Transcript Physics 207, Lecture 8, Oct. 1

Lecture 5

Goals:
 Address systems with multiple accelerations in 2dimensions (including linear, projectile and circular motion)
 Discern different reference frames and understand how
they relate to particle motion in stationary and moving frames
 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)
Read through Chapter 6, Sections 1-4
Physics 207: Lecture 5, Pg 1
Kinematics in 2 D

The position, velocity, and acceleration of a particle moving in
2-dimensions can be expressed as:
r= xi +y j
v = vx i + vy j
a = a x i + ay j
x  x(t )
dx
vx 
dt
d 2x
ax  2
dt
y  y (t )
dy
vy 
dt
d2y
ay  2
dt
Special Cases: 1.
2.
ax=0
ay= -g
Uniform Circular Motion
Physics 207: Lecture 5, Pg 2
Special Case 1: Freefall
x(t )  x0  vx t
vx  const.
y (t )  y0  v y 0t  gt
1
2
2
v y (t )  v y 0  gt
x and y motion are separate and t is common to both
Now: 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 5, 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!
Acceleration may or may not be!
Example Problem
Given


r0 & v0
t=0
x
x vs y
4
How far does the knife travel (if no
air resistance)?
y
Physics 207: Lecture 5, Pg 4
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 5, Pg 5
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 5, Pg 6
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 5, Pg 7
Exercise 1
Trajectories with acceleration
B
From B to C ?
A
A.
B.
C.
D.
E.
A
B
C
D
None of
these
B
C
B
B
C
C
B
C
D
C
Physics 207: Lecture 5, Pg 8
Exercise 2
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 5, Pg 9
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 5, Pg 10
Schematic of the problem
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xB(t) = d = v0 cos q t
yB(t) = hf = v0 sin q t – ½ g t2
xM(t) = d
yM(t) = h – ½ g t2
Does yM(t) = yB(t) = 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 5, Pg 11
hf
Case 2: Uniform Circular Motion
Circular motion has been around a long time
Physics 207: Lecture 5, Pg 12
Generalized motion with only radial acceleration
Uniform Circular Motion


at  a||


ar  a

v

v
a = a + a

a
Changes only in the direction of v
a
=0
A particle doesn’t speed up or slow down!
Physics 207: Lecture 5, Pg 13
Uniform Circular Motion (UCM) is common
so we have specialized terms
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Arc traversed s = q r
Tangential velocity vt
Period, T, and frequency, f
Angular position, q
Angular velocity, w
s
vt
r
q
Period (T): The time required to do one
full revolution, 360° or 2p radians
Frequency (f): 1/T, number of cycles per unit time
Angular velocity or speed w = 2pf = 2p/T, number of
radians traced out per unit time (in UCM average and
instantaneous will be the same)
Physics 207: Lecture 5, Pg 14
Angular displacement and velocity
Arc traversed s = q r
in time t then s = q r
so s / t = (q / t) r
in the limit t  0
vt
one gets
ds / dt = dq / dt r
vt = w r
w ≡ dq / dt
if w is constant, integrating w = dq / dt,
we obtain: q = qo + w t

s
r
q
Counter-clockwise is positive, clockwise is negative
Physics 207: Lecture 5, Pg 15
Circular motion also has a radial (perpendicular) component
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
Centripetal Acceleration
ar
r
ar = vt2/r
Circular motion involves
continuous radial acceleration
Physics 207: Lecture 5, Pg 16
What if w is linearly increasing …
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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 t + ½ at t2 and s = q r
We can relate at to dw/dt
q = qo + wo t +
at
1 at
2 r
t2
w = wo + r t
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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.
Physics 207: Lecture 5, Pg 17
Non-uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar + at (radial and tangential)
at
v2
ar =
r
ar
at =
d| v |
dt
Physics 207: Lecture 5, Pg 18
Angular motion, signs
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If
angular displacement
velocity
accelerations
are counter clockwise then sign is positive.
 If clockwise then negative
Physics 207: Lecture 5, Pg 19
Circular Motion
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UCM enables high accelerations (g’s) in a small space
Comment: In automobile accidents involving rotation severe
injury or death can occur even at modest speeds.
[In physics speed doesn’t kill….acceleration does (i.e., the
sudden change in velocity).]
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Physics 207: Lecture 5, Pg 27
Mass-based separation with a centrifuge
Before
How many g’s?
After
ar=vt2 / r and f = 104 rpm is
typical with r = 0.1 m
and vt = w r = 2p f r
ca. 10000 g’s
Physics 207: Lecture 5, Pg 28
bb5
Relative motion and frames of reference
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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 5, Pg 31
Relative Velocity
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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 t
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 5, Pg 32
Central concept for problem solving: “x” and “y”
components of motion treated independently.
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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 t
3) y = y0 + v0y – g t2/2
x motion: x = vxt
Reference frame
on the ground.
Net motion: R = x(t) i + y(t) j (vector)
Physics 207: Lecture 5, Pg 33
Recap
Assignment: HW2, (Chapters 2 & 3, due
Wednesday)
 Read through Chapter 6, Sections 1-4

Physics 207: Lecture 5, Pg 39