1. Lecture 5
Goals: (finish Ch. 4 and start Ch. 5)
Address systems with multiple accelerations in 2-dimensions (includes linear, projectile and circular motion)
Discern differing reference frames and understand how they relate to particle motion
Recognize different types of forces and know how they act on an object in a particle representation
Identify forces and draw a Free Body Diagram
Assignment: HW2, (Chapters 2 & 3, due Wednesday)
Reading assignment: read through Chapter 6.4 1

2. Kinematics in 2 D
The position, velocity, and acceleration of a particle moving in 2-dimensions can be expressed as:
r = x i + y j  v = vx i + vy j  a = ax i + ay j

3. Kinematics in 2 D
The position, velocity, and acceleration of a particle moving in 2-dimensions can be expressed as:
r = x i + y j  v = vx i + vy j  a = ax i + ay j
Special Case: ax=0 & ay= -g or a = -g j
with an initial velocity v0 q

4. Kinematics in 2 D
The position, velocity, and acceleration of a particle moving in 2-dimensions can be expressed as:
r = x i + y j  v = vx i + vy j  a = ax i + ay j
Special Case: ax=0 & ay= -g or a = -g j
vx(Dt) = v0 cos q
vy(Dt) = v0 sin q – g Dt
x(Dt) = x0 + v0 cos q Dt
y(Dt) = y0 + v0 sin q Dt - ½ g Dt2 q

5. Trajectory with constant acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Velocity vector is always tangent to the curve!
Acceleration may or may not be!
Example Problem: If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel horizontally before it lands on the level ground (if no air resistance)?
Position x
t = 0 4 y
x vs y

6. Example Problem
Example Problem: If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel horizontally before its hit the level ground (if g= -10 m/s2 & no air resistance)?
at t=0 the knife is at (0.0 m, 1.0 m) with vy=0
at Dt the knife is at (x m, 0.0 m)
x = x0 + vx Dt and y = y0 – ½ g Dt2
So x = 10 m/s Dt
and
0.0 m = 1.25 m – ½ g Dt2  2.5/10 s2 = Dt Dt = 0.50 s
x = 10 m/s 0.50 s = 5 .0 m

7. Trajectory with constant acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Velocity vector is always tangent to the curve!
Acceleration may or may not be!
Position
Velocity
x vs y
t = 0 4 y x

8. Trajectory with constant acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Velocity vector is always tangent to the curve!
Acceleration may or may not be!
Position
Velocity
Acceleration
x vs y
t = 0 4 y x

9. x vs y Another trajectory y
Can you identify the dynamics in this picture if each dot represents the position in 1 second intervals ?
How many distinct regimes are there?
Are vx or vy = 0 ? Is vx >,< or = vy ?
x vs y
t = 0 y
t =10 x

10. 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

11. 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 a

12. Exercise 1 Trajectories with accelerationB C B C
From B to C ? A B A B C D
None of these B C B C C D

13. Exercise 2 Trajectories with acceleration
After C ? A B A B C D
None of these C C C D

14. Exercise 3 Relative Trajectories: Monkey and Hunter
All free objects, if acted on by gravity, accelerate similarly.
A hunter sees a monkey in a tree, aims his gun at the monkey and fires. At the same instant the monkey lets go.
Does the bullet …
go over the monkey.
hit the monkey.
go under the monkey.

15. Schematic of the problem
xB(Dt) = d = v0 cos q Dt
yB(Dt) = hf = v0 sin q Dt – ½ g Dt2
xM(Dt) = d
yM(Dt) = h – ½ g Dt2
Does yM(Dt) = yB(Dt) = hf?
(x,y) = (d,h)
Monkey
Does anyone want to change their answer ?
What happens if g=0 ?
How does introducing g change things? g hf v0
Bullet q
(x0,y0) = (0 ,0)
(vx,vy) = (v0 cos q, v0 sin q)

16. Uniform Circular Motion (UCM) is common so we have specialized terms
Angular position q (CCW + CW -)
Radius is r
Arc traversed s = r q
Tangential “velocity” vT = ds/dt = r dq/dt = r w
Angular velocity, w ≡ dq/dt (CCW + CW -) r q vT s

17. v 1. A particle doesn’t speed up or slow down!
Generalized motion with only radial acceleration Uniform Circular Motion (UCM) q v
UCM changes only in the direction of
1. A particle doesn’t speed up or slow down!
2. Velocity is always tangential

18. 1. Arc traversed s’ ≈ Dv= v Dq 2. Divide by Dt Dv/Dt = v Dq/Dt
Generalized motion with only radial acceleration Uniform Circular Motion (UCM)
1. Arc traversed s’ ≈ Dv= v Dq
2. Divide by Dt
Dv/Dt = v Dq/Dt
3. Obtain radial acceleration
ar = v w
4. v is tangential & v = r w
5. Result: ar = vT2 / r = w2 r Dq

19. Again
Uniform circular motion involves only changes in the direction of the velocity vector, thus acceleration is perpendicular to the trajectory at any point, acceleration is only in the radial direction. Quantitatively (see text) vT r ar
Centripetal Acceleration
ar = vT2/r
Circular motion involves
continuous radial acceleration

20. Uniform Circular Motion (UCM) is common so we have specialized terms
New definitions
1. Period (T): The time required to do one full revolution, 360° or 2p radians r q vT s
2. Frequency (f): f ≡ 1/T, number of cycles per unit time
Angular velocity or speed in UCM
w = Dq / Dt = 2p / T = 2p f (radians per unit time)

21. Angular displacement and velocity
Angle traversed q
If w is constant, & w = dq / dt,
Integrating w dt = dq
we obtain: q(Dt) = qo + w Dt
Counter-clockwise is positive, clockwise is negative r q vT s

22. Non-uniform CM: What if w is linearly increasing …
Then angular velocity is no longer constant so dw/dt ≠ 0
Define tangential acceleration as aT = dvT/dt = r dw/dt
So s = s0 + (ds/dt)0 Dt + ½ aT Dt2 and s = q r
We can relate aT to dw/dt
q = qo + wo Dt Dt2
w = wo Dt
Many analogies to linear motion but it isn’t one-to-one
Note: Even if the angular velocity is constant, there is always a radial acceleration. 1 2 aT r aT r

23. Non-uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar + aT (radial and tangential) aT
ar = v2 r ar dt
d| | v
aT =

24. Recap Assignment: HW2, (Chapters 2 & 3, due Wednesday)
Read through Chapter 6, Sections 1-4