Transcript Physics 207: Lecture 2 Notes

Lecture 11

Goals
 Problem solving with Newton’s 1st, 2nd and 3rd Laws
 Forces in circular motion
 Introduce concept of work
 Introduce dot product
Physics 201: Lecture 11, Pg 1
Loop-the-loop 1
The match box car is going to do a loop-the-loop. If the
speed at the bottom is vB, what is the normal force, N, at
that point?
Hint: The car is constrained to the track.
Fr = mar = mvB2/r = N - mg
N = mvB2/r + mg
N
vB
mg
Physics 201: Lecture 11, Pg 2
Another example of circular motion
Loop-the-loop 2
A match box car is going to do a loop-the-loop
of radius r.
What must be its minimum speed vt at the top
so that it can manage the loop successfully ?
Physics 201: Lecture 11, Pg 3
Loop-the-loop 2
To navigate the top of the circle its tangential
velocity vt must be such that its centripetal
acceleration at least equals the force due to
gravity. At this point N, the normal force,
goes to zero (just touching).
Fr = mar = mg = mvt2/r
vt
vt = (gr)1/2
mg
Physics 201: Lecture 11, Pg 4
Loop-the-loop 3
Once again the car is going to execute a loop-the-loop.
What must be its minimum speed at the bottom so
that it can make the loop successfully?
This is a difficult problem to solve using just forces. We
will skip it now and revisit it using energy
considerations later on…
Physics 201: Lecture 11, Pg 5
Orbiting satellites
Net Force:
mar = mg = mvt2 / r
gr = vt2
vt  gr
The only difference is
that g is less because
you are further from
the Earth’s center!
Physics 201: Lecture 11, Pg 6
Example, Circular Motion Forces with Friction
(mar = m vt 2 / r Ff ≤ ms N )

How fast can the race car go?
(How fast can it round a corner with this radius of curvature?)
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2
r
Physics 201: Lecture 11, Pg 7
Navigating a curve

Only one force is in the horizontal direction: static friction
x-dir: Fr = mar = -m vt
2/
y-dir: ma = 0 = N – mg
r = Fs = -ms N (at maximum)
N
N = mg
vt = (ms m g r / m )1/2
vt = (ms g r )1/2 = (0.5 x 10 x 80)1/2
vt = 20 m/s (or 45 mph)
y
Fs
mg
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2
Physics 201: Lecture 11, Pg 8
x
Banked Curves
In the previous car scenario, we drew the following free
body diagram for a race car going around a curve on a
flat track.
n
Ff
mg
What differs on a banked curve?
Physics 201: Lecture 11, Pg 11
Banked Curves
Free Body Diagram for a banked curve.
(rotated x-y coordinates)
Resolve into components parallel and
perpendicular to bank
y
x
N
Ff
No speed
q
mg
Physics 201: Lecture 11, Pg 12
Banked Curves
Free Body Diagram for a banked curve.
(rotated x-y coordinates)
Resolve into components parallel and
perpendicular to bank
y
x
N
mar
Ff
Low speed
q
mg
N
mar
High speed
Ff
q
mg
Physics 201: Lecture 11, Pg 13
Hanging Pink Fuzzy Dice Problem
You are in a car going around a horizontal curve
of radius 40.0 m and a speed of 10 m/s. There is a
0.10 kg pink fuzzy dice at the end of a 0.10 m string.
What is the angle of the die with respect to vertical?
(Little g is 10 m/s2)

x-dir SFx = mar = -m vT2 / r = -Tx = -0.1 x 2.5= - 0.25 N
y-dir SFy = may = 0 = Ty - mg  Ty = 0.1 x 10= 1.0 N
tan q = -0.25  q = 14°
But to you, if you are not paying attention, it may
look like there is a mysterious force pushing the
die out….a so-called centrifugal (center fleeing)
or fictitious force.
Observers in accelerated frames of reference do
not see proper physics and so come to
erroneous conclusions.
T
Physics 201: Lecture 11, Pg 17
mg
Drag forces (forces that oppose motion)

Serway & Jewett describe three models:

R0   f friction vˆ


R1  bv

2
2
1
R2  b' v vˆ   2 DAv vˆ


The velocity dependent model give a general analytic
solution.
Terminal velocity for velocity dependent drag forces
occur where drag force equals applied force
Fnet  0  mg  bvT
vT  mg / b
Physics 201: Lecture 11, Pg 18
Drag at high velocities in a viscous medium

With a cross sectional area, A (in m2), D coefficient of
drag (0.5 to 2.0),  density, and velocity, v (m/s), the
drag force is:
R = ½ D  A v2
Example: Bicycling at 10 m/s (22 m.p.h.), with projected
area of 0.5 m2 and drag coefficient 1 exerts a force of
~30 Newtons
 At low speeds air drag is proportional to v but at high
speeds it is v2
 Minimizing drag is often important
Physics 201: Lecture 11, Pg 19
Energy & Work

Work (Force over a distance) describes energy transfer

No “working” definition for energy….yet

Net forces result in acceleration

Velocity can change direction, magnitude or both

Only net force acting along the path changes the speed

Forces acting over a distance  work….energy changes

Forces acting over a time  impulse…..momentum changes
Physics 201: Lecture 11, Pg 20
Perpendicular Forces





I swing a sling shot over my head. The tension in the rope
keeps the shot moving at constant speed in a circle.
Is there a net force?
Does the force act over a distance?
What if there were a tangential force….what would happen?
Only parallel forces change speed
v
Fc
Physics 201: Lecture 11, Pg 21
Motion along a line

The only net force is along the horizontal
Fy net = 0
F
Fx
Start
F Finish
mg
x



A net force acting along the path, over a distance, induces
changes in speed (magnitude of the velocity)
We call this action “work”
The vector “dot” product simplifies the notation
Physics 201: Lecture 11, Pg 22
Scalar Product (or Dot Product)
   
A  B  A B cosq

Useful for finding parallel components
A  î = Ax
îî=1
îj=0

A
q Ay
î Ax
Calculation can be made in terms of
components.
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
Calculation also in terms of magnitudes and relative angles.
A  B ≡ | A | | B | cos q
You choose the way that works best for you!
Physics 201: Lecture 11, Pg 23
Scalar Product (or Dot Product)
Compare:
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
with A as force F, B as displacement r
Notice if force is constant:
F  r = (Fx )(x) + (Fy )(z ) + (Fz )(z)
Fx x +Fy y + Fz z
So here
Fnet  r ≡ Wnet
A Parallel Force acting Over a Distance does “Work”
Physics 201: Lecture 11, Pg 24
Units:
Force x Distance = Work
Newton x Meter = Joule
[M][L] / [T]2 [L]
[M][L]2 / [T]2
mks
N-m or Joule
cgs
Dyne-cm or erg
= 10-7 J
Other
BTU = 1054 J
calorie= 4.184 J
foot-lb = 1.356 J
eV
= 1.6x10-19 J
Physics 201: Lecture 11, Pg 25
Circular Motion


I swing a sling shot over my head. The tension in the rope
keeps the shot moving at constant speed in a circle.
How much work is done after the ball makes one full revolution?
(A) W > 0
(B) W = 0
v
Fc
(C) W < 0
(D) need more info
Physics 201: Lecture 11, Pg 26
Infinitesimal Work vs speed along a linear path
Let dW  Fx dx
F
Start
F Finish
Fx
 ma x dx
m
dvx
dt
x
dx
 m dvx dxdt  m dvx vx
Physics 201: Lecture 11, Pg 28
Work vs speed along a linear path
dW  Fx dx  m vx dvx
xf
vxf
xi
 If F is constant
vxi
Wnet   Fx dx   mvx dvx
xf
vxf
xi
vxi
Wnet  Fx  dx m  vx dvx
Fx ( x f  xi )  Fx x 

1
2
2
mv xf

1
2
2
mv xi
 K
K is defined to be the change in the kinetic energy
Physics 201: Lecture 11, Pg 29
Work in 3D….

x, y and z with constant F:

Fy ( y f  yi )  Fy y 
1
2
mv xf
2
1
2
mv yf
2
Fx ( x f  xi )  Fx x 
Fz ( z f  zi )  Fz z 
1
2

Fx x  Fy y  Fz z 
with v
2
2
 vx

2
vy

2
vz
1
2
2
mv zf
2
mv f


1
2
1
2
mv xi
2
1
2
mv yi
2
1
2
mv zi
2
2
mvi
 K
 
 v v
Physics 201: Lecture 11, Pg 30
Examples of “Net” Work
(Wnet)
K = Wnet

Pushing a box on a smooth floor with a constant force;
there is an increase in the kinetic energy
Examples of No “Net” Work
Pushing a box on a rough floor at constant speed
 Driving at constant speed in a horizontal circle
 Holding a book at constant height
This last statement reflects what we call the “system”
( Dropping a book is more complicated because it involves
changes in U and K, U is transferred to K. The answer
depends on what we call the system. )

Physics 201: Lecture 11, Pg 31
Net Work: 1-D Example
(constant force)

A force F = 10 N pushes a box across a frictionless floor
for a distance x = 5 m.
Finish
Start
q = 0°
F
x



Net Work is F x = 10 x 5 N m = 50 J
1 Nm ≡ 1 Joule and this is a unit of energy
Work reflects energy transfer
Physics 201: Lecture 11, Pg 32
Net Work: 1-D 2nd Example
(constant force)

A force F = 10 N is opposite the motion of a box across a
frictionless floor for a distance x = 5 m.
Finish
Start
F
q = 180°
x

Net Work is F x = -10 x 5 N m = -50 J

Work again reflects energy transfer
Physics 201: Lecture 11, Pg 33
Work: “2-D” Example
(constant force)

An angled force, F = 10 N, pushes a box across a
frictionless floor for a distance x = 5 m and y = 0 m
Start
F
Finish
q = -45°
Fx
x

(Net) Work is Fx x = F cos(-45°) x = 50 x 0.71 Nm =
35 J

Work reflects energy transfer
Physics 201: Lecture 11, Pg 34
Work and Varying Forces (1D)

Area = Fx x
F is increasing
Here W = F · r
becomes dW = F dx
xf
Consider a varying force F(x)
Fx
W
  F ( x ) dx
x
x
xi
Finish
Start
F
F
q = 0°
x
Work has units of energy and is a scalar!
Physics 201: Lecture 11, Pg 35
Fini

Read all of Chapter 7
Physics 201: Lecture 11, Pg 36