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10/14/2013 - PPT Presentation

PHY 113 C Fall 2013 Lecture 14 1 PHY 113 C General Physics I 11 AM1215 PM MWF Olin 101 Plan for Lecture 14 Chapter 12 Static equilibrium Balancing forces and torques stability Center of gravity ID: 246934

lecture 2013 113 phy 2013 lecture phy 113 fall earth mass assignment webassign equilibrium center iclicker gravity question surface

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Slide1

10/14/2013

PHY 113 C Fall 2013 -- Lecture 14

1

PHY 113 C General Physics I

11 AM-12:15 PM MWF Olin 101

Plan for Lecture 14:

Chapter 12 – Static equilibrium

Balancing forces and torques; stability

Center of gravity

Will discuss elasticity in Lecture 15 (Chapter 15)Slide2

10/14/2013

PHY 113 C Fall 2013 -- Lecture 14

2Slide3

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PHY 113 C Fall 2013 -- Lecture 14

3

Newton’s law of gravitation:

Earth’s gravity:

Stable

circular orbits of gravitational attracted objects:

R

ES

F

a

v

M

sat

Summary of gravity:

R

E

mSlide4

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PHY 113 C Fall 2013 -- Lecture 14

4

From

Webassign

Assignment #12:

When

a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration due to the Earth's gravitation?

m/s

2

towards earth

r=4.4R

E

mSlide5

10/14/2013

PHY 113 C Fall 2013 -- Lecture 14

5

From

Webassign

Assignment #12:

An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.79 m/s

2. Determine the orbital period of the satellite.

rSlide6

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PHY 113 C Fall 2013 -- Lecture 14

6

From

Webassign

Assignment #12:

How much work is done by the Moon's gravitational field as a 1090 kg meteor comes in from outer space and impacts on the Moon's surface?

R

M

iclicker

question

W>0

W<0Slide7

10/14/2013

PHY 113 C Fall 2013 -- Lecture 14

7

From

Webassign

Assignment #12:

A space probe is fired as a projectile from the Earth's surface with an initial speed of 1.74 10

4 m/s. What will its speed be when it is very far from the Earth? Ignore atmospheric friction and the rotation of the Earth.

v

iSlide8

10/14/2013

PHY 113 C Fall 2013 -- Lecture 14

8

From

Webassign

Assignment #12:

Plaskett's

binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v = 190 km/s and the orbital period of each is 10.7 days. Find the mass

M

of each star. (For comparison, the mass of our Sun is 1.99 10

30

kg.)Slide9

10/14/2013

PHY 113 C Fall 2013 -- Lecture 14

9

From

Webassign

Assignment #12:

Plaskett's

binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v = 190 km/s and the orbital period of each is 10.7 days. Find the mass

M

of each star. (For comparison, the mass of our Sun is 1.99 10

30

kg.)

iclicker

exercise:

Who might pose a question like this?

A mean professor.

A puzzle master.

An observational astronomer.Slide10

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PHY 113 C Fall 2013 -- Lecture 14

10

Meanwhile – back on the surface of the Earth:

Conditions for stable equilibriumSlide11

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11

Stability of “rigid bodies”

N

m

i

gSlide12

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PHY 113 C Fall 2013 -- Lecture 14

12

Center-of-mass

Torque on an extended object due to gravity (near surface of the earth) is the same as the torque on a point mass M located at the center of mass.

m

i

r

i

r

CMSlide13

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PHY 113 C Fall 2013 -- Lecture 14

13

Notion of stability:

mg(-

j

)

r

T

q

F

=m

a

T- mg

cos

q = 0

-

mg sin

q = -

ma

q

t

=I

a

r mg sin

q

= mr

2

a =

mraq

Notion of equilibrium:

Example of stable

equilibrium for

q

» 0.Slide14

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PHY 113 C Fall 2013 -- Lecture 14

14

Unstable equilibrium:

mg(-

j

)

r

T

q

Support

above

CM:

Support

below

CM:Slide15

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PHY 113 C Fall 2013 -- Lecture 14

15

Nik

Wallenda

walking on high wire across Grand CanyonSlide16

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16

Analysis of stability:Slide17

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17

**

XSlide18

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18Slide19

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PHY 113 C Fall 2013 -- Lecture 14

19

**

X

F

g1

mg

R

CMSlide20

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20

iclicker

question:

F

1

F

2

Consider the above drawing of the two supports for a uniform plank which has a total weight Mg and has a weight mg at its end. What can you say about F

1

and F

2

?

F

1

and F

2

are both up as shown.

F

1

is up but F

2

is down.

F

1

is down but F

2

is up.

L/3

L

Mg

mgSlide21

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21

F

1

F

2

L/3

L

Mg

mg

**

XSlide22

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22

iclicker

question:

The fact that we found F

1

<0 means:

We set up the problem incorrectly

The analysis is correct, but the direction

of

F

1

is opposite to the arrow

Physics makes no sense

iclicker

question:

What would happen if we analyzed this problem by placing the pivot point

at F1 ?:The answer would be the same.

The answer would be different.

Physics makes no senseSlide23

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PHY 113 C Fall 2013 -- Lecture 14

23

T

Mg

mg

**

XSlide24

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24

d

**

XSlide25

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25

mg

Mg

F

wall

N

T

Mg = 120 N

mg = 98 N

T < 110 N

**

XSlide26

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PHY 113 C Fall 2013 -- Lecture 14

26

x

A ladder of weight Mg and of length L is supported by the ground with static friction force

f

and by a frictionless wall as shown. The firefighter has weight mg and is half-way up the ladder. Find the force that the ladder exerts on the wall.

q

**

X