# Lecture 4

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```					                             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

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
1
2
2
v 2  v x2  2a x (x  x 0 )
x         0
0

v x  v x  a x t                             1
0

a x  const                    v x (avg)    (v x  v x )
2           0

x(displaceme nt )
And     vaverage velocity  
t ( total time )
Physics 207: Lecture 4, Pg 2
The picture
v0                          v1 ( > v0 )
a0=0 m/s2                   a1=0 m/s2

x0   t0                      x1 t1                              x       t2
x x2  x0                             2
   Plus the average velocity v   
    Knowns:                     t t2  t0
 x0 = 0 m
 t0 = 0 s                             x  x  v t
0     x
 v0 = 20 m/s                                             0

 t2 = 10 s                            vx  vx     0

 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

Physics 207: Lecture 4, Pg 3
Using          x  x0  v x t0

v0                              v1 ( > v0 )
a0=0 m/s2                       a1=0 m/s2

x0   t0                        x1 t1                               x       t2
x1  x0  v0 (t1  t0 )       x2  x1  v1 (t2  t1 )
2

 Four
unknowns                                    x x2  x0
 Four                                     v   
relationships                               t t2  t0
x1  ( x2  x0 )
1
2

Physics 207: Lecture 4, Pg 4
Using       x0  0            t0  0
v0                                       v1 ( > v0 )
a0=0 m/s2                                a1=0 m/s2

x0    t0                              x1 t1                                      x       t2
2
1   x1  v0 t1              2   x2  x1  v1 (t2  t1 )
   Eliminate                                                                          x2
unknowns                   3     x1  x2
1
2                                  4     v
t2
x2  1 x2  v1 (t2  t1 )
first x1
2&3                2
next t1
x2  v1 (t2  )   x2
1
1                     2
1   2                     v0             Mult. by 2/ t2
then x2
vt2  v1 (t2  )          vt2
v  v1 (2  vv )
1
4       1                         2

2                             v0                                    0

Physics 207: Lecture 4, Pg 5
Fini
v0                                 v1 ( > v0 )
a0=0 m/s2                          a1=0 m/s2

x0   t0                    x1 t1                                     x       t2
2
   Plus the average velocity   v  v1 (2  vv )        0

    Given:
 v0 = 20 m/s              v1       v v0
2 v0  v
 t2 = 10 s
 vavg = 25 m/s            v1  225 m/s 20 m/s
 20 m/s 25 m/s

 500 m/s
15

 33.3 m/s
Physics 207: Lecture 4, Pg 6

   The sum of two vectors is another vector.
A =B+C
B                         B

C                         A               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
Different direction
A
A =B+C                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.
U = |U| û
    Unit vector u points in the direction of U
Often denoted with a “hat”: u = û                         û

y
Useful examples are the
cartesian unit vectors [ i, j, k ]                 j
 Point in the direction of the
x, y and z axes.                               i       x
k
R = rx i + ry j + rz k        z

Physics 207: Lecture 4, Pg 9

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):
By
C
 Cx = Ax + Bx                                         B
 Cy = Ay + By
Bx
 |C| =[ (Cx)2+ (Cy)2 ]1/2                    A   Ay
Ax

Physics 207: Lecture 4, Pg 10
Example
   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.   {3,-4,2}
B.   {4,-2,5}
C.   {5,-2,4}
D.   None of the above

Physics 207: Lecture 4, Pg 11
Example

   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.   {3,-4,2}
B.   {4,-2,5}
C.   {5,-2,4}
D.   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
y
ry  y  r cos q                                                (x,y)

R  xˆ yˆ
i   j                                   r
ry
r  x2  y 2                                   q
q  tan-1 ( y / x )                           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
   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
   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
( a vector! )
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:

v
                                      
at  a||                      a  0 with a  ar  at
2    2

need both path & time

                a
ar  a                        a = a + a
Two possible options:
Change in the magnitude of v            a      =0

Change in the direction of   v         a     =0

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    (i , j , k unit vectors )
a = a x i + ay j + a z k
x  x(t )        y  y (t )    z  z (t )
dx                dy              dz
vx               vy             vz 
dt                dt              dt
d 2x              d2y             d 2z
ax  2            ay  2          az  2
dt                dt              dt
with, if constant accel., e.g. x(t )  x0  vx t  1 ax t 2
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                 y vs t          t=0           x vs y
x                         y

4
y

0            4       t    0          4         t                    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          3<t<7         7 < t < 10
 I.    vx = constant = v0 ; vy = 0
 II.   vx = vy = v0
 III. vx = 0 ; vy = constant < v0
x vs y
t=0

What can you say about the
y                                   acceleration?

t =10
x
Physics 207: Lecture 4, Pg 24
Exercise 1 & 2
Trajectories with acceleration
   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 ?            B            B

A.   A               A        C
B
C
B.   B
C.   C
D.   D
B                   B
E.   None of these

C        C     D                      C

Physics 207: Lecture 4, Pg 26
Exercise 3
Trajectories with acceleration

After C ?           C               C

A.   A               A               B
B.   B
C.   C
D.   D               C               C
E.   None of these

C              D

Physics 207: Lecture 4, Pg 27
Lecture 4

MP Problem Set 2 due Wednesday

Physics 207: Lecture 4, Pg 28

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