Download PHY 113 C General Physics I 11 AM

10/15/2013
PHY 113 C General Physics I
11 AM-12:15 PM MWF Olin 101
Plan for Lecture 14:
Chapter 12 – Static equilibrium
1. Balancing forces and torques;
stability
2. Center of gravity
3. Will discuss elasticity in Lecture 15
(Chapter 15)
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PHY 113 C Fall 2013 -- Lecture 14
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PHY 113 C Fall 2013 -- Lecture 14
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Summary of gravity:
Newton’s law of gravitation:
Earth’s gravity:
F12 
Gm1m2rˆ12
r122
RE
m
GM E m
F
 mg
RE2
g
GM E 6.67 1011  5.981024

m/s2  9.8m/s2
RE2
(6.37 106 ) 2
Stable circular orbits of gravitational attracted objects:
F
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GM E M sat
v2
 M sat
2
RES
RES
PHY 113 C Fall 2013 -- Lecture 14
RES
F
a
v
Msat
3
1
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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/s2 towards earth
m
r=4.4RE
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GM E m
 ma
r2
GM
GM E
a 2E 
r
4.4RE 2
PHY 113 C Fall 2013 -- Lecture 14
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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/s2.
Determine the orbital period of the satellite.
GM E m
v2
 ma  m
2
r
r
v 2 2r / T   2 

  r
r
r
T 
2
2
r
1/ 4
r
 GM 
 2  3 E 
a
 a 
 T  2
1/ 4
 6.67 1011  5.981024 

T  2 
6.793


PHY 113 C Fall 2013 -- Lecture 14
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 6675s
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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?
W
RM


R
GM M m M
GM M m
dr 
2
r 
r

RM
GM M m
RM
iclicker question
A. W>0
B. W<0
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PHY 113 C Fall 2013 -- Lecture 14
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2
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From Webassign Assignment #12:
A space probe is fired as a projectile from the Earth's surface
with an initial speed of 1.74 104 m/s. What will its speed be
when it is very far from the Earth? Ignore atmospheric friction
and the rotation of the Earth.
Ki  U i  K f  U f
vi
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1 2 GM E m 1 2
mvi 
 mv f
2
2
RE
PHY 113 C Fall 2013 -- Lecture 14
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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 1030 kg.)
v 2 GM 2
M

R 2 R 2
Know v, T ; want t o know M
2R
vT
R
T
2
After some algebra :
v
M 
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PHY 113 C Fall 2013 -- Lecture 14
4Tv 3
2G
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 1030 kg.)
iclicker exercise:
Who might pose a question like this?
A. A mean professor.
B. A puzzle master.
C. An observational astronomer.
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PHY 113 C Fall 2013 -- Lecture 14
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3
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Meanwhile – back on the surface of the Earth:
Conditions for stable equilibrium
Balance of force :
F  0
i
i
t
Balance of torque :
0
i
i
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PHY 113 C Fall 2013 -- Lecture 14
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Stability of “rigid bodies”
N
mig
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PHY 113 C Fall 2013 -- Lecture 14
Center-of-mass
rCM 
11
 mi ri
i
 mi
i
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.
ri
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mi
rCM
t   ri  mi g  j  rCM  Mg  j
i
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 Fi  0
Notion of equilibrium:
q
 ti  0
i
Notion of stability:
F=ma 
r
i
T- mg cos q  0
mg sin q  maq
T
t=I a  r mg sin q = mr2 a  mraq
mg(-j)
Example of stable equilibrium for
q  0
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PHY 113 C Fall 2013 -- Lecture 14
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Unstable equilibrium:
Support above CM:
Support
below CM:
q
r
T
mg(-j)
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PHY 113 C Fall 2013 -- Lecture 14
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Nik Wallenda walking on high wire across
Grand Canyon
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PHY 113 C Fall 2013 -- Lecture 14
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Analysis of stability:
 Fi  0
i
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 ti  0
i
PHY 113 C Fall 2013 -- Lecture 14
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*
X
*
Forces :
Torques :
n  M D g  mc g  m P g  0
M D g (1m )  m c gx  0
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PHY 113 C Fall 2013 -- Lecture 14
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PHY 113 C Fall 2013 -- Lecture 14
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RCM
Fg1
*
X
*
mg
 F g 1 ( 2 m )  mg ( R CM )  0
Torques :
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PHY 113 C Fall 2013 -- Lecture 14
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iclicker question:
F2
F1
L/3
mg
Mg
L
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 F1 and F2?
(a) F1 and F2 are both up as shown.
(b) F1 is up but F2 is down.
(c) F1 is down but F2 is up.
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F1
PHY 113 C Fall 2013 -- Lecture 14
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F2
*
X
*
L/3
mg
Mg
L
Forces :
Torques :
F2 
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F1  F2  Mg  mg  0
F2
3
Mg  3 mg
2
L
L
 Mg
 mgL  0
3
2
F1  
PHY 113 C Fall 2013 -- Lecture 14
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Mg  2 mg
2
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7
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iclicker question:
The fact that we found F1<0 means:
A. We set up the problem incorrectly
B. The analysis is correct, but the
direction of F1 is opposite to the arrow
C. Physics makes no sense
iclicker question:
What would happen if we analyzed this
problem by placing the pivot point at F1 ?:
A. The answer would be the same.
B. The answer would be different.
C. Physics makes no sense
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PHY 113 C Fall 2013 -- Lecture 14
t 0
22

 T sin   0
2
mgx /   Mg / 2
T 
sin 
 mgx  Mg
For   53 o   8 m x  2 m
mg  600 N Mg  200 N
T
T  313 N
*
X
*
x
Mg
mg
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/ 2

PHY 113 C Fall 2013 -- Lecture 14
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*
d
X
*
t 0
mgd  F 2 r  h   0
F
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mgd
 mg
2r  h
PHY 113 C Fall 2013 -- Lecture 14
r 2  r  h 
2r  h
2
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Mg = 120 N
Fwall
mg = 98 N
T < 110 N
T  Fwall  0
N  mg  Mg  0
MgL
 mgxL cos q 
cos q
2
 Fwall L sin q  0
mg
Mg
T 
N *
X
* T
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sin q
PHY 113 C Fall 2013 -- Lecture 14
x
f q
25
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.
Fwall
*
X
*
Fwall  f  0
a
1
x  mg a  0
L
2
ax
a
 f  Mg
 mg
Lh
2h
 Fwall h  Mg
Fwall
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mgx  Mg / 2  cos q
PHY 113 C Fall 2013 -- Lecture 14
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