Transcript 16 inch rim

Lecture 6
Chapter 4
 Discuss circular motion
Discern differing reference frames and understand how
they relate to particle motion
Chapters 5 & 6
 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
 Solve problems with forces in equilibrium (a=0) and nonequilibrium (a≠0) using Newton’s 1st & 2nd laws.
Assignments: HW3 (Chapters 4 & 5), finish reading Chapter 6
Exam 1: Thurs. Oct. 7 from 7:15-8:45 PM Chapters 1-7
Physics 207: Lecture 6, Pg 1
Concept Check
Q1. 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
Q2. A hockey puck slides off the edge of the table, 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 6, Pg 2
Circular Motion (quick “review”)
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Angular position
q
Arc traversed
s=rq
Tangential speed | vT | = Ds / Dt & if Dt  0 ds / dt = r dq /dt
Angular velocity
w = dq /dt = | vT | / r
Radial acceleration | ar | = vT2 / r = w2 r
s
vT
r
q
Physics 207: Lecture 6, Pg 3
Uniform Circular Motion (UCM, a=0)
Period (T): The time required to do one full revolution,
360° or 2p radians
s
vt
r
q
Frequency (f): 1/T, number of “cycles” per unit time
Angular velocity or angular speed w = 2p f = 2p/T
Positive  counter clockwise Negative  clockwise
Physics 207: Lecture 6, Pg 4
Mass-based separation with a centrifuge
Before
After
How many g’s (1 g is ~10 m/s2)?
ar = vT2 / r = w2 r
f = 6000 rpm = 100 rev. per second is typical
with r = 0.10 m
ar = (2p 102)2 x 0.10 m/s2
ar = 4 x 104 m/s2 or ca. 4000 g’s !!!
Physics 207: Lecture 6, Pg 5
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g’s with respect to humans
1 g Standing
1.2 g Normal elevator acceleration (up).
1.5-2g Walking down stairs.
2-3 g Hopping down stairs.
3.5 g Maximum acceleration in amusement park rides (design guidelines).
4 g Indy cars in the second turn at Disney World (side and down force).
4+ g Carrier-based aircraft launch.
10 g Threshold for blackout during violent maneuvers in high performance
aircraft.
11 g Alan Shepard in his historic sub orbital Mercury flight experience a
maximum force of 11 g.
20 g Colonel Stapp’s experiments on acceleration in rocket sleds indicated
that in the 10-20 g range there was the possibility of injury because of
organs moving inside the body. Beyond 20 g they concluded that
there was the potential for death due to internal injuries. Their
experiments were limited to 20 g.
30 g The design maximum for sleds used to test dummies with commercial
restraint and air bag systems is 30 g.
Comment: In automobile accidents involving rotation severe
injury or death can occur even at modest speeds
Physics 207: Lecture 6, Pg 6
Circular Motion
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Radial acceleration | ar | = vT2 / r = w2 r
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Angular acceleration a = d2q /dt = | aT | / r
s = s0+ vT0 Dt + ½ aT Dt2
s
a Tangential
a
(and if constant)
r
w (Dt )  w0  a Dt
q (Dt )  q 0  w0 Dt  a Dt
1
2
2
vT
r
| a | a
2
radial
q
a
2
Tangential
Physics 207: Lecture 6, Pg 7
Example
A horizontally mounted disk 2.0 meters in diameter
(1.0 m in radius) spins at constant angular speed
such that it first undergoes
(1) 10 counter clockwise revolutions in 5.0 seconds
and then, again at constant angular speed,
(2) 2 counter clockwise revolutions in 5.0 seconds.
 1 What is T the period of the initial rotation?
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T = time for 1 revolution = 5 sec / 10 rev = 0.5 s
Physics 207: Lecture 6, Pg 10
Example
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A horizontally mounted disk 2 meters in diameter spins at
constant angular speed such that it first undergoes 10 counter
clockwise revolutions in 5 seconds and then, again at
constant angular speed, 2 counter clockwise revolutions in 5
seconds.
2 What is w the initial angular velocity?
w = dq /dt = Dq /Dt
w = 10 • 2π radians / 5 seconds
= 12.6 rad / s ( also 2 p f = 2 p / T )
Physics 207: Lecture 6, Pg 11
Example
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A horizontally mounted disk 2 meters in diameter spins at
constant angular speed such that it first undergoes 10 counter
clockwise revolutions in 5 seconds and then, again at
constant angular speed, 2 counter clockwise revolutions in 5
seconds.
3 What is the tangential speed of a point on the rim
during this initial period?
| vT | = ds/dt = (r dq) /dt = r w
| vT | = r w = 1 m • 12.6 rad/ s = 12.6 m/s
Physics 207: Lecture 6, Pg 12
Example
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A horizontally mounted disk 1 meter in radius spins at
constant angular speed such that it first undergoes 10 counter
clockwise revolutions in 5 seconds and then, again at
constant angular speed, 2 counter clockwise revolutions in 5
seconds.
4 Sketch the q (angular displacement) versus time
plot.
s
q = q0 + w0 Dt
vt
r
q
Physics 207: Lecture 6, Pg 13
Sketch of q vs. time
q (radians)
30p
q = qo + w Dt
q = 0 + 4p 5 rad
q = qo + w Dt
q = 20p rad + (4p/5) 5 rad
q = 24 rad
20p
10p
0
5
10
time (seconds)
Physics 207: Lecture 6, Pg 14
Example
5
What is the average angular velocity over the 1st 10 seconds?
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Physics 207: Lecture 6, Pg 15
Sketch of q vs. time
q (radians)
30p
q = qo + w Dt
q = 0 + 4p 5 rad
q = qo + w Dt
q = 20p rad + (4p/5) 5 rad
q = 24 rad
20p
10p
0
5
10
time (seconds)
5 Avg. angular velocity = Dq / Dt = 24 p /10 rad/s
Physics 207: Lecture 6, Pg 16
Example
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6 If now the turntable starts from rest and uniformly
accelerates throughout and reaches the same
angular displacement in the same time, what must
be the angular acceleration a ?
Key point …..
a is associated with tangential
acceleration (aT).
Physics 207: Lecture 6, Pg 17
Tangential acceleration?
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6 If now the turntable starts from rest and uniformly
accelerates throughout and reaches the same angular
displacement in the same time, what must the “tangential
acceleration” be?
1 aT 2
q = qo + wo Dt +
Dt
2 r
s
vt
r
q
(from plot, after 10 seconds)
24 p rad = 0 rad + 0 rad/s Dt + ½ (aT/r) Dt2
48 p rad 1m / 100 s2 = aT
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What is the magnitude and direction of the acceleration
after 10 seconds?
Physics 207: Lecture 6, Pg 18
Non-uniform Circular Motion
For an object moving along a curved trajectory, with varying
speed
Vector addition: a = ar + aT (radial and tangential)
aT
a radial
a
ar
2
T
v

r
a Tangential
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d | vT |

dt
2
2
| a | aradial
 aTangential
Physics 207: Lecture 6, Pg 19
Tangential acceleration?
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What is the magnitude and direction of the acceleration
after 10 seconds?
s
aT = 0.48 p m / s2
and
vt
r
q
vT = 0 + aT Dt = 4.8 p m/s = vT
ar = vT2 / r = 23 p2 m/s2
Tangential acceleration is too small to plot!
Physics 207: Lecture 6, Pg 20
Do different observers see the same physics
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 6, Pg 21
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
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r’ = r – vo Dt
The derivative of the position equation will give the velocity
equation
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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 6, Pg 22
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 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 6, Pg 23
No Net Force, No acceleration…a demo exercise
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In this demonstration we have a ball tied to a string
undergoing horizontal UCM (i.e. the ball has only
radial acceleration)
1 Assuming you are looking from above, draw the
orbit with the tangential velocity and the radial
acceleration vectors sketched out.
2 Suddenly the string brakes.
3 Now sketch the trajectory with the velocity and
acceleration vectors drawn again.
Physics 207: Lecture 6, Pg 28
Chaps. 5, 6 & 7
What causes motion?
(What is special about acceleration?)
What are forces ?
What kinds of forces are there ?
How are forces and changes in motion
related ?
Physics 207: Lecture 6, Pg 29
Newton’s First Law and IRFs
An object subject to no external forces moves with constant
velocity if viewed from an inertial reference frame (IRF).
If no net force acting on an object, there is no
acceleration.
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The above statement can be used to define inertial
reference frames.
Physics 207: Lecture 6, Pg 30
IRFs
 An IRF is a reference frame that is not
accelerating (or rotating) with respect to the “fixed
stars”.
 If one IRF exists, infinitely many exist since they
are related by any arbitrary constant velocity vector!
 In many cases (i.e., Chapters 5, 6 & 7) the surface
of the Earth may be viewed as an IRF
Physics 207: Lecture 6, Pg 31
Newton’s Second Law
The acceleration of an object is directly proportional to the
net force acting upon it.
The constant of proportionality is the mass.
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This is a vector expression: Fx, Fy, Fz
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Units
The metric unit of force is kg m/s2 = Newtons (N)
The English unit of force is Pounds (lb)
Physics 207: Lecture 6, Pg 32
Lecture 6
Assignment: Read rest of chapter 6
Physics 207: Lecture 6, Pg 33