Chapter 13
Kinetics of a Particle
Force and Acceleration
Kinematics vs Kinetics
§13.1 Newton’s Second Law of Motion
1. Kinetics.
Kinetics : relationship between change in motion of a body and the
forces that cause it  Newton’s 2nd Law
2. Newton’s Second Law
F  ma
??
(1) A particle accelerates in the direction of the TOTAL force with
a magnitude proportional to the total force
(2) m — property of matter : resistance of the particle to a change
in its velocity
§13.1 Newton’s Second Law of Motion
3. Newton’s Law of Gravitational Attraction.
mutual attraction between two particles with mass:
m1m2
F G 2
r
G  66.73  10 12  m3  kg  s2 
F—Force of attraction between the two particles.
G—Universal constant of gravitation (experimental value)
m1,m2—mass of each particles
r —distance between centers of the two particles
PROCEDURE FOR THE APPLICATION OF THE
EQUATION OF MOTION
1) Select a convenient inertial coordinate system. Rectangular,
normal/tangential, or cylindrical coordinates may be used.
2) Draw a free-body diagram showing all external forces
applied to the particle. Resolve forces into their
appropriate components.
3) Draw the kinetic diagram, showing the particle’s inertial
force, ma. Resolve this vector into its appropriate
components.
4) Apply the equations of motion in their scalar component
form and solve these equations for the unknowns.
5) It may be necessary to apply the proper kinematic relations
to generate additional equations.
CONCEPT MCQ
1. The block (mass = m) is moving upward with a speed v.
Draw the FBD if the kinetic friction coefficient is k.
mg
mg
A)
kN
v
B)
kN
N
N
mg
kmg
C)
N
D) None of the above.
CONCEPT MCQ
2. Packaging for oranges is tested using a machine that exerts forces that
produce ay = 20 m/s2 and ax = 3 m/s2, simultaneously. Select the correct
FBD and kinetic diagram for this condition.
y
A)
may
W
=
•
B)
x
W
=
max
Rx
•
max
Rx
Ry
Ry
C)
D)
may
=
Ry
may
W
•
=
Ry
•
max
ATTENTION MCQ
1. Internal forces are not included in an equation of motion
analysis because the internal forces are_____
A) equal to zero.
B) equal and opposite and do not affect the calculations.
C) negligibly small.
D) not important.
2. A 10 lb block is initially moving down a ramp
with a velocity of v. The force F is applied to
bring the block to rest. Select the correct FBD.
F
10
A)
k10
N
F
10
B)
k10
N
F
10
C)
kN
N
F
v
§13.3 Equation of Motion for a System of Particles
3. For the center of mass G of the particles
If rG is a position vector which locates the center of mass G of the
particles:
mrG   mi ri
Differentiating the equation twice
with respect to time, assuming that
no mass is entering or leaving the
system, yields:
Fi
z
i
fi
maG   mi ai
ri
G
rG
 F  ma
G
y
x
§13.3 Equation of Motion for a System of Particles
3. For the center of mass G:
 F  ma
G
Fi
z
The sum of external forces
acting on the system
is equal to the total mass of the
particles times the acceleration
of its center of mass G.
i
fi
ri
G
rG
y
x
§13.4 Equations of Motion: Rectangular Coordinates
A particle moves relative to an inertial x, y, z frame
The forces and acceleration can be expressed in terms of their i, j, k
components:
 F  ma  F i   F j   F k  m  a i  a j  a k 
G
x
y
x
x
x
z
z
Three scalar equations:
 F  ma
 F  ma
 F  ma
z
Fz
x
y
x
z
z
Fy
Fx
y
x
Example 13.1: 50-kg crate rests on a horizontal surface. The
coefficient of kinetic friction is μk=0.3. P=400N towing force.
Determine the velocity of the crate in 3s starting from rest.
Solution:
(1). Free-Body Diagram .
(2). Equation of Motion.
 F  ma
 F  ma
x
y
x
400cos 300  0.3 N C  50a
N C  290.5 N
y
N C  490.5  400sin 300  0
a  5.185 m s2
a
y
x
Example 13.1: 50-kg crate rests on a horizontal surface. The
coefficient of kinetic friction is μk=0.3. P=400N towing force.
Determine the velocity of the crate in 3s starting from rest.
Solution:
(3). Kinematics.
N C  290.5 N
a  5.185 m s2
v  v0  aC t  0  5.185  3   15.6 m s
a
Example 13.2: 10-kg projectile is fired vertically upward from the
ground. v0=50 m/s. Determine the maximum height. If (a)
atmospheric resistance is neglected; (b) resistance is FD=(0.01v2)N.
Solution: (a)
(1). Free-Body Diagram .
W  mg  10  9.81   98.1 N
(2). Equation of Motion.
2
F

ma
a


9.81
m
s

98.1

10a
 z
z
(3). kinematics.
v 2  v02  2a  z  z0 
0  50 0  2  9.81   h  0 
h  127 m
Example 13.2: 10-kg projectile is fired vertically upward from the
ground. v0=50 m/s. Determine the maximum height. If (a)
atmospheric resistance is neglected; (b) resistance is FD=(0.01v2)N.
Solution: (b)
(1). Free-Body Diagram .
FD   0.01v 2  N
(2). Equation of Motion.
 Fz  maz
(3). kinematics.
 0.01v 2  98.1  10 a
a   0.001v 2  9.81 m s2
adz  v dv
 0.001v  9.81 dz  vdv
2
z0  0, v0  50 m s
Integrating:
Example 13.2: 10-kg projectile is fired vertically upward from the
ground. v0=50 m/s. Determine the maximum height. If (a)
atmospheric resistance is neglected; (b) resistance is FD=(0.01v2)N.
Solution: (b)
(3). kinematics.
 0.001v 2  9.81 dz  vdv
z0  0, v0  50 m s
Integrating:
v dv
0 dz   50m s 0.001v 2  9.81
h
0
 500 ln  v  9810 
2
h  114 m
0
50 m s
Example 13.3: WA=900lb, WB=550lb, WC=325lb. Driving force
FA=40t lb. Truck starts from rest. Determine its speed in 2 seconds.
The horizontal force between A and B. Neglect the size of the truck
and carts.
Solution:
(1). Free-Body Diagram.
(2). Equation of Motion.
 900  550  325 
40
t

F

ma
 x

a
x
32.2


(3). Kinematics.
a  0.7256t
y
x
Example 13.3: WA=900lb, WB=550lb, WC=325lb. Driving force
FA=40t lb. Truck starts from rest. Determine its speed in 2 seconds.
The horizontal force between A and B. Neglect the size of the truck
and carts.
a  0.7256t
Solution:
(3). Kinematics.
v
2s
0
0
2s
 1.45 ft s
dv  adt
 dv   0.7256tdt
v  0.3628t
2
0
Example 13.3: WA=900lb, WB=550lb, WC=325lb. Driving force
FA=40t lb. Truck starts from rest. Determine its speed in 2 seconds.
The horizontal force between A and B. Neglect the size of the truck
and carts.
Solution:
a  0.7256t
(4). Determine the force between the truck and B.
 F  ma
x
x
900
 0.7256  2  
40  2   T 
32.2
T  39.4 lb
Example 13.4: 2-kg collar C. k=3N/m. unstretched length is 0.75m.
Collar is released from rest at A. Determine the acceleration and
the normal force of the rod on the collar at the instant y=1m.
Solution:
(1). Free-Body Diagram.
There are four unknowns: NC, FS, a, θ
(2). Equation of Motion.
 F  ma
 F  ma
x
x
 N C  FS cos   0
y
y
19.62  FS sin   2a
The stretch s of the spring is:
s  CB  AB 
FS  ks  3

tan   y 0.75
y
y 2  0.75 2  0.75
y 2  0.75 2  0.75
 y  1m 
a

x
Example 13.5: 100-kg block A is released from rest. The mass of the
pulley and the cord are neglected. Determine the velocity of the 20kg block B in 2s.
Solution:
(1). Free-Body Diagram.
(2). Equation of Motion.
Block A.
 F  ma
y
y
981  2T  100aA
Block B.
 F  ma
y
y
196.2  T  20aB
(3). Kinematics.
The third equation is obtained
by relating aA to aB using a
dependent motion analysis.
Example 13.5: 100-kg block A is released from rest. The mass of the
pulley and the cord are neglected. Determine the velocity of the 20kg block B in 2s.
Solution:
981  2T  100aA
196.2  T  20aB
(3). Kinematics.
2 s A  sB  l
2a A   a B
Then we can have:
T  327.0 N
aA  3.27 m s2 aB  6.54 m s2
v  v0  a B t
 0   6.54  2 
 13.1 m s
CONCEPT MCQ
1. If the cable has a tension of 3 N,
determine the acceleration of block B.
A) 4.26 m/s2 
B) 4.26 m/s2 
C) 8.31 m/s2 
D) 8.31 m/s2 
10 kg
k=0.4
4 kg
2. Determine the acceleration of the block.
A) 2.20 m/s2 
B) 3.17 m/s2 
C) 11.0 m/s2 
D) 4.26 m/s2 
•
30
60 N
5 kg
ATTENTION MCQ
1. Determine the tension in the cable when the
400 kg box is moving upward with a 4 m/s2
acceleration.
T
60
A) 2265 N
B) 3365 N
a = 4 m/s2
C) 5524 N
D) 6543 N
2. A 10 lb particle has forces of F1= (3i + 5j) lb and
F2= (-7i + 9j) lb acting on it. Determine the acceleration of
the particle.
A) (-0.4 i + 1.4 j) ft/s2
B) (-4 i + 14 j) ft/s2
C) (-12.9 i + 45 j) ft/s2
D) (13 i + 4 j) ft/s2
GROUP PROBLEM SOLVING
Given: The 300-kg bar B, originally at
rest, is towed over a series of
small rollers. The motor M is
drawing in the cable at a rate of
v = (0.4 t2) m/s, where t is in
seconds.
Find: Force in the cable and distance s
when t = 5 s.
Plan:
Since both forces and velocity are involved, this
problem requires both kinematics and the equation of motion.
1) Draw the free-body and kinetic diagrams of the bar.
2) Apply the equation of motion to determine the acceleration
and force.
3) Using a kinematic equation, determine distance.
GROUP PROBLEM SOLVING (continued)
Solution:
1) Free-body and kinetic diagrams of the bar:
W = 300 g
y
x
T
=
N
Note that the bar is moving along the x-axis.
2) Apply the scalar equation of motion in the x-direction
+   Fx = 300 a  T = 300 a
Since v = 0.4 t2, a = ( dv/dt ) = 0.8 t
T = 240 t  T = 1200 N when t = 5s.
300 a
GROUP PROBLEM SOLVING (continued)
3) Using kinematic equation to determine distance;
Since v = (0.4 t2) m/s
t
s = s0 + v dt = 0 + 0 (0.4 t2) dt
s=
0.4 3
t
3
At t = 5 s,
0.4
s = 3 53 = 16.7 m
§13.5 Equations of Motion: Normal and Tangential
Coordinates
1. When a particle moves along a curved path which is known, the
equation of motion for the particle may be written in the tangential,
normal and binormal direction.
 F  ma
2. The Equation of Motion.
Fu  F u  F u
t
t
n n
b b
 mat  man
 F  ma
 F  ma
F  0
t
n
b
t
at  d v d t
n
an  v 2 
PROCEDURE FOR THE APPLICATION OF THE
EQUATION OF MOTION
1) Select a convenient coordinate system. Rectangular,
normal/tangential, or cylindrical coordinates may be used.
2) Draw a free-body diagram showing all external forces
applied to the particle. Resolve forces into their
appropriate components.
3) Draw the kinetic diagram, showing the particle’s inertial
force, ma. Resolve this vector into its appropriate
components.
4) Apply the equations of motion in their scalar component
form and solve these equations for the unknowns.
5) It may be necessary to apply the proper kinematic relations
to generate additional equations.
Example 13.6: Determine the banking angle θ for the racetrack.
Ignore friction, car size, a mass m = 800 kg, and track radius ρ=100
with a constant speed v = 300 km/h.
Example 13.6: Determine the banking angle θ for the race track.
Ignore friction, car size, a mass m = 800 kg, and travel around the
curve of radius ρ=100 with a constant speed v = 300 km/h.
Solution:
(1). Free-Body Diagram.
(2). Equation of Motion.
 Fn  man
F  0
b
N C sin   m v 
2
N C cos   mg  0
tan   v 2 g 
Note:
(1) Result is independent of mass
(2) Force sum in the tangential
direction is zero
Example 13.7: The 3-kg disk is released from rest.
Determine the time to reach a speed enough to break the cord.
The maximum tension is 100N, μk=0.1.
Example 13.7: The 3-kg disk is released from rest.
Determine the time to reach a speed enough to break the cord.
The maximum tension is 100N, μk=0.1.
Solution:
(1). Free-Body Diagram.
F   k N D  0.1N D
W  3  9.81   29.43 N
There are three unknowns: ND, at, v
The force F=μkND gives disk a
tangential component of
acceleration  v increases  T
increases until it reaches 100N
Example 13.7: The 3-kg disk. The disk is released from rest.
Determine the time to reach a speed enough to break the cod. The
maximum tension is 100N. μk=0.1.
Solution:
(2). The equation of Motion.
 Ft  mat
T  3  v 2 1  100 N
 F  ma
F  0
0.1 N D  3at
n
b
n
N D  29.43  0
N D  29.43 N
at  0.981 m s 2
vcr  5.77 m s
Example 13.7: The 3-kg disk. The disk is released from rest.
Determine the time to reach a speed enough to break the cod. The
maximum tension is 100N. μk=0.1.
Solution:
(3). Kinematics.
N D  29.43 N
at  0.981 m s 2
vcr  5.77 m s
at  constant
vcr  v0  at t
5.77  0  0.981t
t  5.89 s
Example 13.8: 150-lb skier. At point A of the jump, find the normal
force on the skier. vA=65ft/s. Find aA
Solution:
(1). Free-Body Diagram.
dy dx  x 100 x  0  0
The slop at A is horizontal.
(2). The equation of Motion.
150
 Ft  mat 0  32.2 at
2
F

ma
150
65
 n
n
N A  150 
32.2 
The radius of curvature ρ must be
determined at point A.
Example 13.8: 150-lb skier. At point A of the jump, find the normal
force on the skier. vA=65ft/s. Find aA
Solution:
150 65 2
150
N A  150 
0
at
32.2 
32.2
The radius of curvature ρ must be
determined at point A.
1 2
y
x  200
200
 x 0
dy
1

x
dx 100
1   y  


y
d2 y
1

2
dx
100
 100ft
N A  347 lb
2
3 2
x 0
Example 13.8: 150-lb skier. At the end point A of the jump which is
the parabola. Determine the normal force on the skier. vA=65ft/s.
and the aA?
Solution:
(3). Kinematics.
at  0
v2
N A  347 lb
 x  0  100ft
65 2
an 

 42.2 ft s 2
 100
Example 13.9: 60-kg skateboarder coasts down the circular track.
Start from rest when θ=0. Determine the magnitude of the normal
reaction the track exerts on him when θ=600.
Example 13.9: 60-kg skateboarder coasts down the circular track.
Start from rest when θ=0. Determine the magnitude of the normal
reaction the track exerts on him when θ=600.
Solution:
(1). Free-Body Diagram.
Three unknowns: NS, at, an(v).
(2). The equation of Motion.
 F  ma
 F  ma
n
n
t
t
v2
N S   60  9.81  N  sin   60 kg
4m
 60  9.81  N  cos    60 kg  at
at  9.81cos 
Example 13.9: 60-kg skateboarder coasts down the circular track.
Start from rest when θ=0. Determine the magnitude of the normal
reaction the track exerts on him when θ=600.
at  9.81cos 
Solution:
v2
N S   60  9.81 N  sin   60 kg
4m
(3). Kinematics.
vdv  at ds
v
600
0
0
 v dv  
2
ds  rd  4d
9.81cos   4d
v
v
600
 39.24 sin  0
2 0
v 2  67.97
N S  1.53kN
CONCEPT MCQ
1. A 10 kg sack slides down a smooth surface. If the normal
force at the flat spot on the surface, A, is 98.1 N () , the
radius of curvature is ____.
A) 0.2 m
B) 0.4 m
v=2m/s
C) 1.0 m
D) None of the above.
A
2. A 20 lb block is moving along a smooth surface. If the
normal force on the surface at A is 10 lb, the velocity is
________.
A
A) 7.6 ft/s
B) 9.6 ft/s
C) 10.6 ft/s
D) 12.6 ft/s
=7 ft
ATTENTION MCQ
1. The tangential acceleration of an object
A) represents the rate of change of the velocity vector’s
direction.
B) represents the rate of change in the magnitude of the
velocity.
C) is a function of the radius of curvature.
D) Both B and C.
2. The block has a mass of 20 kg and a speed of
v = 30 m/s at the instant it is at its lowest point.
Determine the tension in the cord at this instant.
10 m

A) 1596 N
C) 1996 N
B) 1796 N
D) 2196 N
v = 30m/s
GROUP PROBLEM SOLVING I
Given:The boy has a weight of 60 lb.
At the instant  = 60, the boy’s
center of mass G experiences a
speed v = 15 ft/s.
Find: The tension in each of the two
supporting cords of the swing
and the rate of increase in his
speed at this instant.
Plan:
1) Use n-t coordinates and treat the boy as a particle.
Draw the free-body and kinetic diagrams.
2) Apply the equation of motion in the n-t directions.
GROUP PROBLEM SOLVING I (continued)
Solution:
1) The n-t coordinate system can
be established on the boy at
angle 60°. Approximating the
boy as a particle, the free-body
and kinetic diagrams can be
drawn:
Free-body diagram
W
60
n
2T
t
Kinetic diagram
=
60
n
man
mat
t
GROUP PROBLEM SOLVING I (continued)
Free-body diagram
Kinetic diagram
W
60
n
2T
=
60
n
man
t
mat
t
2) Apply the equations of motion in the n-t directions.
Fn = man  2T  W sin 60° = man
Using an = v2/ = 152/10, W = 60 lb,
we get: T = 46.9 lb
Ft = mat  60 cos 60° = (60 / 32.2) at

at = v = 16.1 ft/s2
GROUP PROBLEM SOLVING II
Given: A 800 kg car is traveling over
a hill with the shape of a
parabola. When the car is at
point A, its v = 9 m/s and
a = 3 m/s2. (Neglect the size
of the car.)
Find: The resultant normal force and resultant frictional force
exerted on the road at point A by the car.
Plan:
1) Treat the car as a particle. Draw its free-body and
kinetic diagrams.
2) Apply the equations of motion in the n-t directions.
3) Use calculus to determine the slope and radius of
curvature of the path at point A.
GROUP PROBLEM SOLVING II (continued)
Solution:
1) The n-t coordinate system can
be established on the car at
point A. Treat the car as a
particle and draw the freebody and kinetic diagrams:
W
F
N
n

=

t
man
mat
n
t
W = mg = weight of car
N = resultant normal force on road
F = resultant friction force on road
GROUP PROBLEM SOLVING II (continued)
2) Apply the equations of motion in the n-t directions:
 Fn = man  W cos  – N = man
Using W = mg and an = v2/ = (9)2/
 (800)(9.81) cos  – N = (800) (81/)
 N = 7848 cos  – 64800 / 
Eq. 1
 Ft = mat  W sin  – F = mat
Using W = mg and at = 3 m/s2 (given)
 (800)(9.81) sin  – F = (800) (3)
 F = 7848 sin  – 2400
Eq. (2)
GROUP PROBLEM SOLVING II (continued)
3) Determine  by differentiating y = f(x) at x = 80 m:
y = 20(1 – x2/6400)  dy/dx = (–40) x / 6400
 d2y/dx2 = (–40) / 6400
dy 2 3/2
[1 + (–0.5)2]3/2
[1 + ( ) ]
dx
 =
=
= 223.6 m
2
dy
0.00625
x = 80 m
2
dx
Determine  from the slope of the curve at A:
dy
tan  = dy/dx

dx
x = 80 m
 = tan-1 (dy/dx) = tan-1 (-0.5) = 26.6°
GROUP PROBLEM SOLVING II (continued)
From Eq. (1): N = 7848 cos  – 64800 / 
= 7848 cos (26.6°) – 64800 / 223.6 = 6728 N
From Eq. (2): F = 7848 sin  – 2400
= 7848 sin 26.6° – 2400 = 1114 N
Coordinate systems
1. Cartesian
2. Normal and Tangential
3. Cylindrical
§13.6 Equations of Motion: Cylindrical Coordinates
F u
unit vectors in cylindrical
system: ur , uθ , uz
z
 F u
z
1. Equations of Motion
 F  ma
 F u   F u   F u
r
r
z
 mar ur  ma u  maz uz
 F  ma
 F  ma
 F  ma
r
z
F u
r
z
z

r
r
z
ar  r  r 2
a  2r  r
r
§13.6 Equations of Motion: Cylindrical Coordinates
2. Tangential and Normal Forces - example:
(1). force P makes the particle move along a path r=f(θ) .
(2). The normal force N which the path exerts on the particle is
always perpendicular to the tangent of the path.
§13.6 Equations of Motion: Cylindrical Coordinates
2. Tangential and Normal Forces
(3). Friction force F always acts along the tangent in the opposite
direction of motion
(4). The direction of N and F can be specified relative to the radial
coordinate by using the angle ψ defined between the extended
radial line and the tangent to the curve.
§13.6 Equations of Motion: Cylindrical Coordinates
2. Tangential and Normal Forces
(5). The displacement components.
distance ds along the path:
(a) Component of displacement in radial direction : dr
(b) Component of displacement in transverse direction : rdθ
§13.6 Equations of Motion: Cylindrical Coordinates
2. Tangential and Normal Forces
(6). The determination of angle ψ
The components of dr and rdθ are mutually perpendicular, then:
tan  rd dr  r  dr d 
§13.6 Equations of Motion: Cylindrical Coordinates
2. Tangential and Normal Forces
The sign of angle ψ:
ψ is calculated as a positive quantity
ψ is measured from the extended radial line to the tangent in a
counterclockwise sense or in the positive direction of θ.
Example 13.10: 0.5-kg double-collar. Arm AB rotates with a
constant angular velocity of 3 rad/s. Determine the force the arm
exert on the collar at the instant θ=450. ( all in horizontal plane)
Solution:
(1). Free-Body Diagram.
Four unknowns: NC, F, ar , aθ
(2). The equation of Motion.
 Fr  mar
 F  ma
 N C cos 45 0   0.5 kg  ar
F  N C sin 45 0   0.5 kg  a
(3). Kinematics.
r   0.8 cos   m

  450
 3 rad s

  450
0
Example 13.10: 0.5-kg double-collar. Arm AB rotates with a
constant angular velocity of 3 rad/s. Determine the force the arm
exert on the collar at the instant θ=450. ( all in horizontal plane)
Solution:
(3). Kinematics.

r   0.8 cos   m

 3 rad s
  450
  45
0
r   0.8 sin   
r  0.8   sin      cos    2 
  45 0
r  0.5657 m
r  1.6971 m s
r  5.091 m s2
0
Example 13.10: 0.5-kg double-collar. Arm AB rotates with a
constant angular velocity of 3 rad/s. Determine the force the arm
exert on the collar at the instant θ=450. ( all in horizontal plane)
Solution:
(3). Kinematics.
  45 0   3 rad s
r  0.5657 m
 0
r  1.6971 m s , r  5.091 m s 2
(4). The accelerations ar , aθ
ar  r  r 2  10.18 m s2
a  r  2r  10.18 m s2
(5). The forces of NC, F.
N C  7.2 N,F  0
Example 13.11: 2-kg cylinder C. Arm OA rotates in the vertical
plane at a constant angular velocity of 0.5 rad/s. Determine the
force the arm exerts on the roller at the instant θ=600.
Solution:
(1). Free-Body Diagram.
Four unknowns: NC, FP , ar , aθ
(2). The equation of Motion.
 Fr  mar 19.62sin  NC sin   2ar
 F  ma 19.62cos  F  N cos  2a
P
C
(3). Kinematics.
r  0.4 sin 
r  0.2csc   cot 
r  0.1csc    cot 2   csc 2  
Example 13.11: 2-kg cylinder C. Arm OA rotates in the vertical
plane at a constant angular velocity of 0.5 rad/s. Determine the
force the arm exerts on the roller at the instant θ=600.
Solution:
(3). Kinematics.
r  0.4 sin 
r  0.2csc   cot 
r  0.1csc    cot 2   csc 2  
  60 0
  0.5 rad s ,   0
r  0.462m
r  0.133 m s , r  0.192 m s 2
Example 13.11: 2-kg cylinder C. Arm OA rotates in the vertical
plane at a constant angular velocity of 0.5 rad/s. Determine the
force the arm exerts on the roller at the instant θ=600.
Solution:
(4). The accelerations ar , aθ
ar  r  r 2  0.0770 m s2
a  r  2r  0.133 m s2
(5). The forces of NC, F.
N C  19.4 N,FP  0.356 N
Example 13.12: 0.5-kg cylinder C. Arm OA rotates in the
horizontal plane at a constant angular velocity of 4 rad/s.
Determine the force the arm exerts on the pin at the instant θ=1800.
Solution:
(1). Free-Body Diagram.
r  0.1
dr d  0.1
tan  r  dr d   
1
0
   :   tan   72.3
  900    17.70
Example 13.12: 0.5-kg cylinder C. Arm OA rotates in the
horizontal plane at a constant angular velocity of 4 rad/s.
Determine the force the arm exerts on the pin at the instant θ=1800.
Solution:
(1). Free-Body Diagram.
(2). The equation of Motion.
0
F

ma
N
cos17.7
 0.5ar
 r
r
C
 F  ma
FC  N C sin17.70  0.5a
(3). Kinematics.
r  0.1 , r  0.1 , r  0.1
   r  0.4 m s , r  0
Example 13.12: 0.5-kg cylinder C. Arm OA rotates in the
horizontal plane at a constant angular velocity of 4 rad/s.
Determine the force the arm exerts on the pin at the instant θ=1800.
Solution:
(4). The accelerations ar , aθ
ar  r  r 2  5.03 m s2
a  r  2r  3.2 m s2
(5). The forces of NC, F.
N C  2.64 N,FC  0.800 N
CONCEPT QUIZ
1. When a pilot flies an airplane in a
vertical loop of constant radius r at
constant speed v, his apparent weight
is maximum at
A) Point A
C) Point C
B
C
r
A
D
B) Point B (top of the loop)
D) Point D (bottom of the loop)
2. If needing to solve a problem involving the pilot’s weight at
Point C, select the approach that would be best.
A) Equations of Motion: Cylindrical Coordinates
B) Equations of Motion: Normal & Tangential Coordinates
C) Equations of Motion: Polar Coordinates
D) No real difference – all are bad.
E) Toss up between B and C.
ATTENTION QUIZ
1. For the path defined by r =  2 , the angle  at   05 rad
is
A) 10º
B) 14º
C) 26º
D) 75º
2. If r =  2 and  = 2t, find the magnitude of 𝑟 and 𝜃 when
t = 2 seconds.
A) 4 cm/sec, 2 rad/sec2
B) 4 cm/sec, 0 rad/sec2
8 cm/sec, 16 rad/sec2
D) 16 cm/sec, 0 rad/sec2
C)
GROUP PROBLEM SOLVING I
Given: The smooth can C is lifted
from A to B by a rotating rod.
The mass of can is 3 kg.
Neglect the effects of friction
in the calculation and the size
of the can so that
r = (1.2 cos ) m.
Find:
Forces of the rod on the can when  = 30 and
𝜃 = 0.5 rad/s, which is constant.
Plan:
1) Find the acceleration components using the kinematic
equations.
2) Draw free body diagram & kinetic diagram.
3) Apply the equation of motion to find the forces.
GROUP PROBLEM SOLVING (continued)
Solution:
1) Kinematics:
r = 1.2 cos θ
r = −1.2 (sin θ)θ
r = −1.2 cos θ θ2 − 1.2 (sin θ)θ
When  = 30, θ = 0.5 rad/s and θ = 0 rad/s2.
r = 1.039 m
r = 0.3 m/s
r = 0.2598 m/s2
Accelerations:
ar = r − rθ2 = − 0.2598 − (1.039) 0.52 = − 0.5196 m/s2
a = rθ + 2rθ = (1.039) 0 + 2 (0.3) 0.5 =  0.3 m/s2
GROUP PROBLEM SOLVING (continued)
2) Free Body Diagram

Kinetic Diagram
3(9.81) N
ma
30
r

30
N
mar
F
3) Apply equation of motion:
 Fr = mar  -3(9.81) sin30 + N cos30 = 3 (-0.5196)
 F = ma  F + N sin30  3(9.81) cos30 = 3 (-0.3)
N = 15.2 N, F = 17.0 N