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PHYSICS 231
Lecture 18: equilibrium & revision
Remco Zegers
Walk-in hour: Thursday 11:30-13:30 am
Helproom
PHY 231
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gravitation
Only if an object is near the surface of earth one can use:
Fgravity=mg with g=9.81 m/s2
In all other cases:
Fgravity= GMobjectMplanet/r2 with G=6.67E-11 Nm2/kg2
This will lead to F=mg but g not equal to 9.8 m/s2 (see
Previous lecture!)
If an object is orbiting the planet:
Fgravity=mac=mv2/r=m2r with v: linear velocity =angular vel.
Our solar system!
So: GMobjectMplanet/r2 = mv2/r=m2r
Kepler’s 3rd law: T2=Ksr3 Ks=2.97E-19 s2/m3 r: radius of planet
T: period(time to make one rotation) of planet
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Previously
Torque: =Fd
Center of
Gravity:
m x

m
i
xCG
i
i
i
m y

m
i
i
yCG
i
i
i
i
Translational equilibrium: F=ma=0 The center of gravity
does not move!
Rotational equilibrium:
=0
The object does not
rotate
Mechanical equilibrium: F=ma=0 & =0 No movement!
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examples: A lot more in the book!
Where is the center of gravity?
m x

m
i
xCG
i
i
i
16  0  1  0.1  cos(530 )  1  0.1  cos(530 ) 0.12


 0.0067
16  1  1
18
i
m y

m
i
yCG
i
i
i
16  0  1  0.1  sin(530 )  1  0.1  sin(530 )

0
16  1  1
i
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T1
T2
Weight of board: w
What is the tension in each of the
wires (in terms of w)?
Translational equilibrium
F=ma=0
T1+T2-w=0 so T1=w-T2
Rotational equilibrium
w
=0
T10-0.5*w+0.75*T2=0
T2=0.5/0.75*w=2/3w
0
PHY 231
T1=1/3w
T2=2/3w
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sn
n
w
w
(x=0,y=0)
s=0.5 coef of friction between the wall and
the 4.0 meter bar (weight w). What is the
minimum x where you can hang a weight w
for which the bar does not slide?
Translational equilibrium (Hor.)
T T
Fx=ma=0
y
n-Tx=n-Tcos37o=0 so n=Tcos37o
Translational equilibruim (vert.)
Tx
Fy=ma=0
sn-w-w+Ty=0
sn-2w+Tsin37o=0
sTcos370-2w+Tsin370=0
1.00T=2w
Rotational equilibrium:
=0
xw+2w-4Tsin370=0 so w(x+2-4.8)=0
x=2.8 m
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Tips for study
• Look through the lecture sheets and pick out the
summaries to get a good overview
• Make an overview for yourself (about 1 Letter
size paper)
• Read the chapters in the book to make sure that
your overview contains all the main issues.
• Study the examples given in the lectures
• Study the problems in LON-CAPA
• Study the worked-out examples in the book
• Practice the previous midterm exams.
• Practice problems from the book
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Revision: chapter 5
 Work: W=Fcos()x
 Power: P=W/t
 Potential energy (PE)






Energy transfer
Rate of energy transfer
Energy associated with
position.
Gravitational PE: mgh
Energy associated with
position in grav. field.
PE stored in a spring: 1/2kx2 x is the compression of the spring
k is the spring constant
Kinetic energy KE: 1/2mv2 Energy associated with
motion
Conservative force:
Work done does not depend on path
Non-conservative force:
Work done does depend on path
Mechanical energy ME:
ME=KE+PE
 Conserved if only conservative forces are present
KEi+PEi=KEf+PEf
 Not conserved in the presence of non-conservative forces
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(KEi+PEi)-(KEf+PEf)=Wnc PHY 231
m=1 kg
example
A pendulum is pushed with initial
velocity 0.1 m/s from a height of
1 cm. How far does it compress the
spring? (assume m does not rise
significantly after hitting the spring)
1 cm
k=100 N/m
Conservation of ME:
(mgh+1/2mv2+1/2kx2)initial= (mgh+1/2mv2+1/2kx2)final
1*9.8*0.01+0.5*1*0.12+0.=0.+0.+0.5*100*x2 so x=0.045 m
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Saving electricity
A ‘smart’ student decides to save energy by connecting his
exercise treadmill to his laptop battery. If it takes
70 J to move the belt on the treadmill by 1 meter and
50% of the generated energy is stored in the battery, how
‘far’ must the student run to use his 100 W laptop for free
for 2 hours?
Work done by student: W=70*d J
Energy given to the battery 0.5W=35*d J
100 W laptop for 2 hours: 100 J/s*3600*2 s=7.2E+5 J
720 KJ
7.2E+5=35*d so d=(7.2E+5)/35=2.1E+4 m =21km!!!
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Non-conservative vs conservative case
h
450
A block of 1 kg is pushed up a 45o
slope with an initial velocity of 10 m/s.
How high does the block go if:
a) there is no friction
b) if the coefficient of kinetic friction
is 0.5.
A) Conservation of ME: (mgh+1/2mv2)initial= (mgh+1/2mv2)final
0.+0.5*1*102=1*9.8*h+0. So h=5.1 m
B) Energy is lost to friction:
(mgh+1/2mv2)initial= (mgh+1/2mv2)final+Wfriction
[W=Fx=nx=mgcos(45o)h/sin(45o)=0.5*1*9.8*h=4.9h]
0.+0.5*1*102=1*9.8*h+0.+4.9h so h=3.4 m
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Chapter 6
Momentum p=mv
F=p/t
Impulse (the change in momentum) p= Ft
Inelastic collisions
Elastic collisions
•Momentum is conserved
•Momentum is conserved
•Some energy is lost in the
•No energy is lost in the
collision: KE not conserved
collision: KE conserved
•Perfectly inelastic: the
objects stick together.
Conservation of momentum: Conservation of momentum:
m1v1i+m2v2i=(m1+m2)vf
m1v1i+m2v2i=m1v1f+m2v2f
Conservation of KE:
½m1v1i2+½m2v2i2=½m1v1f2+½m2v2f2
(v1i-v2i)=(v2f-v1f)
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Inelastic collision
0.1 kg V m/s
5 m/s
5 kg
An excellent, but somewhat desperate sharp shooter tries
to stop a cannon ball aimed directly at him by shooting a bullet
from his gun against it. With what velocity does he need to
shoot the bullet to stop the cannon ball assuming that the
bullet gets stuck in the ball? How much energy is released?
Inelastic: only conservation of momentum.
m1v1i+m2v2i=(m1+m2)vf so 0.1V-5*5=0
v=25/0.1=250 m/s
Change in kinetic energy:
Before: ½m1v1i2+½m2v2i2=3125+62.5=3187.5 J After: 0 J
Release: 3187.5 J
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PHY 231
h=100 m
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2
Elastic collision
Two balls (m1=1 kg, m2=2 kg) are
released on a slope and collide in
the valley. How far does each go
back up?
Step 1: calculate their velocities just before the collision:
ball 1: cons. of ME: mgh=0.5mv2 9.8*100=0.5 v2 v=44. m/s
ball 2: cons. of ME:
2*9.8*100=0.5*2*v2 v=-44. m/s
Step 2: Collision, use cons. of P and KE (simplified).
m1v1i+m2v2i=m1v1f+m2v2f so 44-88=v1f+2v2f
(v1i-v2i)=(v2f-v1f)
so 88=v2f-v1f 3v2f=44 v2f=15 v1f=-73
Step 3: Back up the ramp: cons. of ME:
ball 1: 0.5mv2=mgh 0.5*732=9.8h h=272 m
ball 2: 0.5mv2=mgh 0.5*152=9.8h h=11.5 m
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PHY 231
Bouncing ball
A 0.5 kg ball is dropped to the floor from a height of 2 m.
If it bounces back to a height of 1.8 m, what is the
magnitude of its change in momentum?
Some energy is lost in the bounce.
Just before it hits the ground, its velocity is:
(use conservation of ME)
mgh=1/2mv2 so v=(2gh)=(2*9.8*2)= 6.26 m/s
p=m[6.26-(-5.93)]=0.5*12.2=6.1 kgm/s
After the bounce it goes back up 1.8 m.
Just after it bounces back it velocity is:
(use conservation of ME)
mgh=1/2mv2 so v=(2gh)=(2*9.8*1.8)=5.93 m/s
Must be negative!! So -5.93 m/s
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Chapter 7
 f i
 Average angular


t f  ti
t velocity (rad/s)
 Instantaneous
  lim
t 0 t
Angular velocity
 f  i
 Average angular


t f  ti
t acceleration (rad/s2)
 Instantaneous angular
  lim
t 0 t
acceleration
2 rad  3600
10  2/360 rad
1 rad 360/2 deg
Be aware that sometimes rev/s or rev/min is asked
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Angular vs linear/tangential
v

r
linear
r
angular
a

r
See e.g. the bike example (lecture 16)
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Rotational motion
Angular motion
(t)= (0)+(0)t+½t2
(t)= (0)+t
Centripetal acceleration
ac=v2/r directed to the center of the circular motion
Also v=r, so ac=2r
Fto center=mac=mv2/r
This acceleration is caused by ‘known’ force
(gravitation, friction, tension…)
Make sure you understand how to use Kepler’s 3rd law
and the general definition of gravitational PE.
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Whirling ball
A ball of mass 2 kg is
attached to a string of 1m
and whirled around a
smooth horizontal table.
If the tension in the string
exceeds 200 N, it will break.
What will be
the speed at the moment
the string breaks?
v
Fto center=mac=mv2/r
T=2*v2/1=200 N
v=10 m/s
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Consider...
…a child playing on a swing. As she reaches the lowest
point in her swing, which of the following is true?
A) The tension in the robe is equal to her weight
B) The tension in the robe is equal to her mass times
her acceleration
C) Her acceleration is downward and equal to g (9.8 m/s2)
D) Her acceleration is zero
E) Her acceleration is equal to her velocity squared divided
by the length of the swing.
Fto center=T-mg=mac=mv2/L
Lowest point, so no linear acceleration!!!
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Two objects...
Are rotating. One starts with an initial linear velocity
of 1 m/s and rotates with a radius of 2 m. The other starts
from rest and undergoes a constant angular acceleration over
a circle with radius 3 m.
What should its angular acceleration be, so that it overtakes
(for the first time) the first object after 10 revolutions?
1=v1/r1=1/2=0.5 rad/s
1(t)= 1(0)+1(0)t+½1t2=0.5t
1(t)=10*2=0.5t so t=40
2(t)= 2(0)+2(0)t+½2t2=½2t2
20=½2(40)2 so 2=8E-03 rad/s
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Conical motion
Tcos
T


Tsin
mg
If the mass of the swinging object is 1 kg,
and =30o what should the velocity of the
object be so that does not sink or rise?
The length of the robe is 2 m.
Vertical direction:
F=ma
Tcos-mg=0
So T=mg/cos=1*9.8/0.866=11.3 N
Horizontal direction:
F=mac
Tsin=mac
11.3*0.5=mac=1v2/r
v2=11.3*0.5*2 so v=3.4 m/s
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Chapter 8
Summary: see beginning of this lecture!
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Opening a hatch door.
?N
30o
A person is trying to open a
1 meter, 20 kg, trap door by pulling
a rope attached to its non rotating
end at an angle of 30o.
With what force should he at least pull?
1m
Center of gravity of the door: halfway the door’s length: 0.5m
=-mdoorgdCG+Fpull,perpendicularl
0=-20*9.8*0.5+Fpullsin(30o)*1
Fpull=196 N
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