1. 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 1

2. Example of a 1D motion problem
A cart is initially traveling East at a constant speed of 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):
And

3. The picture v0 a0=0 m/s2 v1 ( > v0 ) a1=0 m/s2 x0 t0 x1 t1 x2 t2
Plus the average velocity
Knowns:
x0 = 0 m
t0 = 0 s
v0 = 20 m/s
t2 = 10 s
vavg = 25 m/s
relationship between x1 and x2
Four unknowns x1 v1 t1 & x2 and must find v1 in terms of knowns

4. Four unknowns Four relationships Using t0 t1 v1 ( > v0 ) a1=0 m/s2x0 x1 x2 t2
Four unknowns
Four relationships

5. Eliminate unknowns Using first x1 next t1 then x2 v0 a0=0 m/s2
v1 ( > v0 )
a1=0 m/s2 x0 t0 x1 t1 x2 t2 1 2
Eliminate unknowns
first x1
next t1
then x2 3 4
2 & 3 1
Mult. by 2/ t2 4

6. Plus the average velocity Given: v0 = 20 m/s t2 = 10 s vavg = 25 m/s
Fini t0 t1
v1 ( > v0 )
a1=0 m/s2 v0
a0=0 m/s2 x0 x1 x2 t2
Plus the average velocity
Given:
v0 = 20 m/s
t2 = 10 s
vavg = 25 m/s

7. Vectors and 2D vector addition
The sum of two vectors is another vector.
A = B + C B C A
D = B + 2 C ???

8. 2D Vector subtraction A = B + C
Vector subtraction can be defined in terms of addition.
B - C
= B + (-1)C B C B -C
B - C A
Different direction and magnitude !
A = B + C

9. 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
U = |U| û 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 y j x i k z

10. 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 Bx A By B Ax Ay

11. Example Vector Addition
Vector B = {3,0,2}
Vector C = {1,-4,2}
What is the resultant vector, D, from adding A+B+C?
{3,-4,2}
{4,-2,5}
{5,-2,4}
None of the above

12. Example Vector Addition
Vector B = {3,0,2}
Vector C = {1,-4,2}
What is the resultant vector, D, from adding A+B+C?
{3,-4,2}
{4,-2,5}
{5,-2,4}
None of the above

13. 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: y
(x,y) r ry
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)]

14. Resolving vectors into components A mass on a frictionless inclined plane
A block of mass m slides down a frictionless ramp that makes angle  with respect to horizontal. What is its acceleration a ? m a 

15. Resolving vectors, little g & the inclined planex’ y’ x y g q
g (bold face, vector) can be resolved into its x,y or x’,y’ components
g = - g j
g = - g cos q j’ + g sin q i’
The bigger the tilt the faster the acceleration…..
along the incline

16. 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

17. 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

18. 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 vector! )

19. 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)

20. Generalized motion with non-zero acceleration:
need both path & time a = +
Two possible options:
Change in the magnitude of v a
= 0
Change in the direction of v a
= 0
Animation

21. Kinematics All this complexity is hidden away in
The position, velocity, and acceleration of a particle in
3-dimensions can be expressed as:
r = x i + y j + z k
v = vx i + vy j + vz k (i , j , k unit vectors )
a = ax i + ay j + az k
All this complexity is hidden away in
r = r(Dt) v = dr / dt a = d2r / dt2

22. x vs y x vs t y vs t Special Case x y y t x t
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 y t x 4
x vs t
y vs t
t = 0 y 4 y 4 t x

23. x vs y Another trajectory y
Can you identify the dynamics in this picture?
How many distinct regimes are there?
Are vx or vy = 0 ? Is vx >,< or = vy ?
x vs y
t = 0 y
t =10 x

24. x vs y Another trajectory y x
Can you identify the dynamics in this picture?
How many distinct regimes are there?
0 < t < < 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 acceleration? y
t =10 x

25. 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.
after point C

26. Exercise 1 Trajectories with acceleration
From B to C ? B C B C A B A B C D
None of these B C B C C D

27. Exercise 3 Trajectories with acceleration
After C ? A B A B C D
None of these C C C D

28. Assignment: Read through Chapter 5.4 MP Problem Set 2 due Wednesday
Lecture 4
Assignment: Read through Chapter 5.4
MP Problem Set 2 due Wednesday 1