Dynamics
8-1
Overview
Dynamics—the study of moving objects.
Kinematics— the study of a body’s motion independent of the forces on
the body.
Kinetics—the study of motion and the forces that cause motion.
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Dynamics
8-2
Kinematics—Rectangular Coordinates
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Dynamics
8-3a1
Kinematics—Polar Coordinates
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Dynamics
8-3a2
Kinematics—Polar Coordinates
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Dynamics
8-4a
Kinematics—Circular Motion
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Dynamics
8-4b1
Kinematics—Circular Motion
Angular velocity =
Angular acceleration =
Tangential acceleration =
Normal acceleration =
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Dynamics
8-4b2
Kinematics—Circular Motion
Example (FEIM):
A turntable starts from rest and accelerates uniformly at 1.5 rad/s2. How
many revolutions does it take for the rotational frequency to reach 33.33
rpm?
$ rad '$
rev '$1 min '
rad
" = 2#f = & 2#
)&33.33
)&
) = 3.49
min (% 60 s (
s
% rev (%
!
"=
#
t=
"
#
!
0
2
"=
!
n=
!
tf
#
$dt =
(
( )(
tf
0
%tf2
%tdt =
2
3.49 rads
)
2
#
=
= 4.06 rad
rad
2$ 2 1.5 s
2
)
4.06 rad
= 0.65 revolution
rad
2"
rev
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Dynamics
8-5a1
Kinematics—Projectile Motion
Constant acceleration formulas:
s = s0 + v0t + 21 at 2
v = v0 + a0t
(
v2 = v20 + 2a0 s " s0
)
!
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Dynamics
8-5a2
Kinematics—Projectile Motion
Example 1 (FEIM):
A projectile is launched at 180 m/s at a 30° incline. The launch point
is 150 m above the impact plane. Find the maximum height, flight
time, and range.
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Dynamics
8-5a3
Kinematics—Projectile Motion
h" = v sin# + at = 0
%
m(
'180 * 0.5
s)
sin#
&
t = $v
=$
= 9.18 s
%
(
a
m
'$9.8 *
s)
&
h = h + v t sin# + at
%
%
m(
m(
h = 150 m + '180 *(9.18 s)(0.5) $ '9.8 *(9.18 s) = 563 m
s )
s )
&
&
h = 0 = 150 m + 90t $ 4.9t
max
0
( )
max
0
0
1
2
0
max
2
()
1
2
2
2
2
2
impact
By the quadratic equation, t
impact
= 19.9 s
cos 30° = 0.866
R = R + v t cos #
%
m(
= 0 + '180 *(19.9 s)(0.866) = 3100 m
s)
&
impact
0
0 impact
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Dynamics
8-5b1
Kinematics—Projectile Motion
Example 2 (FEIM):
A bomber flies horizontally at 275 km/h and an altitude of 3000 m. At
what viewing angle from the bomber to the target should the bomb be
dropped?
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Dynamics
8-5b2
Kinematics—Projectile Motion
himpact = at =
1
2
2
()
1
2
#
m& 2
%"9.8 2 (t = "3000 m
s '
$
2himpact
= 24.7 s
g
#
km &#
m &# 1 hr &
%275
(%1000
(%
( = 76.39 m/s
hr
km
3600
s
$
'$
'$
'
#
m&
Rimpact = R0 + v0t = 0 + % 76.39 ( 24.7 s = 1887 m
s'
$
R
1887 m
) = arctan impact = arctan
= 32.2°
himpact
3000 m
t=
(
)
!
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Dynamics
8-6a
Kinetics—Newton’s 2nd Law of Motion
For a constant mass,
One-dimension motion
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Dynamics
8-6b1
Kinetics—Newton’s 2nd Law of Motion
Example (FERM prob. 6, p. 15-5):
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Dynamics
8-6b2
Kinetics—Newton’s 2nd Law of Motion
Example (FERM prob. 6, p. 15-5):
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Dynamics
8-7a
Kinetics—Impulse and Momentum
Impulse (constant mass in one dimension)
Momentum
Impulse-Momentum Principle
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Dynamics
8-7b1
Kinetics—Impulse and Momentum
Example 1 (FEIM):
A 0.046 kg marble attains a velocity of 76 m/s in a slingshot. Contact
with the slingshot is 1/25 of a second. What is the average force on
the marble during the launch?
# m&
(0.046 kg)% 76 (
m"v
$ s'
Fave =
=
= 87.4 N
"t
0.04 s
!
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Dynamics
8-7b2
Kinetics—Impulse and Momentum
Example 2 (FEIM):
A 2000 kg cannon fires a 10 kg projectile horizontally at 600 m/s. It
takes 0.007 s for the projectile to pass through the barrel. What is the
recoil velocity if the cannon is not restrained? What average force
must be exerted on the cannon to keep it from moving?
mprojectile"vprojectile = mcannon"vcannon
m
(10 kg)(600 ) = (2000 kg)(vcannon )
s
m
vcannon = 3 = initial recoil velocity
s
!
!
m
(10 kg)(600 )
m"v
s = 8.57 #105 N
F=
=
"t
0.007 s
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Dynamics
8-8a
Work & Energy
Work
Potential Energy
• Gravity
W = PE2 " PE1 = mg(h2 " h1)
Kinetic Energy of a Mass
•
W = KE2 " KE1 = 21 m(v22 " v12 )
!
Spring (linear)
Fs = kx where the spring is
compressed a distance x
Kinetic Energy of a Rotating Body
!
W = PE2 " PE1 = 21 k(x22 " x12 )
W = KE2 " KE1 = 21 I(#22 " #12 )
!
!
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Dynamics
8-8b
Work & Energy
Conservation of Energy
• For a closed system (no external work), the change in potential
energy equals the change in kinetic energy.
PE1 " PE2 = KE2 " KE1
PE1 + KE1 = PE2 + KE2
• For a system with external work, W equals ΔPE + ΔKE.
!
W1"2 = (PE1 # PE2 ) + (KE2 # KE1)
PE1 + KE1 +W1"2 = PE2 + KE2
!
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Dynamics
8-8c
Work & Energy
Impacts: Momentum is always
conserved.
Elastic Impacts: Kinetic energy is
conserved.
m1v1 + m2 v2 = m1v"1 + m2 v"2
m1 = m2sov1 + v2 = v"1 + v"2 = 0.85 # 0.53 = 0.32
1
2
2
m v12 + 21 m v22 = 21 m v"1 + 21 m v"2
2
v12 + v22 = v"1 + v"2
Example 1 (FEIM):
Two identical balls collide along
!
their centerlines in an elastic
collision. The initial velocity of
ball 1 is 0.85 m/s. The initial
velocity of ball 2 is –0.53 m/s.
!
What is the relative velocity of
each ball after the collision?
(A) –0.53 m/s and 0.85 m/s
(B) –0.72 m/s and 1.2 m/s
(C) –5.1 m/s and 1.2 m/s
(D) 0.98 m/s and 1.8 m/s
2
2
Solving two equations and two unkowns:
v"1 = #0.53 m/s
v"2 = 0.85 m/s
Therefore, (A) is correct.
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Dynamics
8-8d
Work & Energy
Example 2 (FEIM):
Ball A of 200 kg is traveling at 16.7 m/s. It strikes stationary ball B of
200 kg along the centerline. What is the velocity of ball A after the
collision? Assume the collision is elastic.
(A) –16.7 m/s
(B) –8.35 m/s
(C) 0
(D) 8.35 m/s
mA = mBsov A + vB = v"A + v"B = 16.7 m/s
2
v2A + v2B = v"A + v"B
2
There are two possible solutions for these equations.
v"A = 0, v"B = 16.7 m/s
!
or
v"A = 16.7 m/s, v"B = 0
Since there must be a change in the collision, ball A’s velocity must be 0.
Therefore, (C) is correct.
!
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Dynamics
8-8e
Work & Energy
Inelastic Impacts:
Kinetic energy does not have to be
conserved if some energy is converted
to another form.
v" # v" = #e(v # v )
1
2
1
2
where e = coefficient of restitution
m v (1+ e) + (m # em )v
v" =
m +m
2
2
1
2
1
1
1
v" =
2
2
m v (1+ e) + (em # m )v
m +m
1
1
1
1
2
2
2
Example 1 (FEIM):
A ball is dropped from an initial height
ho. If the coefficient of restitution is
0.90, how high will the ball rebound?
(A) 0.45ho
(B) 0.81ho
(C) 0.85ho
(D) 0.9ho
mgh = 21 mv2
v = 2gho
v" = #ev = #e 2gho = 2gh"
h" = e2ho = (0.9)2 ho = 0.81ho
!
Since there must be a change in the
! collision, ball A’s velocity must be 0.
Therefore, (B) is correct.
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Dynamics
8-8f
Work & Energy
!
Example 2 (FEIM):
Two masses collide in a perfectly inelastic collision. What is the
velocity of the combined mass after collision?
m1 = 4m2 v1 = 10 m/s v2 = "20 m/s
(A) 0
(B) 4 m/s
(C) –5m/s
(D) 10 m/s
“Perfectly inelastic” means the masses collide and stick together.
m1v1 + m2 v2 = m3 v3
m3 = m1 + m2 = 5m2
" m%
"
m%
4m2 $10 ' + m2 $(20 ' = 5m2 v3
s&
# s&
#
" m%
" m%
5m2 v3 = $40 'm2 ( $ 20 'm2
# s&
# s&
v3 = 4 m/s
Therefore, (B) is correct.
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Dynamics
8-9a
Kinetics
Friction
Example (FEIM):
A snowmobile tows a sled with a weight of 3000 N. It accelerates up a
15° slope at 0.9 m/s2. The coefficient of friction between the sled and the
snow is 0.1. What is the tension in the tow rope?
Fslope = Frope " (Ffriction + Fgravity ) = maslope
Frope = Ffriction + Fgravity + maslope
= mg sin15° + mgµ cos15° + maslope
= (3000)(0.2588) + (3000 N)(0.1)(0.9659)
"
%
$ 3000 N '"
m%
+$
'$0.9 2 '
m
s &
$ 9.8
'#
#
s2 &
= 1342 N
!
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8-9b1
Kinetics
Plane Motion of a Rigid Body
Similar equations can be written for the y-direction or any other
coordinate direction.
Example (FEIM):
A 2500 kg truck skids with a deceleration of 5 m/s2. What is the
coefficient of sliding friction? What are the frictional forces and normal
reactions (per axle) at the tires?
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Dynamics
8-9b2
Kinetics
The force of deceleration is equal to the friction force.
" m%
Fdeceleration = (2500 kg)$5 2 ' = Ffriction = µmg
# s &
"
m%
= µ(2500 kg)$9.8 2 '
s &
#
!
" m %"
m%
µ = $5 2 '$9.8 2 ' = 0.51
s &
# s &#
!
The moment about the center of gravity, MA, must be equal to zero.
!
"M
A
= (4.5 m)NB # (2 m)mg # (1 m)(Fdeceleration ) = 0
#
# m&
m&
= (4.5 m)NB " (2 m)(2500 kg)%9.8 2 ( " (2500 kg)%5 2 ( = 0
s '
$
$ s '
NB = 13,667 N
!
!
!
#
m&
N A = mg " NB = (2500 kg)%9.8 2 ( " 13 667 N = 10833 N
s '
$
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Dynamics
8-10a1
Rotation
Rotation about Fixed Axis
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Dynamics
8-10a2
Rotation
Example (FEIM):
A mass m is attached to a rope wound around a cylinder of mass mC
and radius r. What is the acceleration of the falling mass? What is the
rope tension?
There are three simultaneous equations for the movement of the mass,
the cylinder, and the relationship between the two:
mg " F = ma
Fr = I#
#r = a
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Dynamics
8-10a3
Rotation
Rearranging,
a
r
mC r 2a mC
F=
=
a
2r 2
2
Fr = I
!
!
mg "
a=g
!
mC
a = ma
2
m
m
m+ C
2
Solving for F,
!
F=
mC
mC m
mmC
a=g
=
g
"
%
2
2m + mC
m
2$m + C '
2 &
#
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Dynamics
8-10b
Rotation
Centripetal Force
• The force required to keep a body rotating about an axis.
(r is the distance from the center of mass to the center of rotation.)
Centrifugal Force
• The “reaction” to centripetal force.
• The centrifugal force, like any inertia force, should not be used in
free-body diagrams.
Example (FEIM):
A 2000 kg car travels 65 km/hr around a curve of radius 60 m.
What is the centripetal force?
65 km/hr = 18.06 m/s
2
"
m%
(2000
kg)
18.06
$
'
s&
m v2
#
Fc =
=
= 10900 N
r
60 m
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Dynamics
8-10c
Rotation
Banking Curves
Example (FEIM):
A 2000 kg car travels at 64 km/hr around a banked curve with a radius
of 150 m. What should the angle between the roadway and the
horizontal be so tire friction is not needed to prevent sliding?
" km %"
m %" 1hr %
m
64
1000
$
'$
'$
' = 17.8
km &# 3600 s &
s
# hr &#
" "
m% %
$ $17.8 ' '
" v2 %
s& '
#
( = arctan$ ' = arctan$
= 12.1°
"
%
$
'
gr
m
# &
9.8
$
'(150 m) '
$
s&
##
&
!
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Dynamics
8-10d
Rotation
Free Vibration
Spring and Mass:
Natural Frequency:
Solution:
For initial conditions x(0) = x0 and x'(0) = v0,
For initial conditions x(0) = x0 and x'(0) = 0,
Torsional Free Vibration:
Natural frequency:
"n =
kt
I
Solution:
!
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Dynamics
8-10e1
Rotation
Example (FEIM):
A 54 kg mass is supported by three springs, as shown. The starting
position is 5.0 cm down from the equilibrium position. No external
forces act on the mass after it is released. What are the maximum
velocity and acceleration?
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Dynamics
8-10e2
Rotation
From the solution to the differential equation,
#v &
x(t) = x0 cos "nt + % 0 ( sin"nt
$ "n '
v(t) = x)(t) = *x0"n sin"nt + v0 cos "nt
But, v0 = 0; so v(t) = "x0#n sin#nt
!
x0 = 0.05 m
!
k = k1 + k2 + k3 = 1750
N
N
N
N
+1750 + 4375 = 7875
m
m
m
m
N
k
m = 12.08 rad/s
"n =
=
m
54 kg
%
rad (
v = x# = ($0.05 m)'12.08
* sin12.08t
s )
&
= "0.604 m/s sin12.08t
7875
!
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Dynamics
8-10e3
Rotation
The maximum velocity is when sin 12.08t = 1. Because the motion is
oscillatory, the maximum velocity occurs in both directions.
vmax = ±0.604 m/s
2
$
rad '
a = v" = (#0.05 m)&12.08
) cos12.08t
s (
%
amax = ±7.30 m/s2
!
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Dynamics
8-11a1
Review
Example:
A yoyo (mass m, inertia I about center of gravity) is placed on a
horizontal surface with a coefficient of friction µ. The string, wrapped
underneath, is pulled with a force F. Determine if the yoyo will slip on the
floor and which direction it will rotate as a function of F.
The two equations for the movement of the center of gravity and the
rotation are
F + R = ma
Fri + Rro = I"
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Dynamics
8-11a2
Review
There are two possibilities for the
third equation: slip or no slip.
No slip: R < µmg " a = #ro $
!
Solving these 3 equations with 3
unknowns (α, a, and R) leads to
r #r
" = #F o i 2 $ " < 0
I + mro
a=
!
!
!
(
Fro ro " ri
I + mro2
) #a > 0
I + mri 2
R = "F
I + mro2
The yoyo is moving to the right while
wrapping itself on the string. The
solution for R gives the condition for
no slip.
I + mr
F < µmg
I + mr
!
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8-11a3
Review
Slip: R = –µmg
Solving these 3 equations lead to
a=
=
!
F " µmg
m
Fri " µmgro
I
The yoyo always moves to the right, but can rotate forward or backard
depending on the value for F.
!
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