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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.

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slide1

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)
slide3

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)

slide4

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

slide5

Room Frequency BA

Clicker Question

Circular motion path shown.

What is the direction of the acceleration when particle is at point X?

Acceleration points toward the center of the circle.

slide6

Demonstration

Float Accelerometer

slide7

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?

slide8

What is the magnitude of centripetal acceleration?

The two triangles are similar triangles!

slide9

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

r = 1 m

T = 3 sec

What is the approximate

centripetal acceleration?

1 m/s2

4 m/s2

0 m/s2

100 m/s2

500 m/s2

slide10

“Wall-of-Death”

But don’t I feel out outward force?

And thus outward acceleration?

I am certainly going in a circle.

slide11

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.

slide12

Room Frequency BA

Clicker Question

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.

slide13

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”

slide14

Room Frequency BA

Clicker Question

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?

Yes

No

Impossible to tell from the information given.

r

v =2πr/τ

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.

slide16

“Wall-of-Death”

But don’t I feel out outward force?

This is a “fictitious” force, not real.

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