Homework: 49, 51, 70 (p. 134-137)
49. In the figure below, a car is driven at constant speed over a
circular hill and then into a circular valley with the same radius. At
the top of the hill, the normal force on the driver from the car seat
is 0. The driver’s mass is 80.0 kg. What is the magnitude of the
normal on the driver from the seat when the car passes through the
bottom of the valley?
FN
At the top of the hill:
FN
v2
Fcentripetal = Fg - FN = m
R
2
v
FN = 0 Þ Fg = m
R
At the bottom of the valley:
Fg
v2
v2
Fcentripetal = FN - Fg = m
Þ FN = m
+ Fg
R
R
FN = 2Fg = 2mg = 2 ´ 80 ´ 9.8 = 1568 (N)
Fg
51. An airplane is flying in a horizontal circle at a speed of 600 km/h.
If its wings are tilted at angle q=400 to the horizontal, what is the
radius of the circle in which the plane is flying? Assume that the
required force is provided entirely by an “aerodynamic lift” that is
perpendicular to the wing surface.
Animation
•According to the Bernoulli’s principle,
the aerodynamic lift appears due to the
air-stream velocity over the top of the
airplane greater than that at the bottom.
Fl,y
Fl
q
Fl,x
Fl, y = Fg
Fl cos q = mg (1)
Fl,x is the
2
Fg
v
centripetal force:
Fl, x = m
R
2
v
Fl sin q = m (2)
R
v = 600 km/h = 166.7 m/s
v2
166.7 2
(1) and (2) Þ R =
=
» 3379 (m) or 3.38 (km)
g tan q 9.8 tan(40)
70. The figure below shows a conical pendulum, in which the bob (the
small object at the lower end of the cord) moves in a horizontal circle
at constant speed. The cord sweeps out a cone as the bob rotates.
The bob has a mass of 0.050 kg, the string has length L=0.90 m and
negligible mass, and the bob follows a circular path of circumference
0.94 m. What are (a) the tension in the string and (b) the period of
the motion?
Animation
Ty - Fg = 0 Þ Tsinq = mg
Tx: the centripetal force
q: the angle between the cord and the
horizontal circle.
2
2
v
v
Tx = m Þ T cos q = m
R
R
R
C 0.94 (m)
cos q = ; R =
=
= 0.15 m
L
2π 2 ´ 3.14
R
q = arccosæç ö÷ = 80.40
L ø
è
mg
Ty
T
q
Tx
Fg
C
T ´ Rcosθ
(a) T =
= 0.5 (m/s); P = = 1.88 (s)
» 0.5 (N) (b) v =
m
v
sinq
Part B Laws of Conservation
Chapter 3 Work and Mechanical Energy
3.1. Kinetic Energy and Work. Power
3.2. Work-Kinetic Energy Theorem
3.3. Work and Potential Energy
3.4. Conservative and Non-conservative Forces. Conservative Forces
and Potential Energy
3.5. Conservation of Mechanical Energy
3.6. Work Done on a System by an External Force. Conservation of
Energy
What is energy?
Energy is the capacity of a system to do work.
There are many forms of energy:
• Mechanical energy:
•Potential energy, stored in a system
•Kinetic energy, from the movement of matter
• Radiant energy (solar energy)
• Thermal energy
• Chemical energy (chemical bonds of molecules)
• Electrical energy (movement of electrons)
• Electromagnetic energy (from X-rays to radio waves)
• Nuclear energy
Energy can be transformed from one type to another and
transferred from one object to another, but the total amount is
always the same (the principle of energy conservation).
What is energy?
Energy is the capacity of a system to do work.
There are many forms of energy:
• Mechanical energy:
•Potential energy, stored in a system
•Kinetic energy, from the movement of matter
• Radiant energy (solar energy)
• Thermal energy
• Chemical energy (chemical bonds of molecules)
• Electrical energy (movement of electrons)
• Electromagnetic energy (from X-rays to radio waves)
• Nuclear energy
Energy can be transformed from one type to another and
transferred from one object to another, but the total amount is
always the same (the principle of energy conservation).
3.1. Kinetic Energy and Work. Power
3.1.1. Kinetic Energy and Work
Kinetic energy is energy associated with the state of motion of an
object.
1
2
K = mv
2
1 joule = 1 J = 1 kg. m2/s2
Work is energy transferred to or from an object by means of a force
acting on the object. Energy transferred to the object is positive work
and energy transferred from the object is negative work.
Unit: joule
3.1.2. Work done by a force
A. Work done by a constant force:
•To establish an expression
for work, we consider a constant force
F that accelerates a bead along a wire:
Fx = ma x
2
2
v = v 0 + 2a x d
1
1
2
2
Þ mv - mv0 = Fx d
2
2
Therefore, the work W done on the bead by F is:
W = Fx d = Fd cosf
!!
W = F d (work done by a constant force)
B. Work done by a general variable force:
One-dimensional analysis:
•Choose Dx small enough, work done by the
force in the jth interval:
ΔWj = Fj,avg Dx
•The total work:
W = å DWj =å Fj,avg Dx
W = lim å Fj,avg Dx
Dx ®0
xf
W = ò F(x)dx (work done by a variable force)
xi
Three-dimensional analysis:
! !
dW = Fd r = Fx dx + Fy dy + Fz dz
rf
xf
yf
zf
ri
xi
yi
zi
W = ò dW = ò Fx dx + ò Fy dy + ò Fz dz
3.1.3. Power
Power is the rate at which work is done.
Average power:
Instantaneous power:
Unit: watt (W)
Pavg
W
=
Dt
dW
P=
dt
1 watt = 1 W = 1J/s
1 horsepower = 1 hp = 550 ft.lb/s = 746 W
1 kilowatt - hour = 1kW.h = (10 W)(3600s) = 3.6 ´10 J = 3.6 MJ
3
dW Fcosfdx
æ dx ö
P=
=
= Fcosf ç ÷
dt
dt
è dt ø
P = Fvcosf
Instantaneous power:
!!
P = Fv
F=constant:
6
3.2. Work-Kinetic Energy Theorem
Let DK be the change in the kinetic
energy of the bead.
ΔK = Kf - Ki = W
This can be read as follows:
æ change in the kinetic
çç
è energy of an object
or
ö æ net work done on ö
÷÷ = çç
÷÷
ø è the object
ø
K f = Ki + W
æ kinetic energy after ö æ kinetic energy
ö æ the net
ö
çç
÷÷ = çç
÷÷ + çç
÷÷
è the net work is done ø è before the net work ø è work done ø
Examples:
E1. Work done by the gravitational force:
W = Fx d = Fd cosf
For a rising object, F=1800:
W = -mgd
For a falling object, F=00:
Kf = Ki = 0 Þ
W = + mgd
Work done in lifting and lowering an object
Gravity and an applied force acting on the
object:
ΔK = K f - K i = Wa + Wg
where Wa is the work done by the
applied force; Wg is the work done by the
gravitational force. If initial and final velocities
are zero:
Kf = Ki = 0 Þ ΔK = 0 Þ Wa = - Wg
Wa = -mgd cosf
E2. Work done by a spring force:
The spring force is computed by:
!
!
Fs = -kd (Hooke' s law)
k: the spring constant (or force constant)
Is an x axis is parallel to the length of
the spring:
Fx = -kx (Hooke's law)
èA spring force is a variable force F=F(x)
To find the work done by the spring force,
We have to make assumptions: (1) the
spring is massless; (2) It is an ideal spring
(it exactly obeys Hooke’s law).
! !
Ws = lim å FxjΔx = lim å FxjΔx cos q = lim å - FxjΔx
Dx®0
Dx®0
Dx®0
Note: q = 1800 and Fxj is the magnitude of the spring force
Ws = òxx f - Fx dx = òxx f - kxdx
i
i
1 2 1 2
Ws = kx i - kx f
2
2
1 2
If x i = 0, x f = x : Ws = - kx
2
Work done by an applied force:
ΔK = K f - K i = Wa + Ws
If the block is stationary before and after the displacement, DK=0:
Wa = -Ws
Homework: 1, 2, 8, 15, 24, 26, 29, 36, 43 (p. 159-163)
Chapter 1:
Review
(All sections of Chapter 1, 2)
Motion in one dimension:
To describe motion, we need to measure:
+ Displacement: Dx = xt – x0 (measured in m or cm)
+ Time interval: Dt = t – t0 (measured in s)
Average velocity:
total distance
Average speed: s avg =
Δt
Δx(t) dx(t)
=
Instantaneous velocity: v(t) = lim
Δt ®0 Δt
dt
Δv v 2 - v1
=
Average acceleration: a avg =
Δt
t 2 - t1
Instantaneous acceleration:
dv(t) d 2 x
a(t) =
= 2
dt
dt
Two basic equations for constant acceleration:
For freely falling objects:
Motion in two dimensions:
a = g = 9.8 (m/s 2 )
Projectile motion:
• Ox: Horizontal motion (no acceleration):
v x = v0cosθ0 = constant
x = x 0 + v0 cosθ0 t
• Oy: Vertical motion (free fall)
v y = v0sinθ 0 - gt
1 2
y = y 0 + v 0 sinθ 0 t - gt
2
•Horizontal range:
R=
2
v0
g
sin2θ 0
Uniform Circular Motion:
The particle is accelerating with a
centripetal acceleration:
2
v
a=
r
Where r is the radius of the circle
v the speed of the particle
2pr
T=
v
(T: period)
Relative Velocity and Relative Acceleration:
v PA = v PB + v BA
!
!
!
v PA = v PB + v BA
Chapter 2:
Newton’s Laws
n !
!
F = 0 or å Fi = 0
i =1
!
!
Fnet = ma
!
!
FBC = - FCB
Some particular forces: gravitational, normal, tension and
frictional forces
Friction and Properties of Friction:
f s,max = µ s FN
f k = µ k FN
µ s is the coefficient of static friction
µ k is the coefficient of kinetic friction
Uniform Circular Motion and Non-uniform Circular Motion:
• Uniform circular motion:
• Non-uniform circular motion: