Lecture 5

Goals: (finish Ch. 4)
 Address 2D motion in systems with acceleration (includes
linear, projectile and circular motion)
 Discern differing reference frames and understand how
they relate to particle motion

Physics 207: Lecture 8, Pg 1
Trajectory with constant
acceleration along the vertical
Special Case:
ax=0 &
ay= -g
vx(t=0) = v0
vy(t=0)
vy(Dt) = – g Dt
Position
t = 0 x vs y
x(Dt) = x0 + v0 Dt
y(Dt) = y0 - ½ g Dt2
y
4
What do the velocity and acceleration
vectors look like?
x
Physics 207: Lecture 8, Pg 3
Trajectory with constant
acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Velocity vector is always tangent to the
curve!
t=0
y
Velocity
x vs y
4
x
Physics 207: Lecture 8, Pg 4
Trajectory with constant
acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Position
Velocity
Velocity vector is always tangent to the
curve!
Acceleration may or may not be!
t=0
y
Acceleration
x vs y
4
x
Physics 207: Lecture 8, Pg 5
Concept Check
E1. You drop a ball from rest, how much of the acceleration
from gravity goes to changing its speed?
A. All of it
B. Most of it
C. Some of it
D. None of it
E2. A hockey puck slides off the edge of a horizontal table,
just at the instant it leaves the table, how much of the
acceleration from gravity goes to changing its speed?
A. All of it
B. Most of it
C. Some of it
D. None of it
Physics 207: Lecture 8, Pg 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 be for it hit the level
ground (if g= -10 m/s2 & no air resistance)?
Physics 207: Lecture 8, Pg 7
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 before it hits the level ground
(assume g= -10 m/s2 & no air resistance)?
 at t=0 the knife is at (0.0 m, 1.0 m) with vy=0
 after Dt the kinfe 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 = Dt2
Dt = 0.50 s
x = 10 m/s 0.50 s = 5 .0 m
Physics 207: Lecture 8, Pg 8
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 …
A.
B.
C.
go over the
monkey.
hit the monkey.
go under the
monkey.
Physics 207: Lecture 8, Pg 9
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
v0
q
Bullet
(x0,y0) = (0 ,0)
(vx,vy) = (v0 cos q, v0 sin q)
Physics 207: Lecture 8, Pg 10
hf
Relative motion and frames of reference



Reference frame S is stationary
Reference frame S’ is moving at vo
This also means that S moves at – vo relative to S’
Define time t = 0 as that time when the origins coincide
Physics 207: Lecture 8, Pg 11
Relative motion and frames of reference



Reference frame S is stationary, A is stationary
Reference frame S’ is moving at vo
A moves at – vo relative to an observer in S’
Define time t = 0 as that time (O & O’ origins coincide)
t later
t=0
-v0
-v0 t
A
A
r’= r-v0t
r
O,O’
r
O’
Physics 207: Lecture 8, Pg 12
Relative Velocity




The positions, r and r’, as seen from the two reference frames
are related through the velocity, vo, where vo is velocity of the
r’ reference frame relative to r

r’ = r – vo Dt
The derivative of the position equation will give the velocity
equation

v’ = v – vo
These are called the Galilean transformation equations
Reference frames that move with “constant velocity” (i.e., at
constant speed in a straight line) are defined to be inertial
reference frames (IRF); anyone in an IRF sees the same
acceleration of a particle moving along a trajectory.

a’ = a
(dvo / dt = 0)
Physics 207: Lecture 8, Pg 13
Central concept for problem solving: “x” and “y”
components of motion treated independently.


Example: Man on cart tosses a ball straight up in the air.
You can view the trajectory from two reference frames:
Reference frame
on the moving cart.
y(t) motion governed by
1) a = -g y
2) vy = v0y – g Dt
3) y = y0 + v0y – g Dt2/2
x motion: x = vxt
Reference frame
on the ground.
Net motion: R = x(t) i + y(t) j (vector)
Physics 207: Lecture 8, Pg 14
Exercise, Relative Motion

You are swimming across a 50. m wide river in which the
current moves at 1.0 m/s with respect to the shore. Your
swimming speed is 2.0 m/s with respect to the water.
You swim perfectly perpendicular to the current,
how fast do you appear to be moving to an observer on shore?

v  1 m/s î + 2 m/s ĵ
2 m/s
50m
1 m/s

2
2
| v |  1.0 + 2.0 m/s  2.2 m/s
Home exercise: What is your apparent speed if you swim across?
Physics 207: Lecture 8, Pg 18
Case 2: Uniform Circular Motion
Circular motion has been around a long time
Physics 207: Lecture 8, Pg 21
Circular Motion (UCM) is common
so specialized terms

Angular position q (CCW + CW -)
q
Physics 207: Lecture 8, Pg 22
Circular Motion (UCM) is common
so specialized terms
Angular position q (CCW + CW -)
 Radius is r

r
q
Physics 207: Lecture 8, Pg 23
Circular Motion (UCM) is common
so specialized terms
Angular position q (CCW + CW -)
 Radius is r
 Arc traversed s = r q (if starting at 0 deg.)

s
r
q
Physics 207: Lecture 8, Pg 24
Circular Motion (UCM) is common
so specialized terms
Angular position q (CCW + CW -)
 Radius is r
 Arc traversed s = r q (if starting at 0 deg.)
 Tangential “velocity” vT = ds/dt

s
vT
r
q
Physics 207: Lecture 8, Pg 25
Circular Motion (UCM) is common
so specialized terms
Angular position q (CCW + CW -)
 Radius is r
 Arc traversed s = r q
 Tangential “velocity” vT = ds/dt
 Angular velocity, w ≡ dq/dt (CCW + CW -)
s
 vT = ds/dt = r dq/dt = r w

vT
r
q
Physics 207: Lecture 8, Pg 26
Uniform Circular Motion (UCM) has only radial acceleration
v
UCM changes only in the direction of
1. Particle doesn’t speed up or slow down!
2. Velocity is always tangential, acceleration perpendicular
!

aT  0

v

v


aradial  a path

a

v
Dq
Dv

v


| v || v |
Dv  vT Dq
dv
dq
2
ar 
 vT
 vT w  vT vT /r  vT /r
dt
dt
Physics 207: Lecture 8, Pg 27
Again
Uniform circular motion involves only changes in the
direction of the velocity vector
Acceleration is perpendicular to the trajectory at any point,
acceleration is only in the radial direction.
vT
Centripetal Acceleration
ar
r
ar = vT2/r = w2 r
Circular motion involves
continuous radial acceleration
Physics 207: Lecture 8, Pg 28
Angular motion, signs

If
angular displacement
velocity
accelerations
are counter clockwise then sign is positive.
 If clockwise then negative
Physics 207: Lecture 8, Pg 30
Recap
Assignment: HW2, (Chapters 2,3 & some 4)
due Wednesday
 Read through Chapter 6, Sections 1-4

Physics 207: Lecture 8, Pg 32