Physics 121,Lecture 16,Pg 1
Physics 121,Sections 9,10,11,and 12
Lecture 15
Today’s Topics:
? Homework 6,
? Chap,6,# 6,12,20,24,29,38,52,57,78,and 83.
? Midterm 1:
? Average,65%
? Chapter 7:
? Collision
? 2-D collisions
? Chapter 8:
? Torque
? C.M.
M i d t e rm 1
0
2
4
6
8
10
12
14
16
5
10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95
1
0
0
M
o
re
S c o r e
N
u
mb
e
r
o
f
s
t
u
d
e
n
t
s
Physics 121,Lecture 16,Pg 2
Impulse-momentum theorem:
? The impulse of the force action on an object equals the change
in momentum of the object:
F?t = ?p = mvf – mvi
Impulse? F?t = ?p
? The impulse imparted by a force during the time interval ?t is
equal to area under the force-time graph from beginning to the
end of the time interval.
F
t
ti tf
?t
Fav?t
F
t
ti tf
?t
Impulse has units of Ns.
Physics 121,Lecture 16,Pg 3
Average Force and Impulse
?t
F
t
F
t
?t
?t big,Fav small
?t small,Fav big
soft spring
stiff spring
Fav
Fav
Physics 121,Lecture 16,Pg 4
Momentum Conservation
? The concept of momentum conservation is one of the most
fundamental principles in physics.
? This is a component (vector) equation.
?We can apply it to any direction in which there is no
external force applied.
? You will see that we often have momentum conservation
even when energy is not conserved.
t
p
E X T ?
??F 0???tp 0?EXTF
Physics 121,Lecture 16,Pg 5
Comment on Energy Conservation
? We have seen that the total kinetic energy of a system
undergoing an inelastic collision is not conserved.
?Energy is lost:
? Heat (bomb)
?Bending of metal (crashing cars)
? Kinetic energy is not conserved since work is done during
the collision !
? Momentum along a certain direction is conserved when
there are no external forces acting in this direction.
?In general,easier to satisfy than energy conservation.
Physics 121,Lecture 16,Pg 6
Ballistic Pendulum
? A projectile of mass m moving horizontally with speed v
strikes a stationary mass M suspended by strings of length
L,Subsequently,m + M rise to a height of H,
Given H,what is the initial speed v of the projectile?
H
L L
L L
m
M
M + mv
V
V=0
Physics 121,Lecture 16,Pg 7
Ballistic Pendulum...
? Two stage process:
1,m collides with M,inelastically,Both M and m
then move together with a velocity V (before
having risen significantly).
2,M and m rise a height H,conserving energy E.
(no non-conservative forces acting after collision)
Physics 121,Lecture 16,Pg 8
Ballistic Pendulum...
? Stage 1,Momentum is conserved
in x-direction,mv m M V? ?( )
V mm M v? ???? ???
? Stage 2,Energy is conserved
( )E EI F?
12 2( ) ( )m M V m M gH? ? ?V gH2 2?
Eliminating V gives,gH2m
M1v ?
?
??
?
? ??
Physics 121,Lecture 16,Pg 9
Ballistic Pendulum
? If we measure the forward displacement d,not H:
H
L L
L L
m
M
M + mv
d
L
d H
L-H
? ?L d L H
H L L d
2 2 2
2 2
? ? ?
? ? ?
Physics 121,Lecture 16,Pg 10
Ballistic Pendulum
L
d H
L-H
??
H ? L ? L 2 ? d 2
? L ? L 1 ?
d 2
L 2
? L ? L 1 ?
d 2
2 L 2
??
??
??
??
??
?? ?
d 2
2 L
d
L ?? 1for
gH2mM1v ?????? ?? LgdmM1v ???????? ??
for d << L
Physics 121,Lecture 16,Pg 11
Example - Elastic Collision
? Suppose I have 2 identical bumper cars,One is motionless
and the other is approaching it with velocity v1,If they
collide elastically,what is the final velocity of each car?
Note that this means,
m1 = m2 = m
v2B = 0
Physics 121,Lecture 16,Pg 12
Example - Elastic Collision
? Let’s start with the equations for conserving momentum and
energy,
mv1,B = mv1,A + mv2,A (1)
1/2 mv21,B = 1/2 mv21,A + 1/2 mv22,A (2)
Initially,v2 = 0,After the collision,v1 = 0.
To satisfy equation (1),v2A = v1B,
2221221 )( AAAA vvvv ??? 02 21 AA vv
v1A=0,v2A=0 or both Both,contradicts (2)
If v2A=0, v1A= v1B from (1) or (2) no collision !
Physics 121,Lecture 16,Pg 13
Lecture 16,ACT 1
Elastic Collisions
? I have a line of 3 bumper cars all touching,A
fourth car smashes into the others from behind,
Is it possible to satisfy both conservation of
energy and momentum if two cars are moving
after the collision?
All masses are identical,elastic collision.
A) Yes B) No
Before
After?
Physics 121,Lecture 16,Pg 14
Glancing Collisions
? x component:
? y component:
? KE:
m1 m
2
x
y
v1i
v1f
v2f
?? c o sc o s0 221111 ffi vmvmvm ???
??
0 ? 0 ? m 1 v 1 f s in ? ? m 2 v 2 f s in ?
222212112121121 ff vmvmvm ??
?
?
Physics 121,Lecture 16,Pg 15
Example of 2-D elastic collisions:
Billiards
? If all we know is the initial velocity of the cue ball,we don’t
have enough information to solve for the exact paths after
the collision,But we can learn some useful things...
Physics 121,Lecture 16,Pg 16
Billiards
? Consider the case where one ball is initially at rest,
pa
pb
F Pa
before afterthe final direction of the
red ball will depend on
where the balls hit.
vcm
Physics 121,Lecture 16,Pg 17
? We know momentum is conserved,pb = pa + Pa
? We also know that energy is conserved:
? Comparing these two equations tells us that,
Billiards
pb2 = (pa + Pa )2 = pa2 + Pa2 + 2 pa Pa cos ?
m2
P
m2
p
m2
p 2a2a2b ??
pa Pa cos ? = 0
and must therefore be orthogonal!
Or … one momentum must be zero.
pa
pb
Pa
Ppp 2a2a2b ??
Physics 121,Lecture 16,Pg 18
Billiards
? The final directions are separated by 90o.
pa
pb
F Pa
before after
vcm
Active Figure
Physics 121,Lecture 16,Pg 19
Rocket Propulsion
? The operation of a rocket depends on the law of conservation of
momentum as applied to a system,where the system is the
rocket plus its ejected fuel.
)()()( evvmvvMvmM ????????
)( evmvM ???
??
? m ? ? ? M ? M ? v ? ? v e ? M ? ? MM ? ? v e ? v
???
?
???
?
??
f
i
eif M
Mvvv ln
And after a little calculus we get:
M?m M?mv v ?v
ve v?m =v-ve
Physics 121,Lecture 16,Pg 20
Rocket Propulsion
? The thrust is the force exerted on the rocket by the ejected
exhaust
? But,we found
? And ?m= - ?M
? So the thrust increases if the exhaust gases go faster and/or the
rate of change of mass (burn rate) gets larger.
??
M ? v ? ? m (v e )
??
F e ? Ma ? M ? v? t
??
Fe ? ve ?M?t
Physics 121,Lecture 16,Pg 21
? In general,the rotation variables are vectors (have
a direction)
? If the plane of rotation is in the x-y plane,then the
convention is
?CCW rotation is in
the + z direction
?CW rotation is in
the - z direction
Chap.8,Rotational Dynamics
Direction of Rotation:
x
y
z
x
y
z
Physics 121,Lecture 16,Pg 22
Direction of Rotation:
The Right Hand Rule
? To figure out in which direction the rotation
vector points,curl the fingers of your right
hand the same way the object turns,and
your thumb will point in the direction of the
rotation vector !
? We normally pick the z-axis to be the rotation
axis as shown.
????= ?z
???= ?z
???= ?z
? For simplicity we omit the subscripts unless
explicitly needed.
x
y
z
x
y
z
Physics 121,Lecture 16,Pg 23
Rotational Dynamics
What makes it spin? Torque
? The tendency of a force to rotate an object about some
axis is measured by a quantity called the torque ?
? = Fd
d
?
L
F
?
L
F
?
F cos ?
F sin ?
Physics 121,Lecture 16,Pg 24
Torque
? The lever arm is the perpendicular distance from the
axis of rotation to a line drawn along the direction of
the force
? = Fd
d
?
L
F
?
L
F
?
F cos ?
F sin ?
The sign of the torque is positive if its turning tendency is counterclockwise
and negative if its turning tendency is clockwise (right-hand rule)
Physics 121,Lecture 16,Pg 25
Torque
o d1
d2
The force F1 tends to rotate the object
counterclockwise about O and F2 tends to
rotate the Object clockwise
F1
F2
?? =?1????2 ??F1d1 - F2d2
Physics 121,Lecture 16,Pg 26
Example,The Spinning Crate
1.0 m
d2 d1
F1=500 N
F2=500 N
The net torque action on a the
crate about O,
?? =?1????2 ??F1d1 - F2d2
= - (500 N)(0.5m)-(500 N)(0.5m)
= 500 N mO
Physics 121,Lecture 16,Pg 27
Example,The Swinging Door
? A top view of a door being pulled by a 300 N force
600
300 N
260 N
150 N
Fx = 300 N cos 600 = 150 N
Fy = 300 N sin 600 = 260 N
?????260 N) (2.0 m) = 520 Nm
2.0 m
Physics 121,Lecture 16,Pg 28
Lecture 16,Act 2
Torque
? In which of the cases shown below is the torque
provided by the applied force about the rotation axis
biggest? In both cases the magnitude and direction
of the applied force is the same.
(a) case 1
(b) case 2
(c) same
L
L
F F
axis
case 1 case 2
Physics 121,Lecture 16,Pg 29
Conditions for Equilibrium
? An object in mechanical equilibrium must satisfy:
1,The net external force must be Zero,SF = 0
2,The net external torque must be Zero,S??= 0
The first condition is translational equilibrium and the second
is rotational equilibrium
If the object is in equilibrium,then the choice of axis of
rotation does not influence the calculation of the net torque;
The location of the axis is arbitrary.
Physics 121,Lecture 16,Pg 30
Objects in Equilibrium
? Walking a Horizontal Beam
A Uniform horizontal 300 N beam,5.00 m long,is
attached to a wall by a pin connection that
allows the beam to rotate,Its far end is supported by
cable that makes an angle of 53.00,If a 600 N person
stand 1.5 m from the wall,find the tension in the cable
and the force exerted by the wall on the beam.
53.00
5.00 m
600 N
Physics 121,Lecture 16,Pg 31
Walking a Horizontal Beam
Fx = Rx - T cos 53.00 = 0
Fy = Ry - T sin 53.00 – 600 N – 300 N = 0
?0???T sin 53.00)(5.00 m)
- (300 N)(2.5 m)
- (600 N)(1.5 m) = 0
? T = 413 N,Rx = 249 N,Ry = 570 N
53.00
300 N600 N
o
300 N600 N
o
Physics 121,Lecture 16,Pg 32
System of Particles:
? Until now,we have considered the behavior of very simple
systems (one or two masses).
? But real life is usually much more interesting !
? For example,consider a simple rotating disk.
? An extended solid object (like a disk) can be thought of as a
collection of parts,The motion of each little part depends on
where it is in the object!
Physics 121,Lecture 16,Pg 33
System of Particles,Center of Mass
? How do we describe the,position” of a system made
up of many parts?
? Define the Center of Mass (average position):
?For a collection of N individual pointlike particles
whose masses and positions we know:
R
r
CM
i i
i
N
i
i
N
m
m
? ?
?
?
?
1
1
y
x
r1
m1
r3
r
2
r
4
m4
m2
m3RCM
(In this case,N = 4)
Physics 121,Lecture 16,Pg 34
System of Particles,Center of Mass
? If the system is made up of only two particles:
R
r
r r
CM
i i
i
N
i
i
N
m
m
m m
m m
? ?
?
?
?
?
?
?
1
1
1 1 2 2
1 2
y
x
r2r
1
m1 m2RCM
r2 - r1
? ? ? ?? ?? ? ? ?
?
m m m
m m
1 2 1 2 2 1
1 2
r r r
? ?R r r rCM mM? ? ?1 2 2 1
where M = m1 + m2
So:
Physics 121,Lecture 16,Pg 35
System of Particles,Center of Mass
? If the system is made up of only two particles:
y
x
r2r
1
m1 m2RCM
r2 - r1
where M = m1 + m2
+
? ?R r r rCM mM? ? ?1 2 2 1
If m1 = m2
the CM is halfway between
the masses.
? ?R r r rCM ? ? ?1 2 112
Physics 121,Lecture 16,Pg 36
System of Particles,Center of Mass
? If the system is made up of only two particles:
y
x
r2r
1
m1
m2
RCM
r2 - r1
where M = m1 + m2
+
If m1 = 3m2
the CM is now closer to
the heavy mass.
? ?R r r rCM mM? ? ?1 2 2 1
? ?R r r rCM ? ? ?1 2 114
Physics 121,Lecture 16,Pg 37
System of Particles,Center of Mass
? The center of mass is where the system is balanced !
?Building a mobile is an exercise in finding centers of mass.
m1 m2+m1 m2+
Physics 121,Lecture 16,Pg 38
System of Particles,Center of Mass
? We can consider the components of RCM separately:
(,,),,X Y Z
m x
M
m y
M
m z
MCM CM CM
i ii i ii i ii? ?
?
??
?
?
??? ? ?
y
x
r1
m1
r3
r
2
r
4
m4
m2
m3RCM
(In this case,N = 4)
Physics 121,Lecture 16,Pg 39
Example Calculation:
? Consider the following mass distribution:
(24,0)
12m4 24m12)m2(0mM xmX i iiCM ??????
6m4 0m12)m2(0mM ymY i iiCM ??????
(0,0)
(12,12)
m
2m
m
RCM = (12,6)
Physics 121,Lecture 16,Pg 40
System of Particles,Center of Mass
The location of the center
of mass is an intrinsic
property of the object !!
(it does not depend on where
you choose the origin or
coordinates when
calculating it).
y
xR
CM
? We find that the Center of Mass is at the,center”
of the object.
Physics 121,Lecture 16,Pg 41
Center of Mass
? We can use intuition to find the location of the center of mass for
symmetric objects that have uniform density:
? It will simply be at the geometrical center !
+
CM
+ +
+ + +
Physics 121,Lecture 16,Pg 42
System of Particles,Center of Mass
? The center of mass for a combination of objects is the average
center of mass location of the objects:
+
+
y
x
R2
R1
RCM
m1
m2
+
21
2211
CM
N
1i
i
N
1i
ii
CM
mm
mm
m
m
?
?
?
?
?
?
?
?
RR
R
R
R
R2 - R1
so if we have two objects:
? ?1221 Mm RRR ???
Physics 121,Lecture 16,Pg 43
System of Particles,Center of Mass
? The center of mass (CM) of an object is where we can freely
pivot that object.
? Gravity acts on the CM of an object (show later)
? If we pivot the object
somewhere else,it will
orient itself so that the
CM is directly below
the pivot.
? This fact can be used to find
the CM of odd-shaped objects.
+
CM
pivot
+ CM
pivotpivot
CM
mg
Physics 121,Lecture 16,Pg 44
System of Particles,Center of Mass
? Hang the object from several pivots and see where the
vertical lines through each pivot intersect !
pivot
pivot pivot
+
CM
? The intersection point must be at the CM.
Physics 121,Lecture 16,Pg 45
Statics:
? Objects are at rest (Static) when,
? ?? 0F ?? 0
When choosing axes about which to calculate torque,we
can be clever and make the problem easy....
No translation No rotation
AND
Physics 121,Lecture 16,Pg 46
Recap of today’s lecture
? Homework 6,
?Chap,6,# 6,12,20,24,29,38,52,57,78,and 83.
? Midterm 1:
?Average,65%
? Chapter 7:
?Linear momentum
?Collision
? Chapter 8:
?Torque
?C.M.