Transcript PPTX

Announcements
• CAPA Set #6 due Friday at 10 pm
• This week in Section 
Lab 2: Acceleration due to gravity
Make sure to do the pre-lab and print materials before lab
* Again note difference in pre-lab questions, just answer
the extra question.
• Read Sections 5.1-5.5 (Circular Motion), Section 5.6-5.9
(Gravity too)
• Advanced reminder  Exam #2 on Tuesday, October 11
• Reminder about office hours …
Nagle (Monday 2-3 in office, Wednesday 1:45-3:45 pm help room)
Kinney (Thursday 4-5 pm help room)
Uzdensky (Tuesday 11am-noon help room)
Circular Motion
Circular Motion: Object moving in a circle of
radius r with a constant speed (meters/second).
|v|=2m/s
|v|=2m/s
|v|=2m/s
r
T = period = time for one revolution (one cycle)
 distance circumfere nce 2 r
speed | v |


time
period
T
Is the object accelerating?
|v|=2m/s
|v|=2m/s
|v|=2m/s
r
The speed |v| is not changing, but the velocity
vector direction is constantly changing.
Recall definition of acceleration
  
 v v2  v1
acceleration  a 

t
t 2  t1
Clicker Question
Room Frequency BA
Circular motion path shown.
What is the direction of the acceleration when particle is at point X?
v2
v1
v1  v  v 2
v  v 2  v1
Acceleration points toward the
center of the circle.
Demonstration
Float Accelerometer
Acceleration points toward the center of the circle
“Centripetal acceleration”  from Latin centrum “center”
and petere “to seek”
Question: What is the magnitude of centripetal acceleration?
What is the magnitude of centripetal acceleration?

v
The two triangles are similar triangles!

base
| v | 
  
height | v |
r
v
v  
r
v


2
v


v
v
 | v |
r

 v 
| a |

r t
r
r
t
t
2
 v
| a |
r
Room Frequency BA
Clicker Question
A merry-go-round has a radius r = 1 meter.
It’s pushed to have a period of T = 3 seconds.
m
 v2
| a |
r
r=1m
T = 3 sec
What is the approximate
centripetal acceleration? A)1 m/s2
2r
2  3 1
v

 2m / s
T
3
2
2r
v
T
2
v
( 2m / s )
a

 4m / s 2
r
1m
B) 4 m/s2
C) 0 m/s2
D)100 m/s2
E) 500 m/s2
“Wall-of-Death”
But don’t I feel out outward force?
And thus outward acceleration?
I am certainly going in a circle.
First, we need a say a little more about inertial frames
(in which Newton’s Three Laws are obeyed).
You are in the passenger seat of the batmobile.
Batman is busy drinking a cup of coffee and
the batmobile is parked (v=0).
You toss a red ball up in the air.
From your viewpoint
(i.e. reference frame)
the ball goes up and then
down following Newton’s Laws.
y=y0+v0yt-1/2gt2
Only gravitational force acting.
Clicker Question
Room Frequency BA
Batman decides he can drink the coffee while driving at a
constant speed of 25 miles per hour.
You toss a red ball up straight up in the air from your
point of view (i.e. your reference frame at rest in the car).
Does the ball land back in your hand?
A) Yes, B) No, C) Depends on the coffee
Newton’s Laws are still obeyed.
Just as you are doing the “red ball”
experiment, the bat signal lights up and
batman accelerates quickly to 100 mph.
In your reference frame, the ball goes up
and then suddenly (as if acted upon by a
mysterious force) moves back and
hits you in the face.
No real force suddenly
acted on the ball.
The car and thus you accelerated,
your reference frame is non-inertial
and Newton’s equations do not
work.
You observe what are called
“fictitious forces”
Room Frequency BA
Clicker Question
r
v =2πr/τ
Imagine a child sitting on a rotating
merry-go-round of radius r who is
rotating with non-zero speed v.
In the child’s frame of reference, do
Newton’s Laws hold?
A) Yes
B) No
C) Impossible to tell from the
information given.
A rotating frame is not an “inertial frame” – so Newton’s
Laws don’t hold in the form stated above. To work they
need to be modified to include a “fictitious” force.
“Wall-of-Death”
But don’t I feel out outward force?
This is a “fictitious” force, not real.