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DYNAMICS Kinematics of Particles Chapter 11
1
Given:
Find:
t when v = 0
v, a and d when x = 0
Assumption:
(a)
Velocity is zero at time 6 s.
(b)
DYNAMICS Kinematics of Particles Chapter 11
2
At
Distance traveled from
Distance traveled from
Distance traveled from
| | | |
When x = 0, ⁄ ⁄
DYNAMICS Kinematics of Particles Chapter 11
3
11.11
At t = 0 s, v=16 in/s
∫
∫
At
⁄
DYNAMICS Kinematics of Particles Chapter 11
4
At t = 1 s, x = 20 in
∫
∫
(
) (
)
At t = 7 s
Distance traveled from
⁄
(
)
Distance traveled from
Distance traveled from
| | | |
DYNAMICS Kinematics of Particles Chapter 11
5
11.19
The packing material can be treated as a spring, and the equipment as a particle in simple harmonic
motion. Its velocity will be maximum when the particle passes through the equilibrium position (x = 0),
and its acceleration will be maximum at the turning points (first turning point at x = 20 mm).
While the box is falling, the packing material is not being compressed, so x = 0. Compression starts when
the box hits the ground at 4 m/s, so we will considered this velocity to be v0 , and x0 = 0 m. Final velocity
is vf = 0 m/s, and xf = 0.02 m. Maximum acceleration of the equipment is reached at the time of
maximum compression. Acceleration is given as a function of x. Positive x direction is up.
Given:
⁄ ⁄
Find:
Maximum acceleration of the equipment.
1. Find value of k.
∫
∫
∫
∫
(
) (
)
2. Use known values of k and xmax to find amax.
⁄ (upwards)
DYNAMICS Kinematics of Particles Chapter 11
6
11.24
Given:
⁄
Find:
∫ ∫
| |
|
(
| |
)
(
| |
)
| |
√
DYNAMICS Kinematics of Particles Chapter 11
7
11.27
(a)
∫
∫
[ ]
[ ]
⁄
⁄
(b)
[ ]
[ ]
(c)
⁄
⁄
DYNAMICS Kinematics of Particles Chapter 11
8
11.28
Given:
Find:
a at x = 2 m when v0 = 3.6 m/s
t from x = 1 to x = 3 m when v0 = 3.6 m/s
(a)
(
)
(
) (
) (
)
For v0 = 3.6 m/s and x = 2:
(b)
(
)
∫
∫ [ ]
[ ]
DYNAMICS Kinematics of Particles Chapter 11
9
11.40
Given:
⁄
⁄
Find: and time when runner B
should begin to run.
(a)
Runner A:
⁄
Velocity of runner A at t = 1.82 s = velocity of runner B
Runner B:
⁄
(b) Find how long it takes runner B to go from 0 m/s to 9.08 m/s:
Therefore, runner B should start running 2.59 seconds before runner A reaches the exchange zone.
DYNAMICS Kinematics of Particles Chapter 11
10
11.45
Given:
⁄
From
From
⁄
At
DYNAMICS Kinematics of Particles Chapter 11
11
At
( )
(
)
(
) (
)
(
)
( )
answer to (b)
(
)
⁄
(c)
(
)
DYNAMICS Kinematics of Particles Chapter 11
12
11.61
Velocity:
⁄
⁄
⁄
⁄
⁄
⁄
⁄
Position:
DYNAMICS Kinematics of Particles Chapter 11
13
(a)
v(t) vs t
x(t) vs t
(b)
⁄
11.101
DYNAMICS Kinematics of Particles Chapter 11
14
Given: ⁄
Find:
Find time when
Find
(a)
Find time when
DYNAMICS Kinematics of Particles Chapter 11
15
Find x when t = 1.2712 s:
(b) Therefore, the ball will land 7.01 m away from the net.
DYNAMICS Kinematics of Particles Chapter 11
16
11.112
Given: ⁄
Find: Largest value (less than 45o) of the angle for which .
Time required for the puck to reach
Time when
Angle when ⁄ :
(
)
(
)
. At this angle, the puck will hit the crossbar.
(a) The largest value (less than 45o) of the angle for which the puck will enter the net is 14.66
o.
(b)
DYNAMICS Kinematics of Particles Chapter 11
17
11.114
Given:
Find: Minimum value for .
(
)
(
)
√
According to the plot to the right of ,
has a minimum value between 0 and 900
at 2.53 m/s when
Therefore:
DYNAMICS Kinematics of Particles Chapter 11
18
11.119
Given:
Velocity of boat:
Velocity of boat relative to river:
DYNAMICS Kinematics of Particles Chapter 11
19
11.125
Given:
Find:
When
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
1
12.4
Given:
Load on spring scale:
Find: Weight of packages.
Load indicated by the spring scale and
the mass needed to balance the lever scale
when the elevator moves upward with an
acceleration of 4 ft/s2.
(a)
Resolve forces on y-axis:
∑
(
)
( )
𝑚 𝑊
𝑔
6
0 5 𝑠𝑙𝑢𝑔
The load on the spring (force 𝐹𝑠) is equal
to the mass of the package times the
acceleration of the elevator and the
acceleration due to gravity combined:
𝐹𝑠 𝑚𝑎 0 5 𝟏𝟒 𝟏 𝒍𝒃
when elevator is going down
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
2
(b)
∑
(
)
6 (
)
(Load indicated by the spring scale when the elevator accelerates
upwards.)
Therefore,
Mass of the weights = Mass of the package
6
𝑚 𝑊
𝑔
6
0 5 𝑠𝑙𝑢𝑔
The load on the spring (force 𝐹𝑠) is
equal to the mass of the package times
the acceleration of the elevator and
the acceleration due to gravity
combined:
𝐹𝑠 𝑚𝑎 0 5 𝟏𝟖 𝟏 𝒍𝒃
when elevator is going up
In order for the lever scale to be in equilibrium, the mass of the weights has to equal the
mass of the package. If the lever scale is in equilibrium, any effects caused by acceleration
will be equal on both sides of the scale, so they can be neglected.
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
3
12.7
Given:
60
50
Find:
x at
F.B.D. of bus on level road.
Forces acting on the bus:
Weight W
Normal N
Traction force P
Level:
∑
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
4
Uphill:
∑
(
)
F.B.D. of bus on the incline
(
)
0
0
0
66
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
5
12.10
Given:
⁄
Find:
F.B.D. of package at point A, with x-axis
along the slope:
Find an expression for the force of friction:
0
0 0
0
0
0
0 0
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
6
0 0
0
0 0
F.B.D. of package at point B:
0
5 0
5
5 0 5
5
5 0 5
5 0 5
5 0 5
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
7
12.13
Given: 0 0 0 0 5 66 0
Find: Shortest distance in which the rig can be brought to a stop if the load is not to shift.
F.B.D. of load while braking (load wants
to move fwd). Sliding is impending.
0
0
0
0
0
0 66
66
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
8
12.14
Given:
60
0
5 000
00
600 and 00
Find: (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.
F.B.D. of tractor and trailer combined:
(a)
600 00 5 000 00
00
00
0
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
9
(b) Replace cab with a coupling. Assume a tensile force on the coupling. Force
is acting on the coupling.
00
00
00
00
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
10
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.
Given: 0 0
Find: The largest allowable acceleration a1 and the largest allowable deceleration a2 if the
bundle is not to slide on the platform.
(a)When the bundle accelerates, the force of
friction on the platform is directed to the right:
F.B.D.
65
65
65
65 [1]
65
65
65
65
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
11
65
65
65 65
65 65
65 65
65 65
0 0
65 0 0 65
(b) When the bundle decelerates, the force of friction on the platform is directed to the
left:
Repeat part (a) changing the sign for a:
65
65 [2]
65
65 65
0 0
65 0 0 65
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
12
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.
Given:
0
Find: The distance the ship travels before coming to a stop.
Since the frictional force is not dependent on the ship’s weight, forces acting on the vertical
plane are omitted, and all calculations occur in the horizontal plane, with the positive
direction being the direction in which the ship is moving.
full steam ahead full steam reverse
When the ship is moving full steam ahead at maximum speed (a constant speed), its
acceleration is zero:
0
0
When the ship’s engines are put on reverse, the ship starts to decelerate until it comes to a
stop:
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
13
(
)
(
)
∫
∫
∫
∫
(
) [
]
(
)
(
)
(
)
(
)
(
) (
)
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
14
12.27 Determine the maximum theoretical speed that a 2700-
lb automobile starting from rest can reach after traveling a
quarter of a mile 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 0 0 , where D and v are
expressed in pounds and ft/s, respectively.
Given: 00 0 0 0
0 6 0 0 0
Find: The maximum theoretical speed.
F.B.D.
=
0 0 0 6 00
( )
0 0
00
6 0 6
00
0
0
0
∫
∫
0
As long as the traction tires are
not skidding, the force of friction
between them and the ground is
equal to the car’s force forward.
The car’s acceleration is a function
of velocity, therefore not constant.
Since the problem involves 𝑥, and
not 𝑡, integrate using 𝑎 𝑣 𝑑𝑣 𝑑𝑥 .
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
15
∫
∫ 0 000
∫
0 000∫
0 6
∫
0 000∫
650 5
0 000 650 5
5 000 650 5
5000 650 5
650 5
√
650 5
Insert value for :
√ 650 5
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
16
12.35 A 500-lb crate B is suspended from a cable attached to a 40-lb
trolley A which rides on an inclined I-beam as shown. Knowing that at
the instant shown the trolley has an acceleration of 1.2 ft/s2 up and to
the right, determine (a) the acceleration of B relative to A, (b) the
tension in cable CD.
Given: 500 0
Find: The acceleration of B relative to A, and the tension in cable CD.
F.B.D. of B
=
(a) For crate B:
0 5
5
5 5
(b) Find tension in cable AB:
5
5
(
5) 500 (
5) 50
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
17
(b)
F.B.D. of A
For trolley A:
5 5
5 5
50 5 0 5 0
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
18
12.37
Given: mass of ball 50 0 5
constant speed
Find:
F.B.D.
0
0
(
)
0 5
6
0 5
0 06
Solving for :
0 5
5
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
19
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.
Given: 5
Find: T when swing is not moving
T immediately after the swing is released F.B.D. when swing is held:
0
5 0
5
5 6
Swing is held by two ropes, so tension of each rope is:
6
F.B.D. when swing is released:
Immediately after the swing is released, the swing’s
acceleration is still zero:
0
5
5
5
6
Swing is held by two ropes, so tension of each rope is:
6
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
20
12.46 During a high-speed chase, a 2400-lb sports car
traveling at a speed of 100 mi/h just loses contact with
the road as it reaches the crest A of a hill. (a) Determine
the radius of curvature of the vertical profile of the
road at A. (b) Using the value of found in part a,
determine the force exerted on a 160-lb driver by the
seat of his 3100-lb car as the car, traveling at a constant
speed of 50 mi/h, passes through A.
Given: 00 00 6 6 60
50
Find:
The force exerted on a 160-lb driver by the seat of his 3100-lb car as the car, traveling
at a constant speed of 50 mi/h, passes through A.
When the car loses contact with the road at point A, Normal and frictional forces are zero:
F.B.D of Maserati Quattroporte as it loses contact with the road:
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
21
(b) Driver is not moving horizontally with respect to her seat, so 0.
F.B.D. of driver at point A
(
)
60 (
66 )
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
22
12.51 A curve in a speed track has a radius of 1000-ft and a rated
speed of 120 mi/h. (See Sample Prob. 12.6 for the definition of rated
speed.) Knowing that a racing car starts skidding on the curve when
traveling at a speed of 180 mi/h, determine (a) the banking angle θ, (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 the curve.
Given: radius of track 000
rated speed 0 6
skidding speed 0 6
Find: (a) the banking angle θ
(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 the curve.
=
(
)
The car travels in a horizontal circular
path of radius 𝜌. The normal component
𝑎𝑛 of the acceleration is directed toward
the center of the path; its magnitude is
𝑎𝑛 𝑣 𝜌, where 𝑣 is the speed of the car
in ft/s. Between 120 mph and 180 mph, 𝐹𝑓
is what keeps the car from skidding.
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
23
(a) At rated speed, 0
0 (
)
(
)
6
000 0 6
(
)
Substituting [1] and [2]:
(
)
(
)
6 000
6 000
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
24
(c) As speed decreases below 120 mph, Ff is what keeps the car from sliding down, so Ff
in this case is pointing in the opposite direction as the force of friction that kept the car from
sliding up the bank. Therefore,
Substituting [1] and [2]:
(
)
(
)
0 000
000
0 5
05 6
0 0 050 0 0 5
0
√
0 5 6
DYNAMICS Kinetics of Particles: Newton’s Second law Chapter 12
25
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.
Given: 5 0 5 0 5
6 6 0
Find: (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 avg mass of a tuft is 1.6
mg.
F.B.D.
(a)
0
(b)
6 0 6 0
6 0 05
√
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
1
13.5 Determine the maximum theoretical speed
that may be achieved over a distance of 360 ft by a
car starting from rest assuming there is no slipping.
The coefficient of static friction between the tires and
pavement is 0.75, and 60 percent of the weight of the
car is distributed over its front wheels and 40 percent
over its rear wheels. Assume (a) front-wheel drive,
(b) rear-wheel drive.
Given:
Find: Maximum theoretical speed.
Principle of work and energy:
Kinetic energy:
Position 1:
Position 2:
Work from 1 to 2:
( ) 𝑁𝑓 𝑊
𝐾 𝑈 𝐾
( 𝑊)( 𝑓𝑡) 𝑊
2𝑔𝑣
𝑣 ( )( )2( 2 2)
1 2 1 𝑓𝑡 𝑠 𝟔𝟗 𝟔 𝒎𝒑𝒉
(a) Front-wheel drive:
𝐹 𝐹𝑓 𝜇𝑁𝑓 𝑁𝑓
( ) 𝑁𝑟 𝑊
𝐾 𝑈 𝐾
( 𝑊)( 𝑓𝑡) 𝑊
2𝑔𝑣
𝑣 ( )( )2( 2 2)
8 𝑓𝑡 𝑠 𝟓𝟔 𝟗 𝒎𝒑𝒉
(b) Rear-wheel drive:
𝐹 𝐹𝑓 𝜇𝑁𝑟 𝑁𝑟
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
2
13.6 Skid marks on a drag race track indicate that the rear (drive) wheels of a car slip for
the first 60 ft of the 1320-ft track. (a) Knowing that the coefficient of kinetic friction is 0.60,
determine the speed of the car at the end of the first 60-ft portion of the track if it starts from
rest and the front wheels are just off the ground. (b) What is the maximum theoretical speed
for the car at the finish line if, after skidding for 60 ft, it is driven without the wheels slipping
for the remainder of the race? Assume that while the car is rolling without slipping, 60
percent of the weight of the car is on the rear wheels and the coefficient of static friction is
0.85. Ignore air resistance and rolling resistance.
Given: Point 1 at
Point 2 at
Point 3 at 1 2
8 rear-wheel drive
Find: and under given conditions.
𝐾
𝑈 𝐹𝑥 𝑊𝑥
( )𝑊( ) 𝑊
𝐾 1
2
𝑊
𝑔𝑣
𝑲𝟏 𝑼𝟏 𝟐 𝑲𝟐
𝑊 𝑊
2𝑔𝑣
𝑣 2𝑔 8 1 𝑓𝑡 𝑠 𝟑𝟐 𝟖 𝒎𝒑𝒉
(a) All weight on rear traction wheels.
Spinning means 𝐹 𝐹𝑓 𝜇𝑘𝑁𝑟 𝜇𝑘𝑊
𝐾 1
2
𝑊
𝑔 8 1 𝑊
𝑈 𝐹𝑥 𝜇𝑠 𝑊𝑥
( 8 )( )𝑊(12 )
2 𝑊
𝐾 1
2
𝑊
𝑔𝑣
𝑲𝟏 𝑼𝟏 𝟐 𝑲𝟐
𝑊 2 𝑊 𝑊
2𝑔𝑣
𝑣 1 𝑔 2 9 𝑓𝑡 𝑠 𝟏𝟒𝟐 𝟓 𝒎𝒑𝒉
(b) 𝐹 𝐹𝑓 𝜇𝑠𝑁𝑟 𝜇𝑠 𝑊
𝑥 1 2 − 12 𝑓𝑡
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
3
13.15 The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully
applied on the wheels of cars B and C, causing them to slide on the track, but are not applied
on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the
wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the
force in each coupling.
Given:
8
1
8
Find: x and the force in each coupling.
(a)
(1
2)
(1
2)2
81 1 9
− ( ) −
81 1 9 −
81 1 9
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
4
(b) Force in coupling AB:
(1
2)
18
2
11 18
− ( ) 12 − (18 )12
12 − 81 8
11 18 12 − 81 8
2
12 19 82 ( )
Force in coupling BC:
(1
2)
8
2 2 9 9
−
12 − (8 )12
12 − 8
2 9 9 12 − 8
1 8 11
12 8 1 ( )
Replace wagon A with a coupling. Assume a tensile
force on the coupling. Force 𝐹𝐴𝐵 is acting on the coupling.
Replace wagons A and B with a coupling. Assume a tensile
force on the coupling. Force 𝐹𝐵𝐶 is acting on the coupling.
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
5
13.17 A trailer truck enters a 2 percent downhill grade traveling at 108 km/h and must slow
down to 72 km/h in 300 m. The cab has a mass of 1800 kg and the trailer 5400 kg. Determine
(a) the average braking force that must be applied, (b) the average force exerted on the
coupling between cab and trailer if 70 percent of the braking force is supplied by the trailer
and 30 percent by the cab.
Given:
1 1 8 18
1 8 2 2
Find: and force on coupling
(a)
(18 )9 81 2
1 12
1 12
1
2
1
2 2 ( ) 2
1
2
1
2 2 (2 ) 1
− 2 −
2 1 2 − 1 1
2 22 1
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
6
(b)
( )9 81 29
1 9 8
1 9 8
1
2
1
2 ( ) 2
1
2
1
2 (2 ) 1 8
− 1 8 − 1 2 − 1 2 8
2 1 − 1 2 8 1 8 1
−111 2
−111 2
− 1
Replace cab with a coupling. Assume a tensile force
on the coupling. Force 𝐹𝐴𝐵 is acting on the coupling.
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
7
13.27 A 10-lb block is attached to an unstretched
spring of constant 12 . The coefficients of
static and kinetic friction between the block and the
plane are 0.60 and 0.40, respectively. If a force F is
applied to the block until the tension in the spring reaches 20 lb and then suddenly removed,
determine (a) how far the block will move to the left before coming to a stop, (b) whether the
block will then move back to the right.
Given: 1 − −2 12
Find: (a) how far the block will move to the left before coming to a stop,
(b) whether the block will then move back to the right.
1. Find how far to the right the spring is stretched by force F:
− −2 −2
−12 1
2. Draw F.B.D. with forces and points:
Point 1: Block is released from rest after spring has been stretched by force F.
Point 2: Block stops after spring has been compressed to its maximum.
Point 3: Block possibly moves back to the right.
The force on the spring is 20 lb, therefore the reaction force by the spring is -20 lb.
The work done by a spring is
negative when the spring is being
stretched or compressed, and
positive when it is released.
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
8
From point 1 to point 2:
−
1
2 (
− ) − ( − )
1
212(1 −
) − ( )(1 )(1 − )
− − 1
(a)
− − 1
− 1
Therefore, the block moves from +1.667 to –1, a distance of 2.667 in or 0.222 ft.
(b) When the block stops at point 2, it will move again to the right if the force exerted by
the spring is greater than the static frictional force exerted by the surface on which the block
stands.
− −(12)(−1) 12
( )(1 )
, therefore the block will move back to the right.
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
9
13.28 A 3-kg block rests on top of a 2-kg block supported by but not
attached to a spring of constant 40 N/m. The upper block is suddenly
removed. Determine (a) the maximum speed reached by the 2-kg block, (b)
the maximum height reached by the 2-kg block.
Given: 2
Find: (a) the maximum speed reached by the 2-kg block,
(b) the maximum height reached by the 2-kg block.
1. Find the force on the spring by the weight of the masses:
( 2)(−9 81) − 9
2. Find the distance that a weight of 49.05 N will compress the given spring:
− −
9
−1 22
3. Define points:
Point 1: 3-kg block has been removed. Speed of 2-kg block is zero and its position is
−1 22 .
Point 2: Spring and block reach their maximum speed at .
Point 3: Spring has been left behind and block reaches its maximum height.
When the 3-kg block is removed, the spring will stretch a certain
distance while pushing the 2-kg block. At x = 0, the spring and the block
will reach their maximum speed. At that point the spring will slow
down until its speed is zero at its maximum stretch, and the 2-kg block
will keep going upwards until the force of gravity makes it reach a
speed of zero before it makes it fall down.
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
10
4. Apply the method of work and energy to find maximum velocity of the spring-mass
system at
1
2
1
22
−
1
2 (
− ) − ( )
1
2 (−1 22 − ) − (2)(9 81)(1 22 )
2 − 2
1 2
1 2
5. Apply the method of work and energy to find height of block at point 3.
−
1
2 (
− ) − ( )
1
2 (−1 22 − ) − (2)(9 81)( ) 2 − 19 2
2 − 19 2
2
19 2 1 29
The spring stops doing
work on the block at y = 0.
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
11
13.44 A small block slides at a speed 8 on a
horizontal surface at a height above the ground.
Determine (a) the angle θ at which it will leave the cylindrical
surface BCD, (b) the distance x at which it will hit the ground.
Neglect friction and air resistance.
Given: 8
Find: and
At point B:
[speed is constant]
−
At point C, using n and t coordinates:
( )
1
2
1
2 8 2
1
2
1
2
( − ) ( − )
2 ( − ) 1
2
2 ( − ) 1
2
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
12
2 2 2( − ) 1
2 2 2
2 9 − 2 2 1 1
8 128
2 2
2
88
88
Find velocity at point C:
2 ( − ) 1
2
2 ( − ) 1
2 2 2 2( − 2 )
1
2
2 2 2( − 2 ) 1
2
9 2
8 21 2
−
2 2 2 −1
2 2 2
1 1 − 2 − 2 2
(8 21 )( )
2
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
13
13.46 (a) A 120-lb woman rides a 15-lb bicycle up a 3-percent slope at a constant speed of
5 ft/s. How much power must be developed by the woman? (b) A 180-lb man on an 18-lb
bicycle starts down the same slope and maintains a constant speed of 20 ft/s by braking. How
much power is dissipated by the brakes? Ignore air resistance and rolling resistance.
Given: 12 1 1 18 18 198
1 18
(a) Forces in the direction of v: −1
(−1 )( )
(a) Forces in the direction of v: 198
(198 )(2 )
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
14
13.49 In an automobile drag race, the rear (drive) wheels of a l000-kg car skid for the first
20 m and roll with sliding impending during the remaining 380 m. The front wheels of the
car are just off the ground for the first 20 m, and for the remainder of the race 80 percent of
the weight is on the rear wheels. Knowing that the coefficients of friction are 9 and
8, determine the power developed by the car at the drive wheels (a) at the end of the
20-m portion of the race, (b) at the end of the race. Give your answer in kW and in hp. Ignore
the effect of air resistance and rolling friction.
Given: 1 9 8
Find: (a) The power developed by the car at the drive wheels at the end of the 20-m portion
of the race,
(b) at the end of the race
(a) All weight on rear traction wheels.
Spinning means ( 8)(1 )(9 81) 8
( 8)(2 ) 1 1
1
2
1
21
1 1
√1 1
1
( 8 )(1 ) 1 8 9
1 kW = 1.34102209 hp
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
15
(b) Roll with sliding impending means:
8 ( 9 )( 8)(1 )(9 81) 2
8
1
2 ( )(1 ) 1 1
( 2)( 8 ) 2 8 1
1
2
1 1 2 8 1
√ 9
( 2)( ) 2 11
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
16
13.69 A spring is used to stop a 50-kg package
which is moving down a 20° incline. The spring
has a constant and is held by
cables so that it is initially compressed 50 mm.
Knowing that the velocity of the package is 2
m/s when it is 8 m from the spring and the
kinetic coefficient of friction between the
package and the incline is 0.2., determine the
maximum additional deformation of the spring
in bringing the package to rest.
Given: 2
Find: The maximum additional deformation of the spring in bringing the package to rest.
1
2
( )( )(2) 1
1
2
( )( ) 2
− ( 2 − 2 )
( 2 − 2 )
(8)( )(9 81)( 1 1) 2
1 2 2 9
2 2 ( 9) 2
2 ( )
1 ( )
1
2 ( )
1 ( )
2 1 ( ) 1 ( )
1 ( ) − 1 ( ) − 2 12
( ) 2281 ( ) 2281
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
17
13.144 The design for a new cementless hip implant is to be studied
using an instrumented implant and a fixed simulated femur. Assuming
the punch applies an average force of 2 kN over a time of 2 ms to the
200 g implant, determine (a) the velocity of the implant immediately
after impact, (b) the average resistance of the implant to penetration if
the implant moves 1 mm before coming to rest.
Given: 2 ( − ) 2 2
Find: (a) the velocity of the implant immediately after impact,
(b) the average resistance of the implant to penetration
if the implant moves 1 mm before coming to rest.
( − ) (2 )( 2)
2
2
1
2
( )( 2)(2 ) 2
− − 1
2 − 1
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
18
13.174 A 1-kg block B is moving with a velocity v0 of
magnitude 2 as it hits the 0.5-kg sphere A,
which is at rest and hanging from a cord attached at O.
Knowing that between the block and the
horizontal surface and 8 between the block and the
sphere, determine after impact (a) the maximum height h
reached by the sphere, (b) the distance x traveled by the
block.
Given: 1
2 8
Find: (a) the maximum height h reached by the sphere,
(b) the distance x traveled by the block.
(1)(2) 1 2
− −
−
−2 8
− −1
1 2} [
−1 1 −1 1 2
] [1 2 1 8
] 2 8
(a) Sphere:
1
2
( )( )(2 ) 1
( )(9 81) 9
1 9
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
19
(b) Block:
1
2
( )(1)( 8) 2
− − ( )(1)(9 81) − 88
2 − 88
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
20
13.180 Two cars of the same mass run head-on into each other at C. After the collision, the
cars skid with their brakes locked and come to a stop in the positions shown in the lower part
of the figure. Knowing that the speed of car A just before impact was 5 mi/h and that the
coefficient of kinetic friction between the pavement and the tires of both cars is 0.30,
determine (a) the speed of car B just before impact, (b) the effective coefficient of restitution
between the two cars.
Given:
−12
−
Find:
(a) the speed of car B just before
impact,
(b) the effective coefficient of restitution between the two cars.
1. Conservation of linear momentum:
[general form]
𝐾 1
2𝑚𝑣 𝐾
𝑈 −𝐹𝑓𝑥 −𝜇𝑘𝑁𝑥
𝑈 −( )𝑚( 2 2)(12) 11 92 𝑚
𝑲𝟏 𝑼𝟏 𝟐 𝑲𝟐
1
2𝑚𝑣 − 11 92 𝑚
𝑣𝐴 (2)(11 92) 1 22 𝑓𝑡 𝑠
CAR A
𝐾 1
2𝑚𝑣 𝐾
𝑈 −𝐹𝑓𝑥 −𝜇𝑘𝑁𝑥
𝑈 −( )𝑚( 2 2)( ) 28 98 𝑚
𝑲𝟏 𝑼𝟏 𝟐 𝑲𝟐
1
2𝑚𝑣 − 28 98 𝑚
𝑣𝐵 (2)(28 98) 1 1 𝑓𝑡 𝑠
CAR B
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
21
− − − [masses are equal, ]
(a)
1 22 1 1 2
(b)
− −
− 1 1 − (−1 22 )
− (− 1 2 )
1 2
1
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
22
Class Example
Vehicles A and B collide as shown.
Given: 1
Find: final position (brakes applied,
both vehicles skid).
1. TANGENTIAL AXIS
Tangential velocities don’t change after impact:
( ) ( ) 2 2 2
( ) ( )
2. NORMAL AXIS
( ) ( ) ( ) ( ) [B has no normal velocity]
( 2 )
( )
( )
( 2 )
( )
1
( )
( 2 ) ( ) 1 ( ) [g cancelled]
( 2 ) ( ) ( ) [divide both sides by 1000]
( ) 2 2 − ( ) [1 equation, 2 unknowns]
3. USE e TO FIND ANOTHER EQUATION WITH SAME VARIABLES
− −
(
) − ( )
( ) − ( )
( ) − ( )
− 2
( ) − ( ) − 2 (
) ( ) 2
2 2 − ( ) ( ) 2 (
) 21 2
DYNAMICS Kinetics of Particles: Energy and Momentum Methods Chapter 13
23
( ) 2 2 9 ( ) 2 2 2
( ) 2 2 − ( 2 2 ) 2 2 ( )
4. FIND RESULTING VELOCITIES:
9 2 2 −
−
5. WITH INITIAL VELOCITIES AFTER COLLISION, FINAL VELOCITIES
(ZERO), AND WORK DONE (FRICTION), THE WORK AND ENERGY METHOD CAN
BE USED TO FIND FINAL POSITIONS.
DYNAMICS Systems of Particles Chapter 14
1
14.3 A 180-lb man and a 120-lb woman stand side by side
at the same end of a 300-lb boat, ready to dive, each with a
16-ft/s velocity relative to the boat. Determine the velocity of
the boat after they have both dived, if (a) the woman dives
first, (b) the man dives first.
Given:
Find: The velocity of the boat after they have both dived, if (a) the woman dives first, (b)
the man dives first.
Principle of Conservation of Linear Momentum L:
(a)
L before anyone dives = L after woman dives:
(
)
( )
L after woman dives = L after man dives
(
) ( )
( )
𝑣𝑑 𝑣𝑏 𝑣𝑑 𝑏
𝑣𝑑 𝑣𝑏
𝑣𝑑 𝑣𝑏
DYNAMICS Systems of Particles Chapter 14
2
(b)
L before anyone dives = L after man dives:
(
)
( )
L after man dives = L after woman dives
(
) ( )
( )
DYNAMICS Systems of Particles Chapter 14
3
14.5 A bullet is fired with a horizontal velocity of
1500 ft/s through a 6-lb block A and becomes
embedded in a 4.95-lb block B. Knowing that blocks A
and B start moving with velocities of 5 ft/s and 9 ft/s,
respectively, determine (a) the weight of the bullet, (b)
its velocity as it travels from block A to block B.
Given:
Find: and between A and B
(a)
L of bullet = L of block A + L of block B and bullet
( )
1491
(b)
Initial L of bullet = L of bullet after going through block A
( ) ( )
( )
( )
( )
DYNAMICS Systems of Particles Chapter 14
4
14.8 At an amusement park there are 200-kg bumper cars A, B, and C that have riders with
masses of 40 kg, 60 kg, and 35 kg respectively. Car A is moving to the right with a velocity
vA = 2 m/s when it hits stationary car B. The coefficient of restitution between each car is 0.8.
Determine the velocity of car C so that after car B collides with car C the velocity of car B is
zero.
Given:
Find: so that after car B collides with car C, the velocity of car B is zero.
( )( ) ( )( )
[
] [
]
( )( ) ( )( )
[
] [
]
DYNAMICS Systems of Particles Chapter 14
5
14.17 A small airplane of mass 1500 kg and a helicopter of mass 3000 kg flying at an
altitude of 1200 m are observed to collide directly above a tower located at O in a wooded
area. Four minutes earlier the helicopter had been sighted 8.4 km due west of the tower and
the airplane 16 km west and 12 km north of the tower. As a result of the collision the
helicopter was split into two pieces, H1 and H2, of mass m1 = 1000 kg and m2 = 2000 kg,
respectively; the airplane remained in one piece as it fell to the ground. Knowing that the two
fragments of the helicopter were located at points H1 (500 m, –100 m) and
H2 (600 m, –500 m), respectively, and assuming that all pieces hit the ground at the same
time, determine the coordinates of the point A where the wreckage of the airplane will be
found.
Given:
( ) ( )
( ) ( )
Find: ( )
1. Find velocity of airplane and helicopter at time of collision:
( ) ( )
( )
( ) ( )
( )
2. Find velocity of mass center G of the fragments after the collision:
( )
( )( ) ( )( ) ( )
The mass center G
of a system of
particles moves as
if the entire mass of
the system and all
the external forces
were concentrated
at that point.
DYNAMICS Systems of Particles Chapter 14
6
3. Find the time it takes the fragments to fall freely from to
√
√
4. Find position of G at time of impact with the ground:
( )
5. Use Equation 14.12 to find position of plane where it hits the ground:
∑
( ) ( ) ( )
( )( ) ( ) ( )( ) ( )( )
( )( ) ( ) ( )( ) ( )( )
( ) ( ) ( )
( )
( )
( ) ( )
DYNAMICS Systems of Particles Chapter 14
7
14.19 Car A was traveling east at high speed when it collided at point O with car B, which
was traveling north at 72 km/h. Car C, which was traveling west at 90 km/h, was 10 m east
and 3 m north of point O at the time of the collision. Because the pavement was wet, the
driver of car C could not prevent his car from sliding into the other two cars, and the three
cars, stuck together, kept sliding until they hit the utility pole P. Knowing that the masses of
cars A, B, and C are, respectively, 1500 kg, 1300 kg, and 1200 kg, and neglecting the forces
exerted on the cars by the wet pavement solve the following problem:
Knowing that the coordinates of the utility pole are xp = 18 m and yp = 13.9 m, determine (a)
the time elapsed from the first collision to the stop at P, (b) the speed of car A.
Given:
( )
Find:
Time from O to and ( )
Use Equation 14.12 to work backwards from final position:
∑
( )
( )( ) ( )
( )( ) ( ) ( )
( ) ( )
DYNAMICS Systems of Particles Chapter 14
8
Equating coefficients of :
Equating coefficients of :
( )
( )
DYNAMICS Systems of Particles Chapter 14
9
14.73 A floor fan designed to deliver air at a maximum
velocity of 6 m/s in a 400-mm-diameter slipstream is supported
by a 200-mm-diameter circular base plate. Knowing that the
total weight of the assembly is 60 N and that its center of
gravity is located directly above the center of the base plate,
determine the maximum height h at which the fan may be
operated if it is not to tip over. Assume ρ = 1.21 kg/m3 for air
and neglect the approach velocity of the air.
Given:
Find: Maximum height h at which the fan may be operated if it is not to tip over.
1. Find mass flow rate dm/dt
( ) (
( )
)
2. Find force of the fluid Ff
( ) ( )( )
3. Find force of thrust Fth
4. Sum forces on an axis or moments.
In this case, we will sum moments about the center of gravity in line with the base of the fan.
When the fan is in “impending tip”, the normal force N is applied at the “tipping hinge”.
( )
( )( )
𝑑𝑚
𝑑𝑡 𝜌𝑄
𝜌𝐴𝑣
𝛾
𝑔𝑄
𝛾
𝑔𝐴𝑣
Mass flow rate
DYNAMICS Systems of Particles Chapter 14
10
14.74 The helicopter shown can produce a maximum
downward air speed of 80 ft/s in a 30-ft-diameter slipstream.
Knowing that the weight of the helicopter and its crew is 3500
lb and assuming lb/ft3 for air, determine the
maximum load that the helicopter can lift while hovering in
midair.
Given:
(specific weight of air)
Find: The maximum load L that the helicopter can lift while hovering in midair.
1. Find mass flow rate dm/dt
(
)
2. Find force of the fluid Ff
( ) ( )( )
3. Find force of thrust Fth
4. Sum forces on an axis or moments.
[acceleration is zero when hovering]
DYNAMICS Systems of Particles Chapter 14
11
14.75 A jet airliner is cruising at a speed of 600 mi/h with each of its three engines
discharging air with a velocity of 2000 ft/s relative to the plane. Determine the speed of the
airliner after it has lost the use of (a) one of its engines, (b) two of its engines. Assume that
the drag due to air friction is proportional to the square of the speed and that the remaining
engines keep operating at the same rate.
Given:
Find: The speed of the airliner after it has lost the use of
(a) one of its engines,
(b) two of its engines.
( )
[acceleration is zero when at cruising speed]
( )
With three engines running:
(
)
( )
(
)
With two engines running:
(
)
(
)
(
)
( )
( )
𝑣𝑖𝑛 𝑣𝑎𝑖𝑟𝑝𝑙𝑎𝑛𝑒
Moving reference frame:
DYNAMICS Systems of Particles Chapter 14
12
With one engine running:
(
)
(
)
(
)
( )
( )
Dynamics Kinematics of Rigid Bodies Chapter 15
1
15.48 In the planetary gear system shown, the radius of gears
A, B, C, and D is 3 in. and the radius of the outer gear E is 9 in.
Knowing that gear E has an angular velocity of 120 rpm
clockwise and that the central gear has an angular velocity of
150 rpm clockwise, determine (a) the angular velocity of each
planetary gear, (b) the angular velocity of the spider connecting
the planetary gears.
Given:
Find:
Gear A: [1]
Spider: [2]
Gear B: [3]
[4]
Gear E: [5]
[ ]
[ ]
( )( ) ( )( )
[ ] ( )( )
( )
[ ]
( )
Dynamics Kinematics of Rigid Bodies Chapter 15
2
15.57 In the engine system shown, l = 160 mm and b = 60 mm.
Knowing that the crank AB rotates with a constant angular velocity of
1000 rpm clockwise, determine the velocity of the piston P and the
angular velocity of the connecting rod when (a) , (b) .
Given:
Find:
(a)
( )( )
( )
⁄
⁄ ⁄
( )
(b)
( )( )
( ) ( ) ( ) ( )
( )
⁄ ( )
( )
Dynamics Kinematics of Rigid Bodies Chapter 15
3
15.111 An automobile travels to the left at a
constant speed of 48 mi/h. Knowing that the diameter
of the wheel is 22 in., determine the acceleration (a) of
point B, (b) of point C, (c) of point D.
Given:
Find:
The wheel is rolling and not sliding, therefore point C is the instantaneous center.
( )( )
( )
( )
( )
Dynamics Kinematics of Rigid Bodies Chapter 15
4
15.125 Knowing that crank AB rotates about point A with a constant
angular velocity of 900 rpm clockwise, determine the acceleration of
the piston P when
Given:
Find:
Rod AB:
( )( )
( ) ( )( )
Rod BD:
( ) ( ) ( )
( )
( ) ( )
( ) ( )( )
( )( )
( ) ( )
Horizontal components
( )
Vertical components
( )( )
DYNAMICS Plane Motion of Rigid Bodies: Forces and Acceleration Chapter 16
1
16.7 A 20-kg cabinet is mounted on casters that allow it to move
freely ( ) on the floor. If a 100-N force is applied as shown,
determine (a) the acceleration of the cabinet, (b) the range of values of h
for which the cabinet will not tip.
Given:
Find: (a) the acceleration of the
cabinet, (b) the range of values of h for
which the cabinet will not tip.
When tipping is impending
When tipping is impending
Therefore, the range of values of h for which the cabinet will not tip is
DYNAMICS Plane Motion of Rigid Bodies: Forces and Acceleration Chapter 16
2
16.33 In order to determine the mass moment of inertia of a flywheel of
radius 600 mm, a 12-kg block is attached to a wire that is wrapped around
the flywheel. The block is released and is observed to fall 3 m in 4.6 s. To
eliminate bearing friction from the computation, a second block of mass
24 kg is used and is observed to fall 3 m in 3.1 s. Assuming that the
moment of the couple due to friction remains constant, determine the
mass moment of inertia of the flywheel.
Given: When
When
Find: the mass moment of inertia of the flywheel.
Case 1:
Case 2:
DYNAMICS Plane Motion of Rigid Bodies: Forces and Acceleration Chapter 16
3
16.62 The 3-oz yo-yo shown has a centroidal radius of gyration of 1.25 in.
The radius of the inner drum on which a string is wound is 0.25 in. Knowing
that at the instant shown the acceleration of the center of the yo-yo is 3 ft/s2
upward, determine (a) the required tension T in the string, (b) the
corresponding angular acceleration of the yo-yo.
Given: ⁄
⁄
⁄
⁄
Find: (a) the required tension T in the string,
(b) the corresponding angular acceleration of the yo-yo.
(
)
DYNAMICS Plane Motion of Rigid Bodies: Forces and Acceleration Chapter 16
4
16.69 A bowler projects an 8-in.-diameter ball weighing 12 lb along an
alley with a forward velocity v0 of 15 ft/s and a backspin of 9 rad/s.
Knowing that the coefficient of kinetic friction between the ball and the
alley is 0.10, determine (a) the time t1 at which the ball will start rolling
without sliding, (b) the speed of the ball at time t1, (c) the distance the
ball will have traveled at time t1.
Given:
At the instant , the ball starts to roll,
point C becomes the instantaneous center of
rotation, and .
At
Kinematics at
DYNAMICS Plane Motion of Rigid Bodies: Forces and Acceleration Chapter 16
5
(
)r
(
)
DYNAMICS Plane Motion of Rigid Bodies: Forces and Acceleration Chapter 16
6
16.82 A turbine disk of mass 26 kg rotates at a constant rate of
9600 rpm. Knowing that the mass center of the disk coincides
with the center of rotation O, determine the reaction at O
immediately after a single blade at A, of mass 45 g, becomes loose
and is thrown off.
Given:
Find: the reaction at O immediately after a single blade at A, of mass 45 g, becomes loose
and is thrown off.
DYNAMICS Plane Motion of Rigid Bodies: Forces and Acceleration Chapter 16
7
16.97 A homogeneous sphere S, a uniform cylinder C,
and a thin pipe P are in contact when they are released
from rest on the incline shown. Knowing that all three
objects roll without slipping, determine, after 4 s of
motion, the clear distance between (a) the pipe and the
cylinder, (b) the cylinder and the sphere. (SP 16.8)
Given:
Find: The clear distance, after 4 s of
motion, between (a) the pipe and the
cylinder, (b) the cylinder and the
sphere.
General case:
For pipe:
For cylinder:
For sphere:
(a)
(b)
DYNAMICS Plane Motion of Rigid Bodies: Forces and Acceleration Chapter 16
8
16.104 A drum of 60-mm radius is attached to a disk of
120-mm radius. The disk and drum have a total mass of 6
kg and a combined radius of gyration of 90 mm. A cord is
attached as shown and pulled with a force P of magnitude
20 N. Knowing that the disk rolls without sliding,
determine (a) the angular acceleration of the disk and the
acceleration of G, (b) the minimum value of the
coefficient of static friction compatible with this motion. (See Sample Problem 16.9)
Given:
Find: (a) the angular acceleration of the disk and the acceleration of G,
(b) the minimum value of the coefficient of static friction compatible with this motion.
DYNAMICS Plane Motion of Rigid Bodies: Energy and Momentum Methods Chapter 17
1
17.9 Each of the gears A and B has a mass of
2.4 kg and a radius of gyration of 60 mm, while
gear C has a mass of 12 kg and a radius of
gyration of 150 mm. A couple M of constant
magnitude 10 N·m is applied to gear C.
Determine (a) the number of revolutions of gear
C required for its angular velocity to increase
from 100 to 450 rpm, (b) the corresponding
tangential force acting on gear A.
Given:
Find: (a) the number of revolutions of gear C required for its angular velocity to increase
from 100 to 450 rpm,
(b) the corresponding tangential force acting on gear A.
All three gears are in mesh, so their contact velocity is the same:
Moment of inertia of the gears:
DYNAMICS Plane Motion of Rigid Bodies: Energy and Momentum Methods Chapter 17
2
Find kinetic energy of system at position 1:
(
)
(
)
(
)
[
]
Find kinetic energy of system at position 2:
(
)
(
)
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Find work from position 1 to position 2 (work of the couple):
From the Principle of Conservation of Energy:
(a) Number of revolutions in 39.898 radians:
(b) Gear A:
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DYNAMICS Plane Motion of Rigid Bodies: Energy and Momentum Methods Chapter 17
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17.24 A 20-kg uniform cylindrical roller, initially at
rest, is acted upon by a 90-N force as shown. Knowing
that the body rolls without slipping, determine (a) the
velocity of its center G after it has moved 1.5 m, (b) the
friction force required to prevent slipping.
Given:
Find: (a) the velocity of its center G after it has moved 1.5 m,
(b) the friction force required to prevent slipping.
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√
DYNAMICS Plane Motion of Rigid Bodies: Energy and Momentum Methods Chapter 17
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17.69 A wheel of radius r and centroidal radius of gyration is
released from rest on the incline shown at time t = 0. Assuming that the
wheel rolls without sliding, determine (a) the velocity of its center at
time t, (b) the coefficient of static friction required to prevent slipping.
The external forces W, F, and N form a system equivalent to the system of effective forces
represented by the vector maG and the couple IGα.
No sliding means aG = r α
(a)
DYNAMICS Plane Motion of Rigid Bodies: Energy and Momentum Methods Chapter 17
5
(b)
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DYNAMICS Plane Motion of Rigid Bodies: Energy and Momentum Methods Chapter 17
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17.99 A 45-g bullet is fired with a velocity of 400 m/s at
θ = 5° into a 9-kg square panel of side b = 200 mm.
Knowing that the panel is initially at rest, determine (a)
the required distance h if the horizontal component of the
impulsive reaction at A is to be zero, (b) the
corresponding velocity of the center of the panel
immediately after the bullet becomes embedded.
Given:
Find: (a) the required distance h if the horizontal component of the impulsive reaction at A
is to be zero,
(b) the corresponding velocity of the center of the panel immediately after the bullet
becomes embedded.
F.B.D
Apply Principle of Impulse and Momentum for the plane motion of a rigid body:
Syst Momenta1 + Syst Ext Imp1→2 = Syst Momenta2
Moments about A :
DYNAMICS Plane Motion of Rigid Bodies: Energy and Momentum Methods Chapter 17
7
[ ]
If the horizontal component of the impulsive reaction at A is to be zero:
Value of into equation [1]:
(a)
(b)