Transcript Mechanics 105 - Georgetown University

Mechanics 105
More applications of Newton’s laws (chapter five)
Forces of friction (static, kinetic)
Uniform circular motion
Nonuniform circular motion
Velocity dependent forces
Numerical methods
Fundamental forces
Gravitational field
Mechanics 105
Friction
Force acting parallel to an interface that
opposes the relative motion
Static – frictional force opposite to applied
force - magnitude fs
fs  sn
where s is the coefficient of static friction
and n is the magnitude of the normal
forces between the surfaces
the equality holds just as the object starts
to slip
Mechanics 105
Friction
Kinetic – frictional force opposite to
relative motion – magnitude fk
fk  k n
where k is the coefficient of static friction
and n is the magnitude of the normal
forces between the surfaces
the kinetic frictional force is constant
s and k are constants that depend on
the nature of the surfaces
Usually, s > k
Mechanics 105
Friction
force of friction (N)
Note: Static friction is not constant – it is whatever is needed to match
the applied force, up to the limit of Sn
static friction:
fs=applied force
kinetic friction:
fs=constant
applied force (N)
As the applied force increases, the static frictional force also
increases, until the limit, then the object begins to slide, and the
frictional force goes to a constant value
Mechanics 105
Friction

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
ConcepTest
Examples
Demo
Mechanics 105
Example
+y
m2
+x

T
m1
m1

m1 g

N

T
m2

m2 g


f s or f k
Mechanics 105
Example
m1 :
( y ) T  m1 g   m1a
m2 :
( x)  T  f s ,k  m2 a
( y ) N-m2 g  0
For the static case (a  0) :
T  m1 g  f s
For the kinetic case (a  0) :
f k   k N   k m2 g
T   k m2 g  m2 a  m1 g  m1a
Mechanics 105
Question
What do you call a broken boomerang?
Mechanics 105
Question
What do you call a broken boomerang?
Answer: A stick.
Mechanics 105
Newton’s 2nd law applied to uniform circular motion
A mass in uniform circular motion (speed v) accelerates
v2
according to
ac 
r
This acceleration must be caused by some force
v2
 F  mac  m r
along a direction towards the center of the radius of
curvature (r)
Mechanics 105
Example: conical pendulum
y:
L

T
y

r

mg
F
T cos

mg
T cos  mg  0

T
T cos  mg
x:
T sin 
v2
 Fx T sin   m r
Combining the two equations :
v2
tan  
 v  rg tan 
rg
Mechanics 105
Uniform circular motion
ConcepTest
Mechanics 105
Nonuniform circular motion
If an object changes its speed while in circular motion, there is both a
radial and a tangential component to the acceleration, therefore,
there will be a radial and tangential force applied.
Example: mass moving in a vertical circle

T

T

mg

mg
R


T

mg
 F  ma
t
t
 mg sin 
v2
 Fr  mar  m R  T  mg cos
From the second equation, we get
 v2

T  m  g cos 
R

Mechanics 105
Words of wisdom

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"If I had only known, I would have been a locksmith."
-Albert Einstein
"There is no clearer manifestation of pure evil than teachers giving
assignments over holiday breaks."
-James Halloran
Mechanics 105
Velocity dependent forces
Two models:
1. Force proprtional to the velocity (viscous, low speed)


R  bv
b is a constant that depends
on the object size and shape
and the medium
2. Force proportional to the square of the magnitude of
the velocity (air, high speed)
D: drag coefficient
1
2
R  DAv
: density of air
2
A: cross sectional area of object
Mechanics 105
Velocity dependent forces
1. Force proprtional to the velocity
In the direction of the motion,
dv
F

ma

mg

bv

m

dt
This can be written
mv(t )  bv(t )  mg  0
or
mx(t )  bx (t )  mg  0

 bv

mg
Mechanics 105
Velocity dependent forces
Can solve differential equation
t
mg
v(t ) 
(1  e  )
b
where  = m/b is a time constant related to
the motion
Or, just find terminal speed (a=0)
mg  bv  0
mg
vT 
b
Mechanics 105
Velocity dependent forces
Force proportional to the square of the
magnitude of the velocity
In the direction of the motion,
1
2
F

ma

mg

D

Av
 ma

2
Nonlinear differential equation
Terminal speed:
1
mg  DAv 2  0  vT 
2
2mg
DA
Mechanics 105
Words of wisdom



"I love deadlines. I like the whooshing sound they make as they fly
by."
-Douglas Adams
"In a survey taken several years ago, all incoming freshman at
MIT were asked if they expected to graduate in the top half of their
class. Ninety-seven percent responded that they did."
-???
"We made too many wrong mistakes."
-Yogi Bera
Mechanics 105
Numerical representations of particle dynamics
Euler method
v(t  t )  v(t )  a(t )t
x(t  t )  x(t )  v(t )t
Mechanics 105
Fundamental forces of nature
Gravitational: force between any two objects

mm
Fg  G 1 2 2 rˆ where G is the universal gravitational constant
r
Electromagnetic: force between two charged objects (q)

qq
Fe  k e 1 2 2 rˆ where ke is the Coulomb constant
r
Nuclear (strong) – short range
Weak – short range
Mechanics 105
Gravitational field
Field: the effect in a region of space that induces a force on an object

g

Fg
GM E
  2 rˆ
m
r
e.g the field (created by a mass) exerts the force on the other masses
Mechanics 105
Last bad joke for this chapter
An atom walking down the street says to its friend “I think I lost an
electron. The friend asks “Are you sure?” to which the first atom
repiles “Yea, I’m positive.”