊM ˆ
= Á E ˜ Á M ˜ gE = Á E ˜ Á M ˜ gE
Ë RM ¯ Ë rE ¯
Ë RM ¯ Ë M E ¯
using the results of Part (b). Then
2
gM
Ê 6370 km ˆ
=Á
(0.01230)(9.81 m/s2 )
Ë 1740 km ˜¯
or
Note:
g moon ª
g moon = 1.62 m/s2 b
1
gearth
6
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133
PROBLEM 12.86
To place a communications satellite into a geosynchronous
orbit (see Problem 12.80) at an altitude of 35800 km above
the surface of the earth, the satellite first is released from
a space shuttle, which is in a circular orbit at an altitude of
300 km, and then is propelled by an upper-stage booster
to its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic
transfer orbit. The booster is again fired at B to insert
the satellite into a geosynchronous orbit. Knowing that
the second firing increases the speed of the satellite by
1440 m/s, determine (a) the speed of the satellite as it
approaches B on the elliptic transfer orbit, (b) the increase
in speed resulting from the first firing at A.
SOLUTION
For earth,
R = 6370 km = 6.37 ¥ 106 m
GM = gR2 = (9.81 m/s2 )(6.37 ¥ 106 )2 = 3.9806 ¥ 1014 m3 /s2
rA = 6370 + 300 = 6670 km = 6.67 ¥ 106 m
rB = 6370 + 35800 = 42,170 km = 42.17 ¥ 106 m
Speed on circular orbit through A.
( v A )circ =
=
GM
rA
3.9806 ¥ 1014
6.67 ¥ 106
= 7.7252 ¥ 103 m/s
Speed on circular orbit through B.
( v B )circ =
=
GM
rB
3.9806 ¥ 1014
42.17 ¥ 106
= 3.07236 ¥ 103 m/s
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134
Problem 12.86 (continued)
(a)
Speed on transfer trajectory at B.
( v B )tr = 3.07236 ¥ 103 - 1440
= 1632.36 ¥ m/s
1632 m/s b
Conservation of angular momentum for transfer trajectory.
rA ( v A )tr = rB ( v B ) tr
r (v )
( v A )tr = B B tr
rA
( 42.17 ¥ 106 )(1632.36)
(6.67 ¥ 106 )
= 10, 320.33 m/s
=
(b)
Change in speed at A.
Dv A = ( v A )tr - ( v A )circ
= 10, 320.33 - 7725.2
= 2595.13 m/s
Dv A = 2600 m/s b
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135
PROBLEM 12.87
A space vehicle is in a circular orbit of 2200-km radius around the moon. To
transfer it to a smaller circular orbit of 2080-km radius, the vehicle is first
placed on an elliptic path AB by reducing its speed by 26.3 m/s as it passes
through A. Knowing that the mass of the moon is 73.49 ¥ 1021 kg, determine
(a) the speed of the vehicle as it approaches B on the elliptic path, (b) the
amount by which its speed should be reduced as it approaches B to insert it
into the smaller circular orbit.
SOLUTION
For a circular orbit,
SFn = man : F = m
F =G
Eq. (12.28):
Then
G
or
v2
r
Mm
r2
v2
Mm
=
m
r
r2
GM
v2 =
r
66.73 ¥ 10 -12 m3 /kg ◊ s2 ¥ 73.49 ¥ 1021 kg
Then
2
( v A )circ
=
or
( v A )circ = 1493.0 m/s
and
( v B )2circ =
or
( v B )circ = 1535.5 m/s
(a)
We have
2200 ¥ 103 m
66.73 ¥ 10 -12 m3 /kg ◊ s2 ¥ 73.49 ¥ 1021 kg
2080 ¥ 103 m
( v A )TR = ( v A )circ + D v A
= (1493.0 - 26.3) m/s
= 1466.7 m/s
Conservation of angular momentum requires that
rA m( v A )TR = rB m( v B )TR
or
2200 km
¥ 1466.7 m/s
2080 km
= 1551.3 m/s
( v B )TR =
or
(b)
Now
or
( v B )TR = 1551 m/s b
( v B )circ = ( v B )TR + Dv B
Dv B = (1535.5 - 1551.3) m/s
Dv B = -15.8 m/s b
or
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136
PROBLEM 12.88
Plans for an unmanned landing mission on the planet Mars called for
the earth-return vehicle to first describe a circular orbit at an altitude
dA = 2200 km above the surface of the planet with a velocity of
2771 m/s. As it passed through point A, the vehicle was to be inserted
into an elliptic transfer orbit by firing its engine and increasing its
speed by Dv A = 1046 m/s. As it passed through point B, at an altitude
dB = 100,000 km, the vehicle was to be inserted into a second transfer
orbit located in a slightly different plane, by changing the direction
of its velocity and reducing its speed by Dv B = -22.0 m/s. Finally,
as the vehicle passed through Point C, at an altitude dC = 1000 km,
its speed was to be increased by DvC = 660 m/s to insert it into its
return trajectory. Knowing that the radius of the planet Mars is
R = 3400 km, determine the velocity of the vehicle after completion
of the last maneuver.
SOLUTION
t A = 3400 + 2200 = 5600 km = 5.60 ¥ 106 m
t B = 3400 + 100, 000 = 103, 400 km = 103.4 ¥ 106 m
t C = 3400 + 1000 = 4400 km = 4.40 ¥ 106 m
First transfer orbit.
v A = 2771 m/s + 1046 m/s = 3817 m/s
Conservation of angular momentum:
t A m v A = t B m vB
6
(5.60 ¥ 10 )(3817) = (103.4 ¥ 106 )v B
v B = 206.7 m/s
Second transfer orbit.
v B¢ = v B + Dv B
= 206.7 - 22.0 = 184.7 m/s
Conservation of angular momentum:
t B mv B¢ = t C mvC
(103.4 ¥ 106 )(184.7) = ( 4.40 ¥ 106 )vC
vC = 4340 m/s
After last maneuver.
v = vC + D vC = 4340 + 660 = 5000 m/s
b
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137
PROBLEM 12.89
A space shuttle S and a satellite A are in the circular orbits shown. In
order for the shuttle to recover the satellite, the shuttle is first placed
in an elliptic path BC by increasing its speed by Dv B = 84 m/s as it
passes through B. As the shuttle approaches C, its speed is increased
by DvC = 78 m/s to insert it into a second elliptic transfer orbit CD.
Knowing that the distance from O to C is 6860 km, determine the
amount by which the speed of the shuttle should be increased as it
approaches D to insert it into the circular orbit of the satellite.
SOLUTION
R = 6370 km = 6.37 ¥ 106 m
First note
rA = (6370 + 610) km = 6.98 ¥ 106 km
rB = (6370 + 290) km = 6660 km = 6.66 ¥ 106 km
SFn = man :
For a circular orbit,
F =G
Eq. (12.28):
Then
or
G
F =m
Mm
r2
v2
Mm
=
m
r
r2
v2 =
GM gR2
=
r
r
using Eq. (12.29).
9.81 ¥ (6.37 ¥ 106 )2
6.98 ¥ 106
Then
( v A )2circ =
or
( v A )circ = 7551.73 m/s
and
( v B )2circ =
or
( v B )circ = 7731.02 m/s
We have
v2
r
9.81 ¥ (6.37 ¥ 106 )2
(6.66 ¥ 10 )
6
( v B )TRBC = ( v B )circ + D v B = (7731.02 + 84) m/s
= 7815.02 m/s
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138
Problem 12.89 (continued)
Conservation of angular momentum requires that
BC : rB m ( v B )TRBC = rC m ( vC )TRBC (1)
CD : rC m ( vC )TRCD = rA m ( vD )TRCD (2)
rB
6.66 ¥ 106
( v B )TRBC =
¥ 7815.02 m/s
rC
6.86 ¥ 106
= 7587.18 m//s
From Eq. (1)
( vC )TRBC =
Now
( vC )TRCD = ( vC )TRBC + DvC = (7587.18 + 78) m/s
= 7665.18 m/s
From Eq. (2)
Finally,
or
rC
(6.86 ¥ 106 )
( vC )TRCD =
¥ 7665.18
rA
(6.98 ¥ 106 )
= 7533.40 m/s
( vD )TRCD =
( v A )circ = ( vD )TRCD + DvD
DvD = (7551.73 - 7533.4) m/s
DvD = 18.33 m/s b
or
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139
PROBLEM 12.90
A 1.5 kg collar can slide on a horizontal rod, which is free to rotate
about a vertical shaft. The collar is initially held at A by a cord
attached to the shaft. A spring of constant 35 N/m is attached to the
collar and to the shaft and is undeformed when the collar is at A.
As the rod rotates at the rate q& = 16 rad/s, the cord is cut and the
collar moves out along the rod. Neglecting friction and the mass
of the rod, determine (a) the radial and transverse components of
the acceleration of the collar at A, (b) the acceleration of the collar
relative to the rod at A, (c) the transverse component of the velocity
of the collar at B.
SOLUTION
Fsp = k ( r - rA )
First note
(a)
Fq = 0 and at A,
Fr = - Fsp = 0
( aA )r = 0 b
( aA )q = 0 b
(b)
+
SFr = mar :
Noting that
-Fsp = m( &&
r - rq& 2 )
acollar/rod = &&
r , we have at A
0 = m[acollar/rod - (0.15 m)(16 rad/s)2 ]
or
( acollar/rod ) A = 38.4 m/s2
(c)After the cord is cut, the only horizontal force acting on the collar is due to the spring. Thus, angular
momentum about the shaft is conserved.
rA m ( v A )q = rB m ( v B )q
Then
( v B )q =
where ( v A )q = rA q&0
0.15m
[(0.15 m)(16 rad/s)] = 0.8 m/s
0.45m
( v B )q = 0.800 m/s b
or
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140
PROBLEM 12.91
For the collar of Problem 12.90, assuming that the rod initially rotates at the rate q = 12 rad/s, determine for
position B of the collar (a) the transverse component of the velocity of the collar, (b) the radial and transverse
components of its acceleration, (c) the acceleration of the collar relative to the rod.
Solution
Fsp = k ( r - rA )
First we note
( Fsp ) B = 35 N m (0.45 - 0.15) m
At B:
= 10.5 N
(a)After the cord is cut, the only horizontal force acting on the collar is due to the spring. Thus, angular momentum
about the shaft is conserved.
rA m( v A )q = rB m( v B )q
where
( v A )q = rAq&0
Then
( v B )q =
0.15m
[(0.15m)(12 rad/s)] = 0.6 m s
0.45m
( v B )q = 0.600 m/s b
or
( FB )q = 0 fi
(b)
+
Now
SFr = mar : - ( Fsp ) B = m( aB ) r
( aB ) r = -
or
( aB )q = 0 b
10.5N
1.5kg
= -7 m/s2
(aB )r = -7.00 m
or
(c)
We have
Then
or
ar = &&
r - rq& 2
acollar/rod = &&
r
Now
at B:
s2 b
and
(v )
q& B = B q
rB
Ê 0.6 ˆ
( acollar/rod ) B = -7 m/s + (0.45 m) Á
Ë 0.45 ˜¯
2
= -6.2 m/s2
2
( acollar/rod )q = -6.20 m s2 b
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141
PROBLEM 12.92
A 200-g ball A and a 400-g ball B are mounted on a horizontal rod
which rotates freely about a vertical shaft. The balls are held in the
positions shown by pins. The pin holding B is suddenly removed
and the ball moves to position C as the rod rotates. Neglecting
friction and the mass of the rod and knowing that the initial speed
of A is v A = 2.5 m/s, determine (a) the radial and transverse
components of the acceleration of ball B immediately after the
pin is removed, (b) the acceleration of ball B relative to the rod
at that instant, (c) the speed of ball A after ball B has reached the
stop at C.
Solution
Let r and q be polar coordinates with the origin lying at the shaft.
Constraint of rod: q B = q A + p radians; q& B = q& A = q&; q&&B = q&&A = q&&.
(a)
Components of acceleration.
Sketch the free body diagrams of the balls showing the radial and
transverse components of the forces acting on them. Owing to
frictionless sliding of B along the rod, ( FB )r = 0.
Radial component of acceleration of B.
Fr = mB ( aB ) r :
( aB ) r = 0 b
Transverse components of acceleration.
( aA )q = rAq&& + 2r&Aq& = raq&&
( aB )q = rBq&& + 2r&Bq&
(1)
Since the rod is massless, it must be in equilibrium. Draw its freebody diagram, applying Newton’s Third Law.
+
SM 0 = 0 : rA ( FA )q + rB ( FB )q = rA mA ( aA )q + rB mB ( aB )q = 0
rA mA rAq&& + rB mB ( rBq&& + 2r&Bq& ) = 0
q&& =
At t = 0,
r&B = 0
From Eq. (1),
-2rB r&Bq&
mA rA2 + mB rB 2
so that
q&& = 0.
( aB )q = 0 b
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142
Problem 12.92 (continued)
(b)
Acceleration of B relative to the rod.
(v )
2.5
(v A )q = 2.5 m/s, q& = A q =
= 10 rad/s
rA
0.25
At t = 0,
&&
rB - rBq& 2 = ( aB ) r = 0
&&
rB = rBq& 2 = (0.2)(10)2 = 20 m/s2
&&
rB = 20.0 m/s2 b
(c)
Speed of A.
Substituting
d
(mr 2q& ) for rFq in each term of the moment equation gives
dt
(
)
(
)
d
d
mA rA2q& +
mB rB2q& = 0
dt
dt
Integrating with respect to time,
(
mA rA2q& + mB rB2q& = mA rA2q&
) + (m r q& )
0
2
B B
0
Applying to the final state with ball B moved to the stop at C,
(m r
2
A A
)
+ mB rC2 q& f = ÈÎmA rA2 + mB ( rB )20 ˘˚ q&0
m r 2 + mB rB2
q& f = A A2
mA rA + mB rC2
=
(0.2)(0.25)2 + (0.4)(0.2)2
(10)
(0.2)(0.25)2 + (0.4)(0.4)2
= 3.7255 rad/s
( v A ) f = rAq& f = (0.25)(3.7255) = 0.93137 m/s
( v A ) f = 0.931 m/s b
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143
PROBLEM 12.93
A small ball swings in a horizontal circle at the end of a cord of length
l1, which forms an angle q1 with the vertical. The cord is then slowly
drawn through the support at O until the length of the free end is l2. (a)
Derive a relation among l1 , l2 , q1 , and q 2 . (b) If the ball is set in motion so
that initially l1 = 0.8 m and q1 = 35∞, determine the angle q 2 when l2 = 0.6 m.
Solution
For State 1 or 2, neglecting the vertical component of acceleration,
+
(a)
SFy = 0 : T cosq - W = 0
T = W cosq
+
But r = l sin q
SFx = man : T sin q = W sin q cos q =
so that
v2 =
mv 2
r
rW
sin 2 q cos q = l g sin q tan q
m
v1 = l1 g sin q1 tan q1
and
v2 = l 2 g sin q 2 tan q 2
SM y = 0 : H y = constant
r1mv1 = r2 mv2 or v1l1 sin q1 = v2 l 2 sin q 2
l31/2 g sin q1 sin q1 tan q1 = l32/2 sin q 2 sin q 2 tan q 2
l31 sin3 q1 tan q1 = l32 sin3 q 2 tan q 2 b
(b)
With q1 = 35∞, l1 = 0.8 m, and l 2 = 0.6 m
(0.8)3 sin3 35∞ tan 35∞ = (0.6)3 sin3 q 2 tan q 2
sin3 q 2 tan q 2 - 0.31320 = 0 q 2 = 43.6∞ b
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144
PROBLEM 12.94
A particle of mass m describes the cardioid r = r0 (1 + cos q ) / 2 under a central
force F directed toward the center of force O. Using Eq. (12.37), show that
F is inversely proportional to the fourth power of the distance r from the
particle to O.
Solution
We have
F
d 2u
+u=
2
dq
mh2 u 2
u=
where
Eq. (12.37)
1
and mh2 = constant
r
Ê d 2u
ˆ
F µ u 2 Á 2 + u˜
Ë dq
¯
u=
Now
1 2
1
=
r r0 1 + cos q
ˆ 2
sin q
du
d Ê2
1
=
=
dq dq ÁË r0 1 + cos q ˜¯ r0 (1 + cos q )2
Then
d 2 u 2 cos q (1 + cos q ) - sin q ÈÎ2 (1 + cos q )˘˚ ( - sin q )
=
dq 2 r0
(1 + cos q )4
2
and
Then
Ê 1ˆ
F µÁ ˜
Ë r¯
2
=
2 1 + cos q + sin 2 q 2 È
1
1 - cos2 q ˘
=
+
Í
˙
r0 (1 + cos q )3
r0 Î (1 + cos q )2 (1 + cos q )3 ˚
=
2 2 - cos q
2Êr ˆ
= Á 0˜
2
r0 (1 + cos q )
r0 Ë 2r ¯
=
r0 Ê
2r ˆ
3- ˜
2 Á
r0 ¯
2r Ë
2
È Ê 2r ˆ ˘
Í2 - Á - 1˜ ˙
¯ ˙˚
ÍÎ Ë r0
È r0 Ê
2r ˆ 1 ˘ 3 r0
Í 2 Á3 - ˜ + ˙ =
r0 ¯ r ˙˚ 2 r 4
ÍÎ 2r Ë
F¥
Note: F 0 implies that F is attractive.
1
r4
Q.E.D. b
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145
PROBLEM 12.95
A particle of mass m is projected from Point A with an initial velocity
v 0 perpendicular to OA and moves under a central force F along
an elliptic path defined by the equation r = r0 /(2 - cos q ). Using
Eq. (12.37), show that F is inversely proportional to the square of the
distance r from the particle to the center of force O.
Solution
u=
d 2u
dq
Solving for F,
2
+u=
F=
1 2 - cos q
=
,
r
r0
F
2
=
r0 mh2 u 2
du sin q
=
,
dq
r0
d 2u
dq
2
=
cos q
r0
by Eq. (12.37).
2mh2 u 2 2mh2
=
r0
r0 r 2
Since m, h, and r0 are constants, F is proportional to
1
r2
, or inversely proportional to r 2 .
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146
PROBLEM 12.96
A particle of mass m describes the path defined by the equation r = r0 sin q under a central force F directed
toward the center of force O. Using Eq. (12.37), show that F is inversely proportional to the fifth power of the
distance r from the particle to O.
Solution
We have
d 2u
dq
2
+u=
u=
where
F
mh2 u 2
Eq. (12.37)
1
and mh2 = constant
r
Ê d 2u
ˆ
F µ u 2 Á 2 + u˜
Ë dq
¯
u=
Now
du
1 Ê 1 ˆ
1 cos q
=
=Á
˜
dq dq Ë r0 sin q ¯
r0 sin 2 q
Then
d 2u
1 È - sin q (sin 2 q ) - cos q (2 sin q cos q ) ˘
=
Í
˙
r0 Î
dq 2
sin 4 q
˚
and
=
Then
1
1
=
r r0 sin q
1 1 + coss2 q
r0 sin3 q
2
2
1 ˆ
Ê 1 ˆ Ê 1 1 + cos q
+
F µÁ ˜ Á
3
Ë r ¯ Ë r0 sin q
r0 sin q ˜¯
=
1 1 Ê 1 + cos2 q sin 2 q ˆ
+ 3 ˜
r0 r 2 ÁË sin3 q
sin q ¯
2 1 1
=
r0 r 2 sin3 q
=
Ê rˆ
sin q = Á ˜
Ër ¯
3
3
0
2r02
5
r
F¥
1
r5
Q.E.D. b
Note: F 0 implies that F is attractive.
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147
PROBLEM 12.97
For the particle of Problem 12.76, and using Eq. (12.37), show that the
central force F is proportional to the distance r from the particle to the
center of force O.
PROBLEM 12.76 A particle of mass m is projected from Point A with
an initial velocity v0 perpendicular to line OA and moves under a central
force F directed away from the center of force O. Knowing that the
particle follows a path defined by the equation r = r0 / cos 2q and using
Eq. (12.27), express the radial and transverse components of the velocity v
of the particle as functions of q.
Solution
We have
F
d 2u
+u=
2
dq
mh2 u 2
u=
where
Eq. (12.37)
1
and mh2 = constant
r
Ê d 2u
ˆ
F µ u 2 Á 2 + u˜
Ë dq
¯
u=
Now
1 1
=
cos2 q
r r0
ˆ
du
d Ê1
1 sin 2q
cos 2q ˜ = =
Á
dq dq Ë r0
r0 cos 2q
¯
Then
d 2u
1 2 cos 2q cos 2q - sin 2q ( - sin 2q / cos 2q )
=2
r0
cos 2q
dq
and
3
4
1 Ê r ˆ È Ê r0 ˆ ˘
1 1 + cos2 2q
=
+
1
=Í
Á ˜ ˙
r0 ÁË r0 ˜¯ ÍÎ Ë r ¯ ˙˚
r0 (cos 2q )3/ 2
4
r3 È Ê r ˆ ˘
= - 4 Í1 + Á 0 ˜ ˙
r0 ÍÎ Ë r ¯ ˙˚
Then
Ê 1ˆ
F µÁ ˜
Ë r¯
2
ÏÔ r 3
Ì- 4
ÓÔ r0
È Ê r0 ˆ 4 ˘ 1 ¸Ô
r
Í1 + Á ˜ ˙ + ˝ = - 4
r0
ÍÎ Ë r ¯ ˙˚ r ˛Ô
F ¥ r Q.E.D. b
Note: F 0 implies that F is repulsive.
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148
PROBLEM 12.98
It was observed that during the Galileo spacecraft’s first flyby of the earth, its minimum altitude was 960 km
above the surface of the earth. Assuming that the trajectory of the spacecraft was parabolic, determine the
maximum velocity of Galileo during its first flyby of the earth.
Solution
First we note
R = 6.37 ¥ 106 m
so that
r0 = (6.37 ¥ 106 + 960 ¥ 103 ) m
= 7.33 ¥ 106 m
Now
vmax = v0
and from page 709 of the text
v0 =
2GM
2 gR2
=
r0
r0
using Eq. (12.30).
1/ 2
Then
vmax
È 2 ¥ 9.81 m/s2 ¥ (6.37 ¥ 106 m)2 ˘
=Í
˙
7.33 ¥ 106 m
Î
˚
= 10, 421.7 m/s
vmax = 10.42 km/s b
or
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149
PROBLEM 12.99
As a space probe approaching the planet Venus on a parabolic trajectory reaches
Point A closest to the planet, its velocity is decreased to insert it into a circular orbit.
Knowing that the mass and the radius of Venus are 4.87 ¥ 1024 kg and 6052 km,
respectively, determine (a) the velocity of the probe as it approaches A, (b) the
decrease in velocity required to insert it into the circular orbit.
Solution
First note
rA = (6052 + 280) km = 6332 km
(a)From page 709 of the text, the velocity at the point of closest approach on a parabolic trajectory is given
by
2GM
v0 =
r0
1/ 2
Thus,
È 2 ¥ 66.73 ¥ 10 -12 m3 /kg ◊ s2 ¥ 4.87 ¥ 1024 kg ˘
( v A ) par = Í
˙
6332 ¥ 103 m
Î
˚
= 10,131.4 m/s
( v A ) par = 10.13 km/s b
or
(b)
We have
( v A )circ = ( v A ) par + Dv A
Now
( v A )circ =
=
Then
GM
r0
Eq. (12.44)
1
( v A ) par
2
1
( v A ) par - ( v A ) par
2
ˆ
Ê 1
=Á
- 1 (10.1314 km/s)
Ë 2 ˜¯
Dv A =
= -2.97 km/s
| Dv A | = 2.97 km/s b
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150
PROBLEM 12.100
It was observed that during its second flyby of the earth, the Galileo spacecraft had a velocity of 13.85 ¥ 103 m/s
as it reached its minimum altitude of 300 km above the surface of the earth. Determine the eccentricity of the
trajectory of the spacecraft during this portion of its flight.
Solution
First we note
R = 6370 km = 6.37 ¥ 106 m
and
r0 = (6370 + 300) km = 6670 km
= 6.67 ¥ 106 km
We have
1 GM
= 2 (1 + e cos q )
r
h
At Point O,
r = r0 , q = 0, h = h0 = r0 v0
Also,
GM = gR2
Eq. (12.30)
Then
gR2
1
=
(1 + e )
r0 ( r0 v0 )2
or
e=
=
r0 v02
gR2
Eq. (12.39)
-1
(6.67 ¥ 106 )(13.85 ¥ 103 m/s)2
-1
(9.81 m/s2 )(6.37 ¥ 106 )2
e = 2.21 b
or
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151
PROBLEM 12.101
It was observed that as the Galileo spacecraft reached the point on its trajectory closest to Io, a moon of the
planet Jupiter, it was at a distance of 2800 km from the center of Io and had a velocity of 15 ¥ 103 m/s. Knowing
that the mass of Io is 0.01496 times the mass of the earth, determine the eccentricity of the trajectory of the
spacecraft as it approached Io.
Solution
First note
r0 = 2800 km = 2.8 ¥ 106 m
Rearth = 6370 km = 6.37 ¥ 106 m
We have
1 GM
= 2 (1 + e cos q )
r
h
At Point O,
r = r0 , q = 0, h = h0 = r0 v0
Also,
GM Io = G(0.01496 M earth )
2
= 0.01496 gRearth
Then
or
Eq. (12.39)
using Eq. (12.30).
2
1 0.01496 gRearth
=
(1 + e )
r0
( r0 v0 )2
e=
=
r0 v02
2
0.01496 gRearth
-1
(2.8 ¥ 106 m)(15 ¥ 103 m/s)2
0.01496(9.81 m/s2 )(6.37 ¥ 106 m)2
-1
e = 104.8 b
or
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152
PROBLEM 12.102
A satellite describes an elliptic orbit about a planet of mass M. Denoting
by r0 and r1 , respectively, the minimum and maximum values of the
distance r from the satellite to the center of the planet, derive the relation
1 1 2GM
+ = 2
r0 r1
h
where h is the angular momentum per unit mass of the satellite.
Solution
Using Eq. (12.39),
1 GM
= 2 + C cos q A
rA
h
and
1 GM
= 2 + C cos q B .
rB
h
But
q B = q A + 180∞,
so that
cos q A = - cos q B .
Adding,
1 1 1 1 2GM
+ = + = 2
rA rB r0 r1
h
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153
PROBLEM 12.103
At main engine cutoff of its thirteenth flight, the space shuttle
Discovery was in an elliptic orbit of minimum altitude 65 km and
maximum altitude 540 km above the surface of the earth. Knowing
that at Point A the shuttle had a velocity v0 parallel to the surface of
the earth and that the shuttle was transferred to a circular orbit as
it passed through Point B, determine (a) the speed v0 of the shuttle
at A, (b) the increase in speed required at B to insert the shuttle
into the circular orbit.
Solution
For earth, R = 6373 km = 6.373 ¥ 106 m
GM = gR2 = (9.81 m s2 )(6.373 ¥ 106 )2 = 3.9843 ¥ 1014 m3 /s2
rA = 6373 + 65 = 6438 km = 6.438 ¥ 106 m
rB = 6373 + 540 = 6913 km = 6.913 ¥ 106 m
Elliptic trajectory.
Using Eq. (12.39),
1 GM
= 2 + C cos q A
rA
h
But
q B = q A + 180∞, so that cos q A = - cos q B
Adding,
and
1 GM
= 2 + C cos q B .
rB
h
1 1 rA + rB 2GM
+ =
= 2
rA rB
rA rB
h
h=
=
2GMrA rB
rA + rB
(2)(3.9843 ¥ 1014 )(6.438 ¥ 106 )(6.913 ¥ 106 )
(6.438 + 6.913) ¥ 106
= 5.15398 ¥ 1010 m2 /s
(a)
Speed v0 at A.
v0 = v A =
( v B )1 =
h 5.15398 ¥ 1010
=
rA
6.438 ¥ 106
v0 = 8.01 ¥ 103 m/s b
h
rB
5.15398 ¥ 1010
6.913 ¥ 106
= 7.4555 ¥ 103 m/s
=
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154
Problem 12.103 (continued)
For a circular orbit through Point B,
( v B )circ =
=
GM
rB
3.9843 ¥ 1014
6.913 ¥ 106
= 7.5918 ¥ 103 m/s
(b)
Increase in speed at Point B.
Dv B = ( v B )circ - ( v B )1
= (7.5918 - 7.4555) ¥ 103 =136.3m/s Dv B = 136.3 m/s b
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155
PROBLEM 12.104
A space probe is describing a circular orbit about a planet of radius R. The altitude of the probe above the
surface of the planet is a R and its speed is v0. To place the probe in an elliptic orbit which will bring it closer
to the planet, its speed is reduced from v0 to bv0 , where b 1, by firing its engine for a short interval of time.
Determine the smallest permissible value of b if the probe is not to crash on the surface of the planet.
Solution
GM
rA
For the circular orbit,
v0 =
where
rA = R + a R = R(1 + a )
Eq. (12.44),
GM = v02 R(1 + a )
Then
From the solution to Problem 12.102, we have for the elliptic orbit,
1 1 2GM
+ = 2
rA rB
h
h = hA = rA ( v A ) AB
Now
= [ R(1 + a )]( b v0 )
Then
2v02 R(1 + a )
1
1
+ =
R(1 + a ) rB [ R(1 + a )b v0 ]2
=
2
b R(1 + a )
2
Now b min corresponds to rB Æ R.
Then
1
1
2
+ = 2
R(1 + a ) R b min R(1 + a )
b min =
or
2
b
2+a
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156
PROBLEM 12.105
As it describes an elliptic orbit about the sun, a spacecraft reaches
a maximum distance of 320 ¥ 106 km from the center of the sun at
Point A (called the aphelion) and a minimum distance of 150 ¥ 106 km
at Point B (called the perihelion). To place the spacecraft in a smaller
elliptic orbit with aphelion at A¢ and perihelion at B ¢, where A¢ and
B ¢ are located 260 ¥ 106 and 135 ¥ 106 km, respectively, from the
center of the sun, the speed of the spacecraft is first reduced as it
passes through A and then is further reduced as it passes through
B ¢. Knowing that the mass of the sun is 332.8 ¥ 103 times the mass
of the earth, determine (a) the speed of the spacecraft at A, (b) the
amounts by which the speed of the spacecraft should be reduced at
A and B ¢ to insert it into the desired elliptic orbit.
Solution
First note
Rearth = 6370 km = 6.37 ¥ 106 m
rA = 320 ¥ 106 km = 320 ¥ 109 m
rB = 150 ¥ 106 km = 150 ¥ 109 m
From the solution to Problem 12.102, we have for any elliptic orbit about the sun
1 1 2GM sun
+ =
r1 r2
h2
(a)
For the elliptic orbit AB, we have
r1 = rA , r2 = rB , h = hA = rA v A
Also,
GM sun = G[(332.8 ¥ 103 ) M earth ]
2
= gRearth
(332.8 ¥ 103 )
Then
or
using Eq. (12.30).
2
(332.8 ¥ 103 )
1 1 2 gRearth
+ =
rA rB
(rA v A )2
R
v A = earth
rA
Ê 665.6 g ¥ 103 ˆ
Á
˜
1
1
Ë
¯
r + r
A
1/ 2
B
(6.37 ¥ 10 m) ÊÁ 665.6 ¥ 9.81 ¥ 10 ˆ˜
=
+
˜¯
(320 ¥ 10 m) ÁË
6
9
3
1
320 ¥ 109
1/ 2
1
150 ¥ 109
= 16, 255.6 m/s
v A = 16.3 ¥ 103 m/s b
or
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157
Problem 12.105 (continued)
(b)
From Part (a), we have
Ê 1 1ˆ
2GM sun = ( rA v A )2 Á + ˜
Ë rA rB ¯
Then, for any other elliptic orbit about the sun, we have
(
2 1
1
1 1 ( rA v A ) rA + rB
+ =
r1 r2
h2
)
For the elliptic transfer orbit AB¢, we have
r1 = rA , r2 = rB ¢ , h = htr = rA ( v A ) tr
Then
or
(
2 1
1
1
1 ( rA v A ) rA + rB
+
=
rA rB ¢
[rA ( v A ) tr ]2
Ê r1 + r1 ˆ
( v A )tr = v A Á 1A 1B ˜
Ë rA + rB¢ ¯
1/ 2
)
Ê 1 + rA ˆ
rB
˜
= vA Á
ÁË 1 + rrA ˜¯
B¢
320 ˆ
Ê 1 + 150
= (16, 255.6 m/s) Á
320 ˜
Ë 1 + 135 ¯
1/ 2
1/ 2
= 15, 673.6 m/s
htr = ( hA ) tr = ( hB ¢ ) tr : rA ( v A ) tr = rB ¢ ( v B ¢ ) tr
Now
Then
( v B¢ ) tr =
320 ¥ 109 km
135 ¥ 109 km
¥ 15673.6 m/s = 37,152.2 m/s
For the elliptic orbit A¢ B ¢, we have
r1 = rA¢ , r2 = rB ¢ , h = rB ¢ v B ¢
Then
or
(
2 1
1
1
1 ( rA v A ) rA + rB
+
=
rA¢ rB ¢
( rB ¢ v B ¢ )2
vB¢
)
1
1
r Ê r +r ˆ
= v A A Á 1A 1B ˜
rB ¢ Ë r + r ¯
A¢
B¢
1/ 2
1
1
Ê 320 ¥ 109 m ˆ Ê 320 ¥ 109 + 150 ¥ 109 ˆ
˜
= (16, 255.6 m/s) Á
˜Á
Ë 135 ¥ 109 m ¯ ÁË 260 1¥ 109 + 135 ¥1 109 ˜¯
1/ 2
= 35, 942.0 m/s
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158
Problem 12.105 (continued)
Finally,
or
( v A )tr = v A + Dv A
Dv A = (15673.6 - 16, 255.6) m/s
|Dv A | = 582 m/s b
or
and
or
v B ¢ = ( v B ¢ )tr + Dv B
Dv B¢ = (35, 942 - 37,152.2) m/s
= -1210.2 m/s
|Dv B | = 1210 m/s b
or
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159
PROBLEM 12.106
A space probe is to be placed in a circular orbit of 9000 km
radius about the planet Venus in a specified plane. As the probe
reaches A, the point of its original trajectory closest to Venus, it
is inserted in a first elliptic transfer orbit by reducing its speed by
Dv A . This orbit brings it to Point B with a much reduced velocity.
There the probe is inserted in a second transfer orbit located in the
specified plane by changing the direction of its velocity and further
reducing its speed by Dv B . Finally, as the probe reaches Point C,
it is inserted in the desired circular orbit by reducing its speed
by DvC . Knowing that the mass of Venus is 0.82 times the mass of
the earth, that rA = 15 ¥ 103 km and rB = 300 ¥ 103 km, and that the
probe approaches A on a parabolic trajectory, determine by how
much the velocity of the probe should be reduced (a) at A, (b) at B,
(c) at C.
Solution
R = 6370 km = 6.37 ¥ 106 m, g = 9.81 m/s2
For Earth,
GM earth = gR2 = (9.81)(6.37 ¥ 106 )2 = 3.98059 ¥ 1014 m3 /s2
GM = 0.82 GM earth = 3.2641 ¥ 1014 m3 /s2
For Venus,
For a parabolic trajectory with
rA = 15 ¥ 103 km = 15 ¥ 106 m
( v A )1 = vesc =
First transfer orbit AB.
(2)(3.2641 ¥ 1014 )
2GM
=
= 6.5971 ¥ 103 m/s
6
rA
15 ¥ 10
rB = 300 ¥ 103 km = 300 ¥ 106 m
At Point A, where q = 180∞
GM
1 GM
= 2 + C cos 180∞ = 2 - C rA hAB
hAB
(1)
GM
1 GM
= 2 + C cos 0 = 2 + C rB
hAB
hAB
(2)
At Point B, where q = 0∞
Adding,
r + rA 2GM
1
1
+
= B
= 2
rA rB
rA rB
hAB
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160
Problem 12.106 (continued)
Solving for hAB ,
hAB =
2GMrA rB
(2)(3.2641 ¥ 1014 )(15 ¥ 106 )(300 ¥ 106 )
=
rB + rA
315 ¥ 106
(
)
= 9.65712 ¥ 1010 m2 /s
Second transfer orbit BC.
( v A )2 =
hAB 9.65712 ¥ 1010
=
= 6.43808 ¥ 103 m/s
rA
15 ¥ 106
(v B )1 =
hAB 9.65712 ¥ 1010
=
= 0.321904 ¥ 103 m/s
6
rB
300 ¥ 10
rC = 9000 km = 9 ¥ 106 m
At Point B, where q = 0
1 GM
GM
= 2 + C cos 0 = 2 + C
rB
hBC
hBC
At Point C, where q = 180∞
1 GM
GM
= 2 + C cos 180∞ = 2 - C
rC
hBC
hBC
Adding,
1 1 rB + rC 2GM
+ =
= 2
rB rC
rB rC
hBC
hBC =
=
2GMrB rC
rB + rC
(2)(3.2641 ¥ 1014 )((300 ¥ 106 )(9 ¥ 106 )
(309 ¥ 106 )
= 7.55265 ¥ 1010 m2 /s
Final circular orbit.
( v B )2 =
hBC 7.55265 ¥ 1010
=
= 251.755 m/s
rB
300 ¥ 106
( vC )1 =
hBC 7.55265 ¥ 1010
= 8391.83 m/s
=
rC
9 ¥ 106
rC = 9 ¥ 106 m
( vC )2 =
=
GM
rC
3.2641 ¥ 1014
9 ¥ 106
= 6.02227 ¥ 103 m/s
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161
Problem 12.106 (continued)
Speed reductions.
(a)
At A:
( v A )1 - ( v A )2 = 6.5971 ¥ 103 - 6.43808 ¥ 103 = 159.02 m s
Dv A = 159.0 m/s b
(b)
At B:
( v B )1 - ( v B )2 = 0.321904 ¥ 103 - 251.755 = 70.149 m s
Dv B = 70.1 m/s b
(c)
At C:
( vC )1 - ( vC )2 = (8391.83 - 6.02227 ¥ 103 )
= 2369.13 m s
DvC = 2370 m/s b
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162
PROBLEM 12.107
For the space probe of Problem 12.106, it is known that
rA = 15 ¥ 103 km and that the velocity of the probe is reduced to
6000 m/s , as it passes through A. Determine (a) the distance from the
center of Venus to Point B, (b) the amounts by which the velocity of
the probe should be reduced at B and C, respectively.
Solution
For Earth,
R = 6370 km = 6.37 ¥ 106 m, g = 9.81 m/s2
GM earth = gR2 = (9.81)(6.37 ¥ 106 )2 = 3.98059 ¥ 1014 m3 /s2
For Venus,
Transfer orbit AB:
GM = 0.82 GM earth = 3.2641 ¥ 1014 m3 /s2
v A = 6000 m/s, rA = 15 ¥ 103 km = 15 ¥ 106 m
hAB = rA v A = (15 ¥ 106 )(6000) = 9 ¥ 1010 m2 /s
At Point A, where q = 180∞
1 GM
GM
= 2 + C cos 180∞ = 2 - C
rA hAB
hAB
At Point B, where q = 0∞
1 GM
GM
= 2 + C cos 0 = 2 + C
rB
hAB
hAB
Adding,
1
1
2GM
+
= 2
rA rB
hAB
1 2GM 1
= 2 rB
rA
hAB
=
(2)(3.2641 ¥ 1014 )
1
10 2
(9 ¥ 10 )
15 ¥ 106
= 1.39284 ¥ 10 -8 m -1
rB = 71.79578 ¥ 106 m
rB = 71.8 ¥ 103 km b
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163
Problem 12.107 (continued)
(a)
rB = 71.79578 ¥ 106 m
Radial coordinate rB .
( v B )1 =
hAB
9 ¥ 1010
=
= 1.25356 ¥ 103 m/s
rB
71.79578 ¥ 106
rC = 9000 km = 9 ¥ 106 m
Second transfer orbit BC.
At Point B, where q = 0
1 GM
GM
= 2 + C cos 0 = 2 + C
rB
hBC
hBC
At Point C, where q = 180∞
1 GM
GM
= 2 + C cos 180∞ = 2 - C
rC
hBC
hBC
Adding,
r + rC 2GM
1
1
+
= B
= 2
rB rC
rB rC
hBC
hBC =
2GMrB rC
=
rB + rC
(2)(3.2641 ¥ 1014 )(71.79578 ¥ 106 )(9 ¥ 106 )
(80.779578 ¥ 106 )
= 7.22555 ¥ 1010 m2 /s
( v B )2 =
hBC 7.22555 ¥ 1010
=
= 1.0064 ¥ 103 m/s
rB
71.79578 ¥ 106
( vC )1 =
hBC 7.22555 ¥ 1010
=
= 8.0284 ¥ 103 m/s
6
rC
9 ¥ 10
rC = 9 ¥ 106 m
Circular orbit with
( vC )2 =
(b)
GM
3.2641 ¥ 1014
=
= 6.0223 ¥ 103 m/s
6
rC
9 ¥ 10
Speed reductions at B and C.
At B:
( v B )1 - ( v B )2 = 1.25356 ¥ 103 - 1.0064 ¥ 103 = 247.16 m s
Dv B = 247 m/s b
At C:
( vC )1 - ( vC )2 = 8.0284 ¥ 103 - 6.0223 ¥ 103
= 2006.1 m s
DvC = 2010 m s b
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164
PROBLEM 12.108
Determine the time needed for the probe of 12.106 to travel from A to B on its first transfer orbit.
Solution
From Problem 12.106 for the first transfer orbit
rA = 15 ¥ 106 m
rB = 300 ¥ 106 m
hAB = 9.65712 ¥ 1010 m2 /s
1
a = ( rA + rB )
2
1
= (15 ¥ 106 + 300 ¥ 106 )
2
= 157.5 ¥ 106 m
b = rA rB
= (15 ¥ 106 )(300 ¥ 106 )
= 67.082 ¥ 106 m
Periodic time for full elliptical orbit.
t=
2p ab
hAB
(2p )(157.5 ¥ 106 )(67.082 ¥ 106 )
9.65712 ¥ 1010
= 687.415 ¥ 103 s
=
Time to travel from A to B.
1
t AB = t = 343.707 ¥ 103 s
2
t AB = 95 hrs.28min 27 sec.
b
ª 95hrs.30 min.
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165
PROBLEM 12.109
The Clementine spacecraft described an elliptic orbit of minimum
altitude hA = 400 km and a maximum altitude of hB = 2940 km
above the surface of the moon. Knowing that the radius of the moon
is 1737 km and the mass of the moon is 0.01230 times the mass of
the earth, determine the periodic time of the spacecraft.
Solution
For earth,
R = 6370 km = 6.370 ¥ 106 m
GM = gR2 = (9.81)(6.370 ¥ 106 )2 = 398.06 ¥ 1012 m3 /s2
For moon,
GM = (0.01230)(398.06 ¥ 1012 ) = 4.896 ¥ 1012 m3 /s2
rA = 1737 + 400 = 2137 km = 2.137 ¥ 106 m
rB = 1737 + 2940 = 4677 km = 4.677 ¥ 106 m
Using Eq. (12.39),
1 GM
= 2 + C cos q A
rA
h
But
q B = q A + 180∞, so that cos q A = - cos q B .
Adding,
and
1 GM
= 2 + C cos q B .
rB
h
1 1 rA + rB 2GM
+ =
= 2
rA rB
rA rB
hAB
hAB =
2GMrA rB
(2)( 4.896 ¥ 1012 )(2.137 ¥ 106 )( 4.677 ¥ 106 )
=
rA + rB
6.804 ¥ 106
= 3.78856 ¥ 109 m2 /s
1
a = ( rA + rB ) = 3.402 ¥ 106 m
2
b = rA rB = 3.16145 ¥ 106 m
Periodic time.
t=
2p ab 2p (3.402 ¥ 106 )(3.16145 ¥ 106 )
=
= 17.837 ¥ 103 s
9
hAB
3.78856 ¥ 10
t = 4.95h b
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166
PROBLEM 12.110
A space probe in a low earth orbit is inserted into an elliptic
transfer orbit to the planet Venus. Knowing that the mass of the
sun is 332.8 ¥ 103 times the mass of the earth and assuming that
the probe is subjected only to the gravitational attraction of the
sun, determine the value of f, which defines the relative position of
Venus with respect to the earth at the time the probe is inserted into
the transfer orbit.
Solution
First determine the time t probe for the probe to travel from the earth to Venus. Now
1
t probe = t tr
2
where t tr is the periodic time of the elliptic transfer orbit. Applying Kepler’s Third Law to the orbits about the
sun of the earth and the probe, we obtain
atr3
t tr2
=
2
3
aearth
t earth
where
1
atr = ( rE + rv )
2
1
= (150 ¥ 106 + 108 ¥ 106 ) km
2
= 129 ¥ 106 km
and
aearth ª rE
( Note: e earth = 0.0167)
Then
1Êa ˆ
t probe = Á tr ˜
2 Ë rE ¯
3/ 2
t earth
1 Ê 129 ¥ 106 km ˆ
= Á
2 Ë 150 ¥ 106 km ˜¯
3/ 2
(365.25 days)
= 145.649 days
= 12.5841 ¥ 106 s
In time t probe , Venus travels through the angle q v given by
v
q v = q&v t probe = v t probe
rv
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167
Problem 12.110 (continued)
Assuming that the orbit of Venus is circular (note: e venus = 0.0068), then, for a circular orbit
vv =
Now
GM sun
rv
[Eq. (12.44)]
GM sun = G(332.8 ¥ 103 M earth )
(
) using Eq. (12.30)
È 332.8 ¥ 10 ( gR ) ˘
˙
Í
2
= 332.8 ¥ 103 gRearth
Then
qv =
t probe
3
rv Í
Î
rv
= t probe Rearth
where
2
earth
1/ 2
˙
˚
(332.8 g ¥ 103 )1/ 2
rv3/ 2
Rearth = 6370 km = 6.37 ¥ 106 m
and
rv = 108 ¥ 106 km = 108 ¥ 109 m
Then
q v = (12.5841 ¥ 106 s)(6.37 ¥ 106 m)
(332.8 ¥ 103 ¥ 9.81 m/s2 )1/ 2
(108 ¥ 109 )3/2
= 4.0809 rad
= 233.818∞
Finally,
f = q v - 180∞ = 233.818∞ - 180∞ = 53.818∞
j = 53.8∞ b
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168
PROBLEM 12.111
Based on observations made during the 1996 sighting of comet Hyakutake, it was concluded that the trajectory
of the comet is a highly elongated ellipse for which the eccentricity is approximately e = 0.999887. Knowing
that for the 1996 sighting, the minimum distance between the comet and the sun was 0.230 RE , where RE is the
mean distance from the sun to the earth, determine the periodic time of the comet.
Solution
For Earth’s orbit about the sun,
v0 =
2p RE 2p RE 3/ 2
GM
, t0 =
=
RE
v0
GM
or
GM =
2p RE3/ 2
t0
(1)
For the comet Hyakutake,
1 GM
= 2 = (1 + e ),
r0
h
1 GM
1+ e
= 2 (1 + e ), r1 =
r0
1- e
r1
h
r
1
1+ e
r0
a = ( r0 + r1 ) = 0 , b = r0 r1 =
2
1- e
1- e
h = GMr0 (1 + e )
t=
=
2p r02 (1 + e )1/ 2
2p ab
=
h
(1 - e )3/ 2 GMr0 (1 + e )
2p r03/ 2
GM (1 - e )3/ 2
Ê r ˆ
=Á 0 ˜
Ë RE ¯
3/ 2
1
(1 - e )3/ 2
= (0.230)3/ 2
Since
=
2p r03/ 2t 0
2p RE3 (1 - e )3/ 2
t0
1
t 0 = 91.8 ¥ 103 t 0
(1 - 0.999887)3/ 2
t 0 = 1 yr, t = (91.8 ¥ 103 )(1.000)
t = 91.8 ¥ 103 yr b
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169
PROBLEM 12.112
Halley’s comet travels in an elongated elliptic orbit for which the minimum distance from the sun is approximately
1
r , where rE = 150 ¥ 106 km is the mean distance from the sun to the earth. Knowing that the periodic time of
2 E
Halley’s comet is about 76 years, determine the maximum distance from the sun reached by the comet.
Solution
We apply Kepler’s Third Law to the orbits and periodic times of earth and Halley’s comet:
2
ÊtH ˆ
Ê aH ˆ
ÁË t ˜¯ = ÁË a ˜¯
E
E
Thus
3
Êt ˆ
aH = aE Á H ˜
Ë tE ¯
2/3
Ê 76 years ˆ
= tE Á
Ë 1 year ˜¯
2/3
= 17.94t E
But
1
aH = (t min + t max )
2
1Ê1
ˆ
17.94t E = Á t E + t max ˜
¯
Ë
2 2
1
t max = 2(17.94t E ) - t E
2
= (35.88 - 0.5)t E
= 35.38 t E
t max = 150 ¥ 106 km = 5.31 ¥ 109 km b
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170
PROBLEM 12.113
Determine the time needed for the space probe of Problem 12.99 to travel from B to
C.
Solution
From the solution to Problem 12.99, we have
( v A ) par = 10,131.4 m/s
and
Also,
( v A )circ =
1
( v A ) par = 7164.0 m/s
2
rA = (6052 + 280) km = 6332 km
For the parabolic trajectory BA, we have
where e = 1. Now
1 GM v
= 2 (1 + e cos q )
r
hBA
at A, q = 0:
1 GM v
= 2 (1 + 1)
rA
hBA
or
rA =
at B, q = -90∞ :
1 GM v
= 2 (1 + 0)
rB
hBA
or
rB =
[Eq. (12.39¢)]
2
hBA
2GM v
2
hBA
GM v
rB = 2rA
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171
Problem 12.113 (continued)
As the probe travels from B to A, the area swept out is the semiparabolic area defined by Vertex A and Point B.
Thus,
2
4
(Area swept out) BA = ABA = ( rA )( rB ) = rA2
3
3
Now
dA 1
= h
dt 2
where h = constant
Then
A=
1
ht
2
or t BA =
=
2 ABA
hBA
2 ¥ 43 rA2
rA v A
hBA = rA v A
=
8 rA
3 vA
8 6332 ¥ 103 m
3 10,131.4 m/s
= 1666.63 s
=
For the circular trajectory AC,
t AC
Finally,
p
rA
p 6332 ¥ 103 m
= 2
=
= 1388.37 s
( v A )circ 2 7164.0 m/s
t BC = t BA + t AC
= (1666.63 + 1388.37) s
=3055.0 s
t BC = 50 min 55 s b
or
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172
PROBLEM 12.114
A space probe is describing a circular orbit of radius nR with a velocity v0
about a planet of radius R and center O. As the probe passes through point A,
its velocity is reduced from v0 to bv0, where b 1, to place the probe on a
crash trajectory. Express in terms of n and b the angle AOB, where B denotes
the point of impact of the probe on the planet.
Solution
For the circular orbit,
r0 = rA = nR
v0 =
GM
GM
=
r0
nR
The crash trajectory is elliptic.
v A = b v0 =
b 2GM
nR
h = rA v A = nRv A = b 2 nGMR
GM
1
= 2
b nR
h2
At Point A, q = 180∞
1 + e cos q
1 GM
= 2 (1 + e cos q ) =
r
h
b 2 nR
1
1
1- e
=
= 2
rA nR b nR
or b 2 = 1 - e
or e = 1 - b 2
At impact Point B, q = p - f
1 1
=
rB R
1 1 + e cos (p - f ) 1 - e cos f
=
=
R
b 2 nR
b 2 nR
e cos f = 1 - nb 2
or cos f =
1 - nb 2 1 - nb 2
=
e
1- b2
f = cos -1[(1 - nb 2 ) / (1 - b 2 )] b
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173
PROBLEM 12.115
Prior to the Apollo missions to the moon, several Lunar Orbiter
spacecraft were used to photograph the lunar surface to obtain
information regarding possible landing sites. At the conclusion of
each mission, the trajectory of the spacecraft was adjusted so that the
spacecraft would crash on the moon to further study the characteristics
of the lunar surface. Shown is the elliptic orbit of Lunar Orbiter 2.
Knowing that the mass of the moon is 0.01230 times the mass of the
earth, determine the amount by which the speed of the orbiter should
be reduced at Point B so that it impacts the lunar surface at Point C.
(Hint: Point B is the apogee of the elliptic impact trajectory.)
Solution
From the solution to Problem 12.102, we have for the elliptic orbit AB
1 1 2GM moon
+ =
2
rA rB
hAB
where
and
hAB = ( hB ) AB = rB ( v B ) AB
GM moon = G(0.01230 M earth )
2
= 0.01230 gRearth
Then
or
using Eq. (12.30)
2
)
1 1 2(0.01230 gRearth
+ =
2
rA rB
[rB ( v B ) AB ]
( v B ) AB
R
= earth
rB
Ê 0.0246 g ˆ
Á 1 1 ˜
Ë rA + rB ¯
1/ 2
6.37 ¥ 106 m Ê 0.0246 ¥ 9.81 m/s2 ˆ
˜
Á
=
3600 ¥ 103 m ÁË 1790 ¥1103 m + 3600 ¥1 103 m ˜¯
1/ 2
= 950.43 m/s
For the elliptic impact trajectory, we have
1 GM moon
=
+ C cosq
2
r
hBC
where
[Eq. (12.39)]
hBC = ( hB ) BC = rB ( v B ) BC
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174
Problem 12.115 (continued)
Noting that Point B is the apogee of this trajectory, we have
1 GM moon
at B, q = 180∞ :
=
-C
2
rB
hBC
GM moon
C=
at C, q = -70∞ :
1 GM moon
=
+ C cos ( -70∞)
2
R
hBC
or
C=
Then
or
GM moon
2
hBC
-
2
hBC
-
1
rB
or
1 Ê 1 GM moon ˆ
˜
2
cos 70∞ ÁË R
hBC
¯
1
1 Ê 1 GM moon ˆ
=
˜
2
rB cos 70∞ ÁË R
hBC
¯
2
hBC
=
GM moon (1 + cos 70∞)
1
R
+
cos 70∞
rB
1/ 2
or
( v B ) BC
R
= earth
rB
È 0.012309(1 + cos 70∞) ˘
Í
˙
cos 70∞
1
Í
˙
+ r
R
Î
B
˚
1/ 2
È
˘
6.37 ¥ 10 m Í 0.01230(9.81 m/s2 )(1 + cos 70∞) ˙
=
cos 70∞
1
˙
3600 ¥ 103 m Í
+ 3600 ¥ 103 m
1737 ¥ 103 m
ÎÍ
˚˙
= 869.43 m/s
6
( v B ) BC
Finally,
or
( v B ) BC = ( v B ) AB + Dv B
Dv B = (869.43 - 950.43) m/s
|Dv B | = 81.0 m/s b
or
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175
PROBLEM 12.116
As a spacecraft approaches the planet Jupiter, it releases a probe which is to
enter the planet’s atmosphere at Point B at an altitude of 450 km above the
surface of the planet. The trajectory of the probe is a hyperbola of eccentricity
e = 1.031. Knowing that the radius and the mass of Jupiter are 71.492 ¥ 103 km
and 1.9 ¥ 1027 kg, respectively, and that the velocity vB of the probe at B forms
an angle of 82.9° with the direction of OA, determine (a) the angle AOB, (b) the
speed v B of the probe at B.
Solution
rB = (71.492 ¥ 103 + 450) km = 71.942 ¥ 103 km
First we note
(a)
1 GM j
= 2 (1 + e cos q )
r
h
We have
1 GM j
= 2 (1 + e )
rA
h
At A, q = 0:
h2
= rA (1 + e )
GM j
or
At B, q = q B = R AOB :
or
Then
or
[Eq. (12.39¢)]
1 GM j
= 2 (1 + e cos q B )
rB
h
h2
= rB (1 + e cos q B )
GM j
rA (1 + e ) = rB (1 + e cos q B )
cos q B =
=
˘
1 È rA
Í (1 + e ) - 1˙
e Î rB
˚
˘
1 È 70.8 ¥ 103 km
(1 + 1.031) - 1˙
Í
3
1.031 Î 71.942 ¥ 10 km
˚
= 0.96873
or
R AOB = 14.37∞ b
q B = 14.3661∞
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176
Problem 12.116 (continued)
(b)
From above
where
Then
h2 = GM j rB (1 + e cos q B )
1
|rB ¥ mv B | = rB v B sin f
m
f = (q B + 82.9∞) = 97.2661∞
h=
( rB v B sin f )2 = GM j rB (1 + e cos q B )
1/ 2
or
˘
1 È GM j
(1 + e cos q B )˙
vB =
Í
sin f Î rB
˚
=
1
sin 97.2661∞
1/ 2
ÏÔ 66.73 ¥ 10 -12 m2 /kg ◊ s2 ¥ 1.9 ¥ 1027 kg
¸Ô
¥ [1 + (1.031)(0.96873)]˝
Ì
6
71.942 ¥ 10 m
ÓÔ
˛Ô
v B = 59.8 km/s b
or
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177
PROBLEM 12.117
A space shuttle is describing a circular orbit at an altitude of 600 km above
the surface of the earth. As it passes through Point A, it fires its engine for
a short interval of time to reduce its speed by 150 m/s and begin its descent
toward the earth. Determine the angle AOB so that the altitude of the shuttle
at Point B is 120 km. (Hint: Point A is the apogee of the elliptic descent
trajectory.)
Solution
First we note
R = 6370 km = 6.37 ¥ 106 m
rA = (6370 + 600) km = 6970 km
= 6.97 ¥ 106 m
rB = (6370 + 125) km = 6495 km
For the circular orbit, we have
vcirc =
gR2
rA
[Eq. (12.44)]
Ê 9.81 m/s2 ˆ
= 6.37 ¥ 10 m Á
˜
Ë 6.97 ¥ 106 m ¯
1/ 2
6
= 7557.14 m/s
Now
( v A ) AB = vcirc + Dv A = (7557.14 - 150) m/s
= 7407.14 m/s
For the elliptic descent trajectory, we have
1 GM
= 2 + C cosq
r
h
[Eq. (12.39)]
Noting that Point A is at the apogee of this trajectory, we have
1 GM
at A, q = 180∞ :
= 2 -C
rA
h
or
C=
GM 1
rA
h2
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178
Problem 12.117 (continued)
at B, q = q B = 180∞ -R AOB :
1 GM
= 2 + C cos q B
rB
h
or
C=
Then
or
Now
1 Ê 1 GM ˆ
- 2 ˜
cos q B ÁË rB
h ¯
GM 1
1
- =
2
r
cos
qB
h
A
cos q B =
Ê 1 GM ˆ
ÁË r - h2 ˜¯
B
GM
1
rB - h2
GM
- r1
h2
A
h = ( hA ) AB = rA ( v A ) AB
and
GM = gR2
[Eq. (12.30)]
From above,
gR2 = rA ( vcirc )2
[Eq. (12.44)]
Then
˘
rA ( vcirc )2
GM
1 È v
=
= Í circ ˙
2
2
rA Î (v A ) AB ˚
[rA ( v A ) AB ]
h
so that
2
2
2
v
rA
È vcirc ˘
- r1 È (v circ) ˘
rB - Î (v A ) AB ˚
A Î A AB ˚
=
cos q B =
2
1 È vcirc ˘2 1
È vcirc ˘ - 1
)
(
v
(
)
v
r
A
Î A AB ˚
rA Î A AB ˚
1
rB
=
.14 2
( 7557
7407.14 )
.14 2
( 7557
7407.14 ) - 1
6970 km
6495 km
-
= 0.78758
or
Finally,
q B = 38.04∞
R AOB = 180∞ - 38.04∞
R AOB = 142.0∞ b
or
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179
PROBLEM 12.118
A satellite describes an elliptic orbit about a planet. Denoting by r0
and r1 the distances corresponding, respectively, to the perigee and
apogee of the orbit, show that the curvature of the orbit at each of
these two points can be expressed as
1 1Ê 1 1ˆ
=
+
r 2 ÁË r0 r1 ˜¯
Solution
Using Eq. (12.39),
1 GM
= 2 + C cos q A
rA
h
and
1 GM
= 2 + C cos q B .
rB
h
But
q B = q A + 180∞,
so that
cos q A = - cos q B
Adding,
1 1 2GM
+ = 2
rA rB
h
At Points A and B, the radial direction is normal to the path.
v2
h2
an =
= 2
r rr
But
Fn =
GMm
mh2
=
ma
=
n
r2
r 2r
1 GM 1 Ê 1 1 ˆ
= 2 = Á + ˜
r
2 Ë rA rB ¯
h
1 1Ê 1 1ˆ
=
+
b
r 2 ÁË r0 r1 ˜¯
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180
PROBLEM 12.119
(a) Express the eccentricity e of the elliptic orbit described
by a satellite about a planet in terms of the distances r0 and r1
corresponding, respectively, to the perigee and apogee of the orbit.
(b) Use the result obtained in Part a and the data given in Problem
12.111, where RE = 149.6 ¥ 106 km, to determine the approximate
maximum distance from the sun reached by comet Hyakutake.
Solution
(a)
We have
1 GM
= 2 (1 + e cos q )
r
h
At A, q = 0:
1 GM
= 2 (1 + e )
r0
h
or
At B, q = 180∞ :
or
Then
Eq. (12.39¢)
h2
= r0 (1 + e )
GM
1 GM
= 2 (1 - e )
r1
h
h2
= r1 (1 - e )
GM
r0 (1 + e ) = r1 (1 - e )
e=
or
(b)
1+ e
r0
1- e
From above,
r1 =
where
r0 = 0.230 RE
Then
r1 =
1 + 0.999887
¥ 0.230(149.6 ¥ 109 m)
1 - 0.999887
or
Note: r1 = 4070 RE
r1 - r0
b
r1 + r0
r1 = 609 ¥ 1012 m b
or r1 = 0.064 lightyears.
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181
PROBLEM 12.120
Show that the angular momentum per unit mass h of a satellite describing an elliptic orbit of semimajor axis a
and eccentricity e about a planet of mass M can be expressed as
h = GMa(1 - e 2 )
Solution
By Eq. (12.39¢ ),
1 GM
= 2 (1 + e cos q )
r
h
At A, q = 0∞ :
1 GM
= 2 = (1 + e )
rA
h
or
rA =
h2
GM (1 + e )
At B, q = 180∞ :
1 GM
= 2 = (1 - e )
rB
h
or
rB =
h2
GM (1 - e )
h2
1 ˆ
2 h2
Ê 1
=
=Á
+
˜
GM Ë 1 + e 1 - e ¯ GM (1 - e 2 )
Adding,
rA + rB =
But for an ellipse,
rA + rB = 2a
2a =
2 h2
GM (1 - e 2 )
h = GMa(1 - e 2 ) b
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182
PROBLEM 12.121
Derive Kepler’s third law of planetary motion from Eqs. (12.39) and (12.45).
Solution
For an ellipse,
2a = rA + rB
and b = rA rB
Using Eq. (12.39),
1 GM
= 2 + C cosq A
rA
h
and
1 GM
= 2 + C cos q B .
rB
h
But
q B = q A + 180∞,
so that
cos q A = - cos q B .
Adding,
1 1 rA + rB 2a 2GM
+ =
= 2 = 2
rA rB
rA rB
b
h
h=b
By Eq. (12.45),
t=
t2 =
GM
a
2p ab 2p ab a 2p a3/ 2
=
=
h
b GM
GM
4p 2 a3
GM
For Orbits 1 and 2 about the same large mass,
and
t 12 =
4p 2 a13
GM
t 22 =
4p 2 a23
GM
2
3
Ê t1 ˆ
Ê a1 ˆ
ÁË t ˜¯ = ÁË a ˜¯ b
2
2
Forming the ratio,
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183
PROBLEM 12.122
A 1000-kg automobile is being driven down a 5° incline at a speed of 80 km/h when the brakes are applied,
causing a total braking force of 6000 N to be applied to the automobile. Determine the distance traveled by the
automobile before it comes to a stop.
Solution
+
We have
SFx = ma : W sin 5∞ - ( FF + FR ) =
W
a
g
FF + FR = Fbrake
where
Ê
6000 N ˆ
a = (9.81 m/s2 ) Á sin 5∞ 1000 ¥ 9.81˜¯
Ë
Then
= -5.145 m/s2
For this uniformly decelerated motion, we have
v 2 = v02 + 2a( x - 0)
where
Then when v = 0,
v0 = 80 km/h =
200
m/s
9
2
Ê 200 ˆ
+ 2( -5.145 m/s2 ) x
0=Á
Ë 9 ˜¯
x = 47.991 m
x = 48.0 m b
or
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184
PROBLEM 12.123
A 6-kg block B rests as shown on a 10-kg bracket A. The coefficients
of friction are m s = 0.30 and m k = 0.25 between block B and bracket
A, and there is no friction in the pulley or between the bracket and
the horizontal surface. (a) Determine the maximum mass of block
C if block B is not to slide on bracket A. (b) If the mass of block C is
10% larger than the answer found in a, determine the accelerations
of A, B and C.
Solution
Kinematics. Let x A and x B be horizontal coordinates of A and B measured from a fixed vertical line to the left
of A and B. Let yC be the distance that block C is below the pulley. Note that yC increases when C moves
downward. See figure.
The cable length L is fixed.
L = ( x B - x A ) + ( xP - x A ) + yC + constant
Differentiating and noting that x& P = 0,
v B - 2v A + vC = 0
-2aA + aB + aC = 0 (1)
Here, aA and aB are positive to the right, and aC is positive downward.
Kinetics. Let T be the tension in the cable and FAB be the friction force between blocks A and B. The free body
diagrams are:
Bracket A:
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185
Problem 12.123 (continued)
Block B:
Block C:
+
(2)
+
Block B:
SFx = max : 2T - FAB = mA a A
SFx = max : FAB - T = mB aB
(3)
+
Bracket A:
SFy = ma y : N AB - mA g = 0
N AB = mA g
or
+
Block C:
SFy = ma y : mC g - T = maC (4)
Adding Eqs. (2), (3), and (4), and transposing,
mA aA + mB aB + mC aC = mC g (5)
Subtracting Eq. (4) from Eq. (3) and transposing,
mB aB - mC aC = FAB - mC g (a)
(6)
No slip between A and B. aB = aA
From Eq. (1),
From Eq. (5),
For impending slip,
aA = aB = aC = a
a=
mC g
mA + mB + mC
FAB = m s N AB = m s mB g
Substituting into Eq. (6),
( mB - mC )( mC g )
= m s mB g - mC g
mA + mB + mC
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186
Problem 12.123 (continued)
Solving for mC ,
mC =
=
m s mB ( mA + mB )
mA + 2mB - m s mC
(0.30)(6)(10 + 6)
10 + (2)(6) - (0.30)(6)
mC = 1.426 kg b
(b)
mC increased by 10%.
mC = 1.568 kg
Since slip is occurring,
FAB = m k N AB = m k mB g
Eq. (6) becomes
mB aB - mC aC = ( m k mB - mC ) g
or
6aB - 1.568aC = [(0.25)(6) - 1.568](9.81)
(7)
With numerical data, Eq. (5) becomes
10a A + 6aB + 1.568aC = (1.568)(9.81)
(8)
Solving Eqs. (1), (7), and (8) gives
aA = 1.053 m/s2 , aB = 0.348 m/s2 , aC = 1.759 m/s2
a A = 1.503 m/s2
b
a B = 0.348 m/s2
b
aC = 1.759 m/s2 b
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187
PROBLEM 12.124
Block A 10 kg, and blocks B and C weigh 5 kg each. Knowing that the blocks are
initially at rest and that B moves 2.5 in 2 s, determine (a) the magnitude of the force
P, (b) the tension in the cord AD. Neglect the masses of the pulleys and axle friction.
Solution
Let the position coordinate y be positive downward.
Constraint of cord AD: y A + yD = constant
v A + vD = 0,
Constraint of cord BC:
a A + aD = 0
( y B - yD ) + ( yC - yD ) = constant
v B + vC - 2vD = 0,
aB + aC - 2aD = 0
2aA + aB + aC = 0
Eliminate aD .
(1)
We have uniformly accelerated motion because all of the forces are
constant.
1
y B = ( y B ) 0 + ( v B ) 0 t + aB t 2 , ( v B ) 0 = 0
2
Pulley D:
+
aB =
2[ y B - ( y B )0 ]
t
2
=
2(2.5)
= 1.25 m/s2
2
(2)
SFy = 0 : 2TBC - TAD = 0
Block A:
or
+
TAD = 2TBC
SFy = ma y : mA g - TAD = mA aA
aA = g -
TAD
mA
(2)
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188
Problem 12.124 (continued)
+
Block C:
SFy = ma y : mc g - TBC = mc ac
TBC
(3)
mC
Substituting the value for aB and Eqs. (2) and (3) into Eq. (1), and solving for
TBC ,
Ê
Ê
T ˆ
2T ˆ
2 Á g - BC ˜ + aB + Á g - BC ˜ = 0
mC ¯
mA ¯
Ë
Ë
aC = g -
or
2g -
4TBC
T
+ 1.25 + g - BC = 0
10
5
Block B:
(a)
+
3g + 1.25 =
3TBC
5
TBC = 51.133 N
SFy = ma y : P + mB g - TBC = mB aB
Magnitude of P.
P = TBC - mB g + mB a B
P = 51.133 - 5 g + 5(1.25) = 8.333 N
(b)
P = 8.33 N b
Tension in cord AD.
TAD = 2TBC = (2)(51.133)
= 102.266 N TAD = 102.3 N b
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189
PROBLEM 12.125
A 6-kg block B rests as shown on the upper surface of a 15 kg wedge A.
Neglecting friction, determine immediately after the system is released from
rest (a) the acceleration of A, (b) the acceleration of B relative to A.
Solution
Acceleration vectors:
a A = aA
30∞, a B /A = aB /A
a B = a A + a B/ A
SFx = max : mB aB /A - mB a A cos 30∞ = 0
+
Block B:
aB /A = aA cos 30∞
(1)
+
SFy = ma y : N AB - WB = - mB aA sin 30∞
N AB = WB - (WB sin 30∞)
(2)
aA
g
a
a
W A sin 30∞ + WB sin 30∞ - (WB sin 2 30∞) A = WA A
g
g
+
Block A:
aA
g
SF = ma : WA sin 30∞ + N AB sin 30∞ = WA
aA =
(W A + WB )sin 30∞
2
g=
( mA + mB )sin 30∞
W A + WB sin 30∞
mA + mB sin 2 30∞
21sin 30∞
¥ (9.81)
=
15 + 6 sin 2 30∞
g
a A = 6.24 m/s2
(a)
30∞ b
aB/ A = (6.24) cos 30∞ = 5.40 m/s2
a B /A = 5.40 m/s2
(b)
b
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190
PROBLEM 12.126
The roller-coaster track shown is contained in a vertical plane.
The portion of track between A and B is straight and horizontal,
while the portions to the left of A and to the right of B have
radii of curvature as indicated. A car is traveling at a speed of
72 km/h when the brakes are suddenly applied, causing the
wheels of the car to slide on the track ( m k = 0.25). Determine
the initial deceleration of the car if the brakes are applied as the
car (a) has almost reached A, (b) is traveling between A and B,
(c) has just passed B.
Solution
v = 72 km/h = 20 m/s
(a)
r = 30 m
Almost reached Point A.
v 2 (20)2
=
= 13.333 m/s2 ≠
r
30
SFy = ma y : N R + N F - mg = man
an =
N R + N F = m( g + an )
F = m k ( N R + N F ) = m k m( g + an )
SFx = max : - F = mat
+
at = -
F
= -m k ( g + an )
m
| at | = m k ( g + an ) = 0.25(9.81 + 13.33)
(b)
Between A and B.
r=•
an = 0
| at | = m k g = (0.25)(9.81)
(c)
Just passed Point B.
| at | = 5.79 m/s2 b
| at | = 2.45 m/s2 b
r = 40 m
v 2 (20)2
=
= 8.8889 m/s2 Ø
r
45
SFy = ma y : N R + N F - mg = - man
an =
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191
Problem 12.126 (continued)
or
N R + N F = m( g - an )
F = m k ( N R + N F ) = m k m( g - an )
+
SFx = max : - F = mat
at = -
F
= - m k ( g - an )
m
| at | = m k ( g - an ) = (0.25)(9.81 - 8.8889)
| at | = 0.230 m/s2 b
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192
PROBLEM 12.127
A small 200-g collar C can slide on a semicircular rod which is made to rotate about
the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value
of the coefficient of static friction between the collar and the rod if the collar is not to
slide when (a) q = 90∞, (b) q = 75∞, (c) q = 45∞. Indicate in each case the direction of
the impending motion.
Solution
vC = ( r sin q )f& AB
First note
= (0.6 m)(6 rad/s)sin q
= (3.6 m/s)sin q
(a)
With q = 90∞,
vC = 3.6 m/s
+
SFy = 0 : F - WC = 0
or
F = mC g
Now
F = ms N
1
N=
mC g
ms
or
+
or
or
SFn = mC an : N = mC
vC2
r
v2
1
mC g = mC C
ms
r
ms =
gr (9.81 m/s2 )(0.6 m)
=
vC2
(3.6 m/s)2
( m s ) min = 0.454 b
or
The direction of the impending motion is downward.
b
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193
Problem 12.127 (continued)
(b) and (c)
First observe that for an arbitrary value of q, it is not known whether the impending motion will be upward
or downward. To consider both possibilities for each value of q, let Fdown correspond to impending motion
downward, Fup correspond to impending motion upward, then with the “top sign” corresponding to Fdown,
we have
+
SFy = 0 : N cos q ± F sin q - WC = 0
F = ms N
Now
N cos q ± m s N sin q - mC g = 0
Then
or
N=
mC g
cos q ± m s sin q
and
F=
m s mC g
cos q ± m s sin q
+
SFn = mC an : N sin q m F cos q = mC
vC2
r
r = r sin q
Substituting for N and F
m s mC g
mC g
v2
sin q m
cos q = mC C
r sin q
cos q ± m s sin q
cos q ± m s sin q
ms
vC2
tan q
m
=
1 ± m s tan q 1 ± m s tan q gr sin q
or
ms = ±
or
1+
vC2
gr sin q
vC2
gr sin q
tan q
vC2
[(3.6 m/s)sin q ]2
=
= 2.2018 sin q
gr sin q (9.81 m/s2 )(0.6 m)sin q
Now
Then
(b)
tan q -
ms = ±
tan q - 2.2018 sin q
1 + 2.2018 sin q tan q
ms = ±
tan 75∞ - 2.2018 sin 75∞
= ± 0.1796
1 + 2.2018 sin 75∞ tan 75∞
q = 75∞
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194
Problem 12.127 (continued)
Then
downward:
m s = + 0.1796
upward:
ms 0
not possible
( m s ) min = 0.1796 b
The direction of the impending motion is downward.
(c)
b
q = 45∞
ms = ±
tan 45∞ - 2.2018 sin 45∞
= ± ( - 0.218)
1 + 2.2018 sin 45∞ tan 45∞
Then
downward:
ms 0
upward:
m s = 0.218
not possible
( m s ) min = 0.218 b
The direction of the impending motion is upward.
Note: When
or
b
tan q - 2.2018 sin q = 0
q = 62.988∞,
m s = 0. Thus, for this value of q, friction is not necessary to prevent the collar from sliding on the rod.
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195
PROBLEM 12.128
Pin B weighs 100 g and is free to slide in a horizontal plane along the
rotating arm OC and along the circular slot DE of radius b = 500 mm.
Neglecting friction and assuming that q& = 15 rad/s and q&& = 250 rad/s2
for the position q = 20∞, determine for that position (a) the radial and
transverse components of the resultant force exerted on pin B, (b)
the forces P and Q exerted on pin B, respectively, by rod OC and the
wall of slot DE.
Solution
Kinematics.
From the drawing of the system, we have
r = 2b cosq
Then
r& = - (2b sin q )q&
and
&&
r = -2b(q&& sin q + q& 2 cos q )
ar = &&
r - rq& 2 = -2b(q&& sin q + q& 2 cos q ) - (2b cos q )q& 2
= - 2b(q&& sin q + 2q& 2 cos q )
Now
= - 2(0.5 m)[(250 rad/s2 )sin 20∞ + 2(15 rad/s)2 cos 20∞]
= - 508.367 m/s2
aq = rq&& + 2r&q& = (2b cos q )q&& + 2( -2bq& sin q )q&
= 2b(q&& cos q - 2q& 2 sin q )
and
= 2(0.5 m)[(250 rad/s2 ) cos 20∞ - 2(15 rad/s)2 sin 20∞]
= 81.014 m/s2
Kinetics.
(a)
We have
Fr = mar = 0.1 kg ¥ ( -508.367 m/s2 ) = -50.8367 N
Fr = -50.8 N b
or
and
Fq = maq = 0.1 kg ¥ (81.014 m/s2 ) = 8.1014 N
Fq = 8.10 N b
or
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196
Problem 12.128 (continued)
+
(b)
or
+
or
SFr : -Fr = -Q cos 20∞
1
(50.8367 N)
Q=
cos 20∞
= 54.099 N
SFq : Fq = P - Q sin 20∞
P = (8.1014 + 54.099 sin 20∞) N
= 26.604 N
P = 26.6 N
70∞ b
Q = 54.1 N
40∞ b
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197
PROBLEM 12.129
A particle of mass m is projected from Point A with an initial
velocity v0 perpendicular to OA and moves under a central force F
directed away from the center of force O. Knowing that the particle
follows a path defined by the equation r = r0 / cos 2q , and using
Eq. (12.27), express the radial and transverse components of the velocity v
of the particle as functions of the angle q.
Solution
We have
r=
r0
cos 2q
Then
r& =
2r0 sin 2q &
q
cos2 2q
Now
v = r&e r + rq& eq
so that at t = 0,
From Eq. (12.27):
v0 = r0q&0
r 2q& = r02q&0 = r0 v0
2
or
Ê cos 2q ˆ
rv
v
q& = 0 20 = r0 v0 Á
= 0 cos2 2q
˜
r0
Ë r0 ¯
r
Then
r& =
Now
vr = r&
ˆ
2r0 sin 2q Ê v0
cos2 2q ˜ = 2v0 sin 2q
Á
2
¯
cos 2q Ë r0
vr = 2v0 sin 2q b
or
and
r
v
vq = rq& = 0 = 0 cos2 2q
cos 2q r0
vq = v0 cos 2q b
or
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198
PROBLEM 12.130
Show that the radius r of the moon’s orbit can be determined from the radius R of the earth, the acceleration of
gravity g at the surface of the earth, and the time t required for the moon to complete one full revolution about
the earth. Compute r knowing that t = 27.3 days.
Solution
Mm
r2
We have
F =G
and
F = Fn = man = m
Then
or
Now
so that
G
Mm
r2
=m
v2 =
[Eq. (12.28)]
v2
r
GM
r
GM = gR2
v2 =
v2
r
[Eq. (12.30)]
gR2
g
or v = R
r
r
2p r 2p r
=
g
v
R r
For one orbit,
t=
or
Ê gt 2 R2 ˆ
r=Á
˜
Ë 4p 2 ¯
Now
t = 27.3 days = 2.35872 ¥ 106 s
1/ 3
Q.E.D.
b
R = 3960 mi = 20.9088 ¥ 106 ft
1/ 3
SI:
È 9.81 m/s2 ¥ (2.35872 ¥ 106 s)2 ¥ (6.37 ¥ 106 m)2 ˘
r=Í
˙
4p 2
Î
˚
= 382.81 ¥ 106 m
r = 383 ¥ 103 km b
or
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199
PROBLEM 12.131*
Disk A rotates in a horizontal plane about a vertical axis at the
constant rate q&0 = 10 rad/s. Slider B weighs 240 g and moves in a
frictionless slot cut in the disk. The slider is attached to a spring
of constant k, which is undeformed when r = 0. Knowing that the
slider is released with no radial velocity in the position r = 400 mm,
determine the position of the slider and the horizontal force exerted
on it by the disk at t = 0.1 s for (a) k = 24 N/m, (b) k = 40 N/m.
Solution
First we note
r = 0,
when
X sp = 0 fi Fsp = kr
r0 = 400 mm = 0.4 m
and
q& = q&0 = 10 rad/s
then
q&& = 0
+
SFr = mB ar : - Fsp = mB ( &&
r - rq&02 )
Ê k
ˆ
&&
- q&02 ˜ r = 0
r +Á
Ë mB
¯
or
+
SFq = mB aq : FA = mB (0 + 2r&q&0 )
(1)
(2)
k = 24 N/m
(a)
Substituting the given values into Eq. (1)
È 24
˘
&&
- (10)2 ˙ r = 0
r+Í
0
.
24
Î
˚
&&
or
r=0
dr&
Then
r = 0 and at t = 0, r& = 0 :
= &&
dt
r&
0.1
Ú0 dr& = Ú0
or
(0)dt
r& = 0
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200
Problem 12.131* (continued)
and
dr
= r& = 0 and at t = 0, r0 = 0.4 m
dt
0.1
r
Úr dr = Ú0
(0) dt
0
r = r0
or
r = 0.4 m b
Note: r& = 0 implies that the slider remains at its initial radial position.
With r& = 0, Eq. (2) implies
FA = 0 b
k = 40 N/m
(b)
Substituting the given values into Eq. (1)
È 40
˘
&&
- 100 ˙ r = 0
r+Í
Î 0.24
˚
or
&&
r+
200
r=0
3
Now
&&
r=
d
( r&) r& = vr
dt
Then
&&
r = vr
so that
vr
d dr d
d
=
= vr
dt dt dr
dr
dvr
dr
dvr 200
+
r=0
3
dr
vr
Ú 0 vr d vr = -
At t = 0, vr = 0, r = r0 :
200 r
r dr
3 Ú r0
200 2
( r - r02 )
3
or
vr2 = -
or
vr =
10 6 2
r0 - r 2
3
Now
vr =
dr 10 6 2
=
r0 - r 2
dt
3
At t = 0, r = r0 :
r
Úr
0
dr
r02
-r
2
=Ú
t 10 6
0
3
dt =
10
6
t
3
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201
Problem 12.131* (continued)
r = r0 sin f ,
Let
sin -1 ( r/r0 )
Úp /2
Then
dr = r0 cos f df
r0 cos f df
=
10 6
t
3
df =
10 6
t
3
r02 - r02 sin 2 f
sin -1 ( r/r0 )
Úp/2
or
Ê r ˆ p 10 6
t
sin -1 Á ˜ - =
3
Ë r0 ¯ 2
or
or
Ê 10 6
pˆ
10 6
10 6
r = r0 sin Á
t + ˜ = r0 cos
t = (0.4 m) cos
t
2¯
3
3
Ë 3
Then
r& = -
Finally,
10 6
4 6
sin
t
3
3
at t = 0.1 s:
Ê 10 6
ˆ
r = (0.4 m)cos Á
¥ 0.1˜
Ë 3
¯
r = 0.274 m b
or
Eq. (2)
È 4 6
Ê 6ˆ˘
FH = 0.24 kg ¥ 2 ¥ Ísin Á
˜ ˙ (10 rad/s)
Ë 3 ¯ ˙˚
ÍÎ 3
FH = -11.42 N b
or
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202
PROBLEM 12.132
It was observed that as the Voyager I spacecraft reached the point of its trajectory closest to the planet Saturn,
it was at a distance of 185 ¥ 103 km from the center of the planet and had a velocity of 21.0 km/s. Knowing
that Tethys, one of Saturn’s moons, describes a circular orbit of radius 295 ¥ 103 km at a speed of 11.35 km/s,
determine the eccentricity of the trajectory of Voyager I on its approach to Saturn.
Solution
For a circular orbit,
Eq. (12.44)
v=
GM
r
For the orbit of Tethys,
GM = rT vT2
For Voyager’s trajectory, we have
1 GM
= 2 (1 + e cos q )
r
h
where h = r0 v0
At O,
Then
or
r = r0 , q = 0
1
GM
=
(1 + e )
r0 ( r0 v0 )2
e=
=
r0 v02
r v2
- 1 = 0 02 - 1
GM
rT vT
185 ¥ 103 km
2
Ê 21.0 km/s ˆ
¥Á
˜ -1
3
295 ¥ 10 km Ë 11.35 km/s ¯
e = 1.147 b
or
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203
PROBLEM 12.133
At engine burnout on a mission, a shuttle had reached Point A at an
altitude of 80 km above the surface of the earth and had a horizontal
velocity v0. Knowing that its first orbit was elliptic and that the shuttle
was transferred to a circular orbit as it passed through Point B at an
altitude of 270 km, determine (a) the time needed for the shuttle to travel
from A to B on its original elliptic orbit, (b) the periodic time of the
shuttle on its final circular orbit.
Solution
For Earth,
R = 6370 km = 6.37 ¥ 106 m, g = 9.81 m/s2
GM = gR2 = (9.81)(6.37 ¥ 106 )2 = 3.98059 ¥ 1014 m3 /s2
(a)
For the elliptic orbit,
rA = 6370 + 80 = 6450 km = 6.45 ¥ 106 m
rB = 6370 + 270 = 6640 km = 6.64 ¥ 106 m
1
a = ( rA + rB ) = 6.545 ¥ 106 m
2
b = rA rB =
Using Eq. 12.39,
and
(6.45 ¥ 10 ) (6.545 ¥ 10 ) = 6.49733 ¥ 10
6
6
6
m
1 GM
= 2 + C cosq A
rA
h
1 GM
= 2 + C cosq B
rB
rB
But q B = q A + 180∞, so that cos q A = - cos q B
Adding,
1 1 rA + rB 2a 2GM
+ =
= 2 = 2
rA rB
rA rB
b
h
or
h=
Periodic time.
t=
t=
GMb2
a
2p ab 2p ab a 2p a3/ 2
=
=
h
GM
GMb2
2p (6.545 ¥ 106 )3/ 2
3.98059 ¥ 1014
= 5273.156 s = 1 hr 27 min 53sec
The time to travel from A to B is one half the periodic time
t AB = 2636.58 s
t AB = 43 min 56.6 sec b
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204
Problem 12.133 (continued)
(b)
For the circular orbit,
a = b = rB = 6.64 ¥ 106 m
t circ =
2p a3/ 2 2p (6.64 ¥ 106 )3/2
=
= 5388 s
GM
3.98059 ¥ 1014
t circ = 1 h 29 min 48sec
t circ = 1 h 29 min 48 sec b
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205