Transcript Transparancies for Motion & Newton`s Laws

EE1 Particle Kinematics :
Newton’s Legacy
"If I see further, it is because
I stand on the shoulders of giants,"
Motion
Forces
Energy & Momentum
Conservation
Circular Motion
Gravity
October 2005
http://ppewww.ph.gla.ac.uk/~parkes/teaching/PK/PK.html
Chris Parkes
Motion
• Position [m]
• Velocity [ms-1]
x
e.g
dx
dx
v
dt
– Rate of change of position 0
2
• Acceleration [ms-2] a  dv  d
dt
– Rate of change of velocity
dt
x
v
2
0
a
0
dt
t
t
Equations of motion in 1D
– Initially (t=0) at x0
– Initial velocity u,
– acceleration a,
x  x0  ut  at
1
2
Differentiate w.r.t. time:
dx
 v  u  at
dt
2
d x
aa
2
dt
s=ut+1/2 at2,
2
where s is displacement from
initial position
v=u+at
v 2  (u  at ) 2  u 2  2uat  a 2t 2
v 2  u 2  2a(ut  12 at 2 )
v2=u2+2 as
2D motion: vector quantities
• Position is a vector
– r, (x,y) or (r,  )
– Cartesian or
cylindrical polar coordinates
– For 3D would specify
z also
Scalar: 1 number
Vector: magnitude & direction,
>1 number
Y
• Right angle triangle
x=r cos , y=r sin 
r2=x2+y2, tan  = y/x
r
0

x
y
X
vector addition
• c=a+b
y
cx= ax +bx
cy= ay +by
b
can use unit vectors i,j
a
c
i vector length 1 in x direction
x
j vector length 1 in y direction
scalar product
finding the angle between two vectors
a  b  ab cos   a x bx  a y by a,b, lengths of a,b
Result is a scalar
a xbx  a y by
a b
cos 

2
2
2
2
ab
a x  a y  bx  by
a

b
Velocity and acceleration vectors
• Position changes with time
Y
• Rate of change of r is
velocity
– How much is the change in a
very small amount of time t
d r r (t  t )  r (t ) Limit at t0
v

dt
t
0
yd
xd
 yv ,
 xv
td
td
d v v(t  t )  v(t ) d 2 r
a

 2
dt
t
dt
vd
y vd
 ya , x
 xa
td
td
r(t)
x
r(t+t)
X
We described the motion, position, velocity, acceleration,
now look at the underlying causes
Newton’s laws
• First Law
– A body continues in a state of rest or uniform
motion unless there are forces acting on it.
• No external force means no change in velocity
• Second Law
– A net force F acting on a body of mass m [kg]
produces an acceleration a = F /m [ms-2]
• Relates motion to its cause
F = ma
units of F: kg.m.s-2, called Newtons [N]
• Third Law
– The force exerted by A on B is equal and
opposite to the forceFexerted by B on A
b
•Force exerted by
block on table is Fa
Block on table
Fa=-Fb
Weight
Fa
(a Force)
Examples of Forces
For this course:
weight of body from gravity (mg),
tension, compression
Friction,
•Force exerted by
table on block is Fb
(Both equal to weight)
Tension & Compression
• Tension
– Pulling force - flexible or rigid
• String, rope, chain and bars
mg
• Compression
– Pushing force
• Bars
mg
mg
• Tension & compression act in BOTH
directions.
– Imagine string cut
– Two equal & opposite forces – the tension
Friction
• A contact force resisting sliding
– Origin is chemical forces between atoms in the two
surfaces.
• Static Friction (fs)
– Must be overcome before an objects starts to move
• Kinetic Friction (fk)
– The resisting force once sliding has started
• does not depend on speed
N
fs or fk
F
mg
fs  s N
fk  k N
•
Questions
Topics Covered
From Benson, University
Physics, Revised Addition
– Vector addition and dot product, descriptions of motion, Newton’s 3 laws, Friction.
Chapter
Page
Exercise
Topic
2
27
9
Vector addition
3
50
12
Motion
3
51
34
Motion
5
96
19
Newton’s 2nd law
Chapter
Page
Exercise
Topic
2
28
26
Vector addition
2
29
46
Vector dot product
3
50
20
Motion
3
52
42
Motion
3
53
62
Motion
5
96
25
Newton’s 2nd law
6
119
4
Friction
6
119
10
Friction
6
124
3
Newton’s laws