FE Review - Kinetics &
Dynamics
Kinematics
Kinematics- The study of the geometry of motion given a position vector r(t)
The velocity vector v̄(t) =
The acceleration vector ā(t) =
dr̄
dt
dv̄
d2 r̄
= 2
dt
dt
Coordinate Systems
The kinetics (and dynamics) of an object must be described relative to a coordinate system.
Rectangular Coordinates
r̄
= x î + y ĵ + z k̂
dx
dy
dz
v̄ =
î +
ĵ +
k̂
dt
dt
dt
= ẋ î + ẏ ĵ + ż k̂
ā = ẍ î + ÿ ĵ + z̈ k̂
k z
i
j
y
x
Figure 1: Rectangular coordinates
1
eθ
er
r
θ
Figure 2: Polar coordinates
Polar Coordinates
r̄
= r êr
v̄ = ṙ êr + rθ̇ êθ
ā = (r̈ − r θ̇2 ) êr + (rθ̈ + 2ṙθ̇) êθ
where r
θ
:
:
radial position
angular position
ṙ
r̈
:
:
radial velocity
radial acceleration
θ̇
θ̈
:
:
angular velocity
angular acceleration
Normal-Tangent Coordinates
eT
en
ρ
Path of
motion
Figure 3: Normal-Tangent coordinates
v̄
= v êτ
v2
dv
ā =
êτ +
ên
dt
ρ
where v : tangential velocity
dv
: rate of change of the tangential velocity
dt
ρ : radius of curvature.
2
(1)
Motions
Straight Line Motion
Constant Acceleration (with initial velocity vo and initial position so )
dv
dt
ds
dt
= ao
= v
→
→
v=
s=
Z
Z
t
ao dt
→
to
t
v = v o + ao t
(vo + ao τ ) dτ
to
1
s = s o + v o t + a o t2
2
→
Projectile Motion
y
g
vo
θ
x
Figure 4: Projectile Motion
Acceleration in the x direction
ax
vx
= 0
= vo cos θ
x = xo + vo t cos θ
Acceleration in the y direction
ay
vy
y
= −g
= vo sin θ − gt
= yo + vo t sin θ −
gt2
2
Plane Circular Motion
r
= constant radius
θ̇
α
= ω = angular velocity
= ω̇ = angular acceleration
Tangential acceleration
Normal acceleration
:
:
aτ = αr
an = rω 2
Tangential velocity
Position (arc length)
:
:
vτ = ωr
s = θr
3
vt
r
s
θ
Figure 5: Plane Circular Motion
Newton’s Laws of Motion
(1) An object moving in a straight line at constant velocity will remain in motion unless
acted on by an external force. An object at rest will remain at rest unless acted on
by an external force.
(2) The net external force acting on an object is equal to the rate of change of linear
momentum, i.e,
where
F̄
=
F̄
mv̄
:
:
d
(mv̄)
dt
net external force
linear momentum
For constant mass,
F̄ = m
dv̄
= mā
dt
(3) For two object in contact the mutual force on interaction is equal and opposite.
One Dimensional Motion
F
v(t)
= ma → a(t) = F (t)/m
Z t
= vo +
(F (τ )/m) dτ
to
x(t)
= x o + vo t +
4
Z
t
v(τ ) dτ
to
Impulse and Momentum
The impulse of a force is equal to the change in linear momentum
Z t1
F̄ dt = mv̄(t1 ) − mv̄(t0 )
I¯ =
t0
If the net external force is zero the linear momentum is conserved
Z t1
F̄ dt = 0 = mv̄(t1 ) − mv̄(t0 )
I¯ =
t0
mv̄(t1 ) = mv̄(t0 )
This called the ‘Principle of Conservation of Linear Momentum’.
Work and Energy
Work done by a force,
W =
Z
F̄ dr̄
Work done by a force in moving an object from point 1 to point 2 is the change in kinetic
energy
W
= T 2 − T1
1
1
=
mv22 − mv12
2
2
Work done by a conservative force in moving an object from point 1 to point 2 is
W = U 1 − U2
where U1 is the potential energy at point 1 and U2 is the potential energy at point 2.
Principle of Conservation of Energy
If all the forces acting on a system are conservative then the total energy is constant, i.e.,
E=
=
T1 + U1
| {z }
T otal energy at 1
T2 + U 2
| {z }
T otal energy at 2
Potential energy due to gravity
g
h
Figure 6: Potential energy due to gravity
Ugravity = mgh
5
Potential energy due to a spring
x
k
Change in length of the spring
from the undeformed length
Figure 7: Potential energy due to a spring
Uspring =
1 2
kx
2
Power
Power is the rate at which work is done
P =
dW
dt
Impact
m1
v1
m2
v2
Common normal
Common normal
v1
v2
m1
m2
Figure 8: Direct Impact and Oblique Impact
Direct Impact
All velocities before and after impact are along the common normal.
Oblique Impact
There is a velocity component along the common tangent.
Newton’s Collision Rule
v10 n − v20 n
| {z }
=
relative veloctiy along the common normal af ter impact
6
−e(v1n − v2n )
|
{z
}
relative velocity along the common normal bef ore impact
where e is coefficient of restitution.
m1 v̄ 0 + m2 v̄20
| 1 {z
}
=
linear momentum af ter impact
m1 v̄1 + m2 v̄2
|
{z
}
linear momentum bef ore impact
Plane Motion of a Rigid Body
Fx
= max
Fy
Mc
= may
= Ic α
Mc - net external moment about the center of mass.
Ic - Mass moment of inertia about the center of mass.
α - Angular acceleration.
Rotation about a fixed point
Mo
o
Figure 9: Rotation about a fixed point
Mo = I o α
where Io is a moment of inertia about point O.
Kinetic energy of a plane rigid body
For general motion,
T
=
1
1
m(vx2 + vy2 ) + Ic ω 2
2
2
If the rotation is about a fixed point, then
T =
1
Io ω 2
2
Free Vibration
The equation of motion for the system shown here is
mẍ + kx = 0.
7
k
m
x
Figure 10: Free vibration
The solution to this differential equation is
x(t) = C1 cos ωt + C2 sin ωt,
where C1 , and C2 are constants that depend on the initial conditions, and
r
k
,
ω=
m
is called the natural frequency. The period of oscillation is given by
T =
2π
.
ω
If x(0) = x0 , ẋ(0) = v0 , then
x(t) = x0 cos ωt +
v0
sin ωt.
ω
If x(0) = x0 , ẋ(0) = 0, then
x(t) = x0 cos ωt.
Torsional Vibration
kt : Torsional spring
o
θ
Io
Figure 11: Torsional vibration
In this case the equation of motion is given by
θ̈ + ω 2 θ = 0,
where
r
kt
,
I0
is the natural frequency. Here, kt is the torsional spring stiffness, and Io is the moment of
inertia about O.
ω=
8
Problems
1. The velocity of a particle at time t is
v(t) = 12t4 + 7/t
what is the total distance travelled between t = 0.2 and t = 0.3?
s =
Z
t2
vdt =
t1
Z
0.3
(12t4 + 7/t)dt
0.2
12 5
t + 7 ln t |0.3
0.2
5
12
12
5
5
=
(0.3) + 7 ln(0.3) −
(0.2) + 7 ln(0.2)
5
5
= 2.84.
=
2. How far will an object under earth’s gravity drop in 13 s, starting from rest and
neglecting air friction?
s = s o + vo t +
ao 2
1
m
t = (9.81 2 )(13 s)2
2
2
s
= 829 m.
3. The position of an object is given by
s = 4t3 + 2t2 − t + 3
What is the acceleration at t = 2 ?
ds
= 12t2 + 4t − 1
dt
dv
a =
= 24t + 4
dt
a(2) = 52.
v
=
4. An object weights 2 pounds is travelling in a circular path of radius 5 feet at a constant
speed of 10 feet per second. The acceleration of the object is most nearly:
Since the object is moving in a circular path and the speed along the path is given we
can use the normal-tangent coordinate system to get
aτ =
dv
,
dt
an =
v2
.
ρ
Since v = 10 ft/s is constant we get aτ = 0. Using ρ = 5 ft gives
an =
(10 ft/s)2
ft
= 20 2 .
5 ft
s
5. A projectile is launched from a level plane at 30◦ from the horizontal with an initial
velocity of 1250 m/s. (a) What is the maximum height above the plane the projectile
will reach? (b) What is the maximum range of the projectile at the maximum range
y = 0?
9
(a)
vy
y
vo
sin θ
g
v 2 sin2 θ
vo2 sin2 θ vo2 sin2 θ
−
= o
g
2g
2g
2
m 2
◦
(1250 s ) (sin 30 )
= 19.9 km.
m
)
(2)(9.81 sm2 )(1000 km
= vo sin θ − gt = 0 → t =
=
=
(b)
y
= 0 → t(vo sin θ − gt/2) = 0
2vo sin θ
therefore
t=
g
(vo )(2vo ) sin θ cos θ
v 2 sin 2θ
x =
= o
g
g
m 2
◦
(1250 s ) sin 60
.
= 137938 m = 138 km.
=
m
9.81 s2
6. The force F is gradually increased until the 20 kg block begins moving to the right.
The 5 kg block is prevented from moving by a cord. What is the minimum force F
for which movement is possible?
g
µ = 0.15
5 kg
F
µ = 0.3
20 kg
The free body diagram for thw system is as follows,
m1g
Fc
1
F1 = µ1m1g
N1
F1 = µ1m1g
2
m2g
F2 = µ2(m1+m2)g
F
N2
Since body 1 is in equilibrium we have
2
Fc = F1 = µ1 N1 = µ1 m1 g = (0.15)(5 kg)(9.81 m/s ) = 7.36 N.
10
If body 2 is in equilibrium then,
F
= F1 + F2
= 7.36 + (µ2 (m1 + m2 )g)
= 7.36 + (0.3)(25 kg)(9.81 m/s2 )
= 80.9 N.
7. What is the frictional force between the 80 kg block and the ramp. Given µs = 0.2,
and µf = 0.15.
g
80 kg
40
The free body diagram for the system is shown below.
Y
X
F1
θ = 40
mg
Summing the forces along the Y -axis gives,
X
FY = 0 → N = mg cos θ
N
N = (80)(9.81
m
) cos 40◦ = 601.2 N.
s2
First assume that the system is in static equilbrium. Then, the static friction force
along the X-axis satisfies
F1static = µs N = (0.2)(601.2) = 120.2 N.
The force due to gravity along the X-axis is
Fx = mg sin θ = (80)(9.81
m
) sin 40◦ = 504.5 N.
s2
Since, Fx is greater than F1static we can conclude that the block is sliding. Thus, the
sliding friction force is given by
Ff = µf N = (0.15)(601.2 N) = 90.17 N.
8. The slender homogeneous rod shown below has just been released from rest. If g is
the acceleration due to gravity, the magnitude of the angular acceleration of the rod
is:
11
L
g
The free body diagram below shows the reaction forces at the pivot (i.e., H and V ),
as well as the weight of the rod.
L/2
O
H
mg
V
Free body diagram
Since the pivot O is a fixed point on the rod the equation of motion is
MO = IO α,
where MO is the sum of the moments about O, IO is the moment of inertia about O,
and α is the angular acceleration. Here,
1
L
MO = mg , and IO = mL2 .
2
3
Therefore,
mg
L
2
=
α
=
1
mL2 α
3
3g
.
2L
9. A rocket is moving through a vacuum. It changes its velocity from 9, 020 m/s to
5, 100 m/s in 48 s. How much power is required to accomplish this if the rocket mass
is 213, 000 kg?
P
T1
P
dW
∆E
T1 − T 2
=
=
dt
∆t
∆t
1
1
2
=
mv , T2 = mv22
2 1
2
2
1
(213000 kg)[(9020)2 − (5100)2 ] ms2
= 2
48 s
.
= 122.8 × 109 W = 123 GW.
=
10. A 60 kg ball is dropped from a height at 48 m above a table. (a) What is the velocity
of the ball just before impact? (b) If e = 0.9, what is the kinetic energy of the ball
immediately after impact?
(a) Applying the Principle of Conservation of Energy gives,
T1 + U 1
= T2 + U 2
1
mgh =
mv 2
2 2
12
m
g
v2 =
p
2gh =
r
(2)(9.81
h
m
)(48 m) = 30.7 m/s.
s2
(b) The subscript of ‘b’ is related the ball, and the subscript of ‘g’ is related the
ground. From Newton’s collision rule we have
vb0 − vg0 = −e(vb − vg ).
Therefore, the velocity of the ball after impact is vb0 = −evb = −(0.9)(30.7) m/s.
Hence the kinetic energy after impact is
T =
1
1
2
m vb0 = ( )(60)(−(0.9)(30.7))2 = 22900 J = 22.9 kJ.
2
2
11. What is the component of velocity perpendicular to the wall after impact, if e = 0.8
and the velocity before impact has magnitude 50 m/s?
X
3
4
The X direction defines the common normal therefore, Newton’s collision rule gives
vx0
vx
vx0
= −evx
m
3
= 50 √
s 32 + 4 2
= −24 m/s.
12. A spring has a spring constant k = 10 N/cm. It is compressed 5 cm. The spring is
released and pushes against a free projectile with a mass 1 kg. What is the projectile
velocity immediately after losing contact with the spring?
Apply the Principle of Conservation of Energy to get,
T1 + U 1
1 2
kx
2
v
= T2 + U 2
1
=
mv 2
2
s
r
N
10 cm
100 cm
k
1m
m
= x
= 5 cm
= 1.5 m/s.
m
100 cm
1 kg
13
g
5 cm
m
k
13. A projectile is launched at 52◦ from the horizontal with an initial velocity of 3600 m/s.
If the mass of the projectile is 32 kg what is the total kinetic energy and potential
energy when t = 13 s? Neglect all forms of the friction.
Apply the Principle of Conservation of energy to get
E
T1
= T 1 + U 1 = T2 + U 2
1
m
1 MJ
1
= 207 MJ = E.
= T2 + U2 = mv 2 = (32 kg)(3600 )2
2
2
s 106 kg 2m2
s
14. The block rests at a distance of 2 m from the center of the rotating platform. If the
coefficient of static friction between the block and the platform is µs = 0.3, determine
the maximum speed the block can attain before it begins to slip.
2m
ω
g
mg
n^
µs N
N
Freebody diagram
In the freebody diagram n̂ denotes the normal direction of the normal-tangent coordinate system. (The tangent is directed out of the paper.) Applying Newton’s second
law in the z and n̂ directions gives
X
Fz = maz → N − mg = 0 → N = mg
(i)
v2
(ii)
ρ
In equation (ii) v is the velocity of the block tangent to the path of motion, and ρ is
the radius of curvature. Also, Fn̂ = µs N therefore, equation (ii) yields
X
Fn̂ = man̂ = m
v2
ρ
v2
(0.3)(m 9.81) = m
2p
→ v = (2)(0.3)(9.81) = 2.43 m/s
µs N
=
m
14
15. A pendulum bob is released from rest when it is at A.
Determine the speed of the bob and the tension in the
cord when the bob passes through its lowest position B.
The length of the cord is 1.5 m, and the bob has mass 2
kg.
Using the lowest point, B, as our reference we can apply
the Prinicple of Conservation of Energy at points A and B
to get
L = 1.5 m
A
g
m = 2 kg
B
TA + U A
= TB + U B
1
2
0 + mghA =
mvB
+0
2
1
2
(2)vB
+0
0 + (2)(9.81)(1.5) =
2
→
vB = 5.42 m/s
The figure on the left shows the free body diagram of the system when the bob passes through the point B. Here, T denotes
n
the tension in the cord and mg is the weight of the bob.
Using the normal-tangent coordinate system Newton’s second
T
law, in the normal direction, gives
X
B
Fn = man
τ
2
vB
.
ρ
Free body diagram
Here, the radius of curvature is ρ = L = 1.5 m. Therefore, the tension in the cord at
B is
v2
(5.42)2
= 58.79N.
T = mg + m B = (2)(9.81) + (2)
ρ
1.5
mg
T − mg
= m
16. Disk A weights 2 lb and is sliding on a smooth horizontal plane with a velocity of 3
ft/s. Disk B weights 11 lb and is initially at rest. If after the impact A has a velocity
of 1 ft/s directed along the positive x axis, determine the speed of B after impact.
Apply the Priniciple of Conservation of Linear Momentum in
the x-direction to get
y
B
mA vAx1 + mB vBx1
= mA vAx2 + mB vBx2
2
11
vBx2
0+0 =
(1) +
32.2
32.2
→
vBx2 = −0.1818 ft/s.
x
3 ft/s
A
Apply the Priniciple of Conservation of Linear Momentum in the y-direction to get
mA vAy1 + mB vBy1 = mA vAy2 + mB vBy2
2
11
(3) + 0 = 0 +
vBy2
32.2
32.2
→
vBy2 = 0.545 ft/s.
Therefore the speed of B after impact is
p
|vB2 | = (−0.1818)2 + (0.545)2 = 0.575 ft/s.
15
17. The robot arm is programmed so that point P follows the path r = 1 − 0.5 cos 2πt
m, θ = 0.5 − 0.2 sin 2πt radians. What is the velocity of P at time t = 0.8 s.
Using a polar coordinate system, we have from kinematics
that
y
P
r
v̄
θ
= ṙêr + rθ̇êθ
dr
=
= π sin 2πt
dt
dθ
=
= −0.4π cos 2πt
dt
ṙ
x
θ̇
Therefore,
v̄ = (π sin 2πt)êr + (1 − 0.5 cos 2πt)(−0.4π cos 2πt)êθ
At t = 0.8 s, we get
v̄ = −2.99 êr − 0.328 êθ m/s.
18. A motorcycle starts at t = 0 on a circular track of radius ρ = 400 m. The tangential
component of the acceleration is aτ = 2 + 0.2t m/s2 . Determine the magnitude of the
acceleration at t = 10 s.
Using the normal-tangent coordinate system we have
s
aτ
ρ
Z
2
Z
=
dv
dt
aτ dt = v
(2 + 0.2t) dt = v
Therefore, v = 2t + 0.1t m/s. In addition, the acceleration vector is
ā =
v2
n̂ + aτ τ̂ .
ρ
At t = 10 s we get v = (2)(10) + 0.1(102 ) = 30 m/s, v 2 /ρ = (302 )/400 = 2.25 m/s2 ,
and aτ = 2 + (0.2)(10) = 4 m/s2 . Therefore,
p
kāk = 42 + 2.252 = 4.59 m/s2 .
16