Transcript Physics 207: Lecture 2 Notes

Lecture 4

Goals for Chapter 3 & 4
 Perform vector algebra
• (addition & subtraction) graphically or by xyz components
• Interconvert between Cartesian and Polar coordinates
 Work with 2D motion
• Distinguish position-time graphs from particle trajectory plots
• Trajectories
 Obtain velocities
 Acceleration: Deduce components parallel and
perpendicular to the trajectory path
 Solve classic problems with acceleration(s) in 2D
(including linear, projectile and circular motion)
 Discern different reference frames and understand how
they relate to motion in stationary and moving frames
Assignment: Read thru Chapter 5.4
MP Problem Set 2 due this Wednesday
Physics 207: Lecture 4, Pg 1
Example of a 1D motion problem



A cart is initially traveling East at a constant speed of
20 m/s. When it is halfway (in distance) to its destination
its speed suddenly increases and thereafter remains
constant. All told the cart spends a total of 10 s in transit
with an average speed of 25 m/s.
What is the speed of the cart during the 2nd half of the trip?
Dynamical relationships (only if constant acceleration):
x  x0  v x t  a x t
0
v x  v x  a x t
0
a x  const
And
1
2
2
v 2x  v x2  2a x (x  x 0 )
0
v x (avg)
1
 (v x  v x )
2
0
x(displaceme nt )
vaverage velocity  
t ( total time )
Physics 207: Lecture 4, Pg 2
The picture
x0


t0
v0
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
x1 t1
x
x x2  x0
v

t t2  t0
t2
2
Plus the average velocity
Knowns:
 x0 = 0 m
 t0 = 0 s
0
x
 v0 = 20 m/s
vx  vx
 t2 = 10 s
 vavg = 25 m/s
ax  0
 relationship between x1 and x2
Four unknowns x1 v1 t1 & x2 and must find v1 in terms of knowns
x  x  v t
0
0

Physics 207: Lecture 4, Pg 3
x  x0  v x t
Using
0
x0
v0
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
t0
x1  x0  v0 (t1  t0 )
Four
unknowns
 Four
relationships
x1 t1
x
x2  x1  v1 (t2  t1 )
t2
2

x x2  x0
v

t t2  t0
x1  ( x2  x0 )
1
2
Physics 207: Lecture 4, Pg 4
x0  0
Using
x0
v0
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
t0
x1 t1
1

t0  0
x1  v0 t1
Eliminate
unknowns
3
x2  x1  v1 (t2  t1 )
x1  x2
1
2
4
x2  12 x2  v1 (t2  t1 )
first x1
2&3
next t1
2
x
1
1
2
x2  v1 (t2  )
4
1
2
vt2  v1 (t2  )
then x2
x2
v0
t2
2
x2
v
t2
1
2
1
2
vt2
v0
Mult. by 2/ t2
v  v1 (2  vv )
0
Physics 207: Lecture 4, Pg 5
Fini
x0


v0
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
t0
Plus the average velocity
Given:
 v0 = 20 m/s
 t2 = 10 s
 vavg = 25 m/s
x1 t1
x
v  v1 (2  vv )
t2
2
0
v1 
v v0
2 v0  v
v1  225 20m/sm/s2025m/sm/s
 50015m/s
 33.3 m/s
Physics 207: Lecture 4, Pg 6
Vectors and 2D vector addition

The sum of two vectors is another vector.
B
A =B+C
B
A
C
C
D = B + 2 C ???
Physics 207: Lecture 4, Pg 7
2D Vector subtraction

Vector subtraction can be defined in terms of addition.
B-C
= B + (-1)C
B
B-C
-C
C
B
A =B+C
A
Different direction
and magnitude !
Physics 207: Lecture 4, Pg 8
Reference vectors: Unit Vectors



A Unit Vector is a vector having length 1
and no units
It is used to specify a direction.
Unit vector u points in the direction of U
Often denoted with a “hat”: u = û
Useful examples are the
cartesian unit vectors [ i, j, k ]
 Point in the direction of the
x, y and z axes.
R = rx i + ry j + rz k
U = |U| û
û
y

j
k
i
x
z
Physics 207: Lecture 4, Pg 9
Vector addition using components:
Consider, in 2D, C = A + B.
(a) C = (Ax i + Ay j ) + (Bx i + By j ) = (Ax + Bx )i + (Ay + By )
(b) C = (Cx i + Cy j )


Comparing components of (a) and (b):
 Cx = Ax + Bx
 Cy = Ay + By
 |C| =[ (Cx)2+ (Cy)2 ]1/2
C
B
A
Ay
By
Bx
Ax
Physics 207: Lecture 4, Pg 10
Example
Vector Addition


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Vector A = {0,2,1}
Vector B = {3,0,2}
Vector C = {1,-4,2}
What is the resultant vector, D, from adding A+B+C?
A.
B.
C.
D.
{3,-4,2}
{4,-2,5}
{5,-2,4}
None of the above
Physics 207: Lecture 4, Pg 11
Example
Vector Addition



Vector A = {0,2,1}
Vector B = {3,0,2}
Vector C = {1,-4,2}
What is the resultant vector, D, from adding A+B+C?
A.
B.
C.
D.
{3,-4,2}
{4,-2,5}
{5,-2,4}
None of the above
Physics 207: Lecture 4, Pg 12
Converting Coordinate Systems

In polar coordinates the vector R = (r,q)

In Cartesian the vector R = (rx,ry) = (x,y)

We can convert between the two as follows:
rx  x  r cos q
ry  y  r cos q
R  x ˆi  y ˆj
r  x2  y 2
qtan-1 ( y / x )
y
ry
(x,y)
r
q
rx
x
• In 3D cylindrical coordinates (r,q,z), r is the same as the
magnitude of the vector in the x-y plane [sqrt(x2 +y2)]
Physics 207: Lecture 4, Pg 13
Resolving vectors into components
A mass on a frictionless inclined plane

A block of mass m slides down a frictionless ramp
that makes angle q with respect to horizontal.
What is its acceleration a ?
m
a
q
Physics 207: Lecture 4, Pg 14
Resolving vectors, little g & the inclined plane
gq
y
q

x
g (bold face, vector) can be resolved into its x,y or x’,y’
components
g=-gj
 g = - g cos q j’ + g sin q i’

The bigger the tilt the faster the acceleration…..
along the incline
Physics 207: Lecture 4, Pg 15
Dynamics II: Motion along a line but with a twist
(2D dimensional motion, magnitude and directions)

Particle motions involve a path or trajectory

Recall instantaneous velocity and acceleration

These are vector expressions reflecting x, y & z motion
r = r(t)
v = dr / dt
a = d2r / dt2
Physics 207: Lecture 4, Pg 16
Instantaneous Velocity
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But how we think about requires knowledge of the path.
The direction of the instantaneous velocity is along a line
that is tangent to the path of the particle’s direction of
motion.
The magnitude of the
instantaneous velocity
vector is the speed, s.
(Knight uses v)
s = (vx2 + vy2 + vz )1/2
v
Physics 207: Lecture 4, Pg 17
Average Acceleration
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
The average acceleration of particle motion reflects
changes in the instantaneous velocity vector (divided
by the time interval during which that change occurs).
The average
acceleration is a
vector quantity
directed along ∆v
( a vector! )
a
Physics 207: Lecture 4, Pg 18
Instantaneous Acceleration

The instantaneous acceleration is the limit of the average
acceleration as ∆v/∆t approaches zero

The instantaneous acceleration is a vector with components
parallel (tangential) and/or perpendicular (radial) to the
tangent of the path

Changes in a particle’s path may produce an acceleration
 The magnitude of the velocity vector may change
 The direction of the velocity vector may change
(Even if the magnitude remains constant)
 Both may change simultaneously (depends: path vs time)
Physics 207: Lecture 4, Pg 19
Generalized motion with non-zero acceleration:


at  a||


ar  a

v


2
2
a  0 with a  ar  at

a
need both path & time
a = a + a
Two possible options:
Change in the magnitude of v
a
=0
v
a
=0
Change in the direction of
Animation
Physics 207: Lecture 4, Pg 20
Kinematics

The position, velocity, and acceleration of a particle in
3-dimensions can be expressed as:
r= xi +y j+z k
v = v x i + v y j + vz k
a = a x i + ay j + a z k
(i , j , k unit vectors )
x  x(t )
y  y (t )
z  z (t )
dx
vx 
dt
d 2x
ax  2
dt
dy
vy 
dt
d2y
ay  2
dt
dz
vz 
dt
d 2z
az  2
dt
with, if constant accel., e.g. x(t )  x0  vx t  12 ax t 2
0

All this complexity is hidden away in
r = r(t)
v = dr / dt
a = d2r / dt2
Physics 207: Lecture 4, Pg 21
Special Case
Throwing an object with x along the
horizontal and y along the vertical.
x and y motion both coexist and t is common to both
Let g act in the –y direction, v0x= v0 and v0y= 0
x vs t
x
y
y vs t
t=0
y
0
4
t
0
4
t
x vs y
4
x
Physics 207: Lecture 4, Pg 22
Another trajectory
Can you identify the dynamics in this picture?
How many distinct regimes are there?
Are vx or vy = 0 ? Is vx >,< or = vy ?
t=0
x vs y
y
t =10
x
Physics 207: Lecture 4, Pg 23
Another trajectory
Can you identify the dynamics in this picture?
How many distinct regimes are there?
0<t<3
 I.
3<t<7
7 < t < 10
vx = constant = v0 ; vy = 0
 II. vx = vy = v0
t=0
 III. vx = 0 ; vy = constant < v0
x vs y
What can you say about the
acceleration?
y
t =10
x
Physics 207: Lecture 4, Pg 24
Exercise 1 & 2
Trajectories with acceleration
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
A rocket is drifting sideways (from left to right) in deep
space, with its engine off, from A to B. It is not near any
stars or planets or other outside forces.
Its “constant thrust” engine (i.e., acceleration is constant) is
fired at point B and left on for 2 seconds in which time the
rocket travels from point B to some point C
 Sketch the shape of the path
from B to C.
At point C the engine is turned off.
 Sketch the shape of the path
after point C
Physics 207: Lecture 4, Pg 25
Exercise 1
Trajectories with acceleration
From B to C ?
A.
B.
C.
D.
E.
A
B
C
D
None of these
B
A
B
C
B
B
C
C
B
C
D
C
Physics 207: Lecture 4, Pg 26
Exercise 3
Trajectories with acceleration
After C ?
A.
B.
C.
D.
E.
A
B
C
D
None of these
C
C
A
B
C
C
C
D
Physics 207: Lecture 4, Pg 27
Lecture 4
Assignment: Read through Chapter 5.4
MP Problem Set 2 due Wednesday
Physics 207: Lecture 4, Pg 28