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M
OMENTUM!
Momentum
Impulse
Conservation of Momentum in
1 Dimension
Conservation of Momentum in
2 Dimensions
Angular Momentum
Torque
Moment of Inertia
Chapter 9: Linear Momentum and
Collisions
The linear momentum of a particle of mass m and velocity
v is defined as


p  mv
The linear momentum is a vector quantity.
It’s direction is along v.
The components of the momentum of a particle:
px  m  vx
py  m  vy
pz  m  vz
Momentum Facts
• p = mv
• Momentum is a vector quantity!
• Velocity and momentum vectors point in the same direction.
• SI unit for momentum: kg·m/s (no special name).
• Momentum is a conserved quantity (this will be proven later).
• A net force is required to change a body’s momentum.
• Momentum is directly proportional to both mass and speed.
• Something big and slow could have the same momentum as
something small and fast.
Momentum Examples
10 kg
3 m/s
10 kg
30 kg · m /s
Note: The momentum vector does not have to be drawn 10 times
longer than the velocity vector, since only vectors of the same
quantity can be compared in this way.
26º
5g
p = 45 kg · m /s
at 26º N of E
Equivalent Momenta
Car: m = 1800 kg; v = 80 m /s
p = 1.44 ·105 kg · m /s
Bus: m = 9000 kg; v = 16 m /s
p = 1.44 ·105 kg · m /s
Train: m = 3.6·104 kg; v = 4 m /s
p = 1.44 ·105 kg · m /s
continued on next slide
Equivalent Momenta
(cont.)
The train, bus, and car all have different masses and
speeds, but their momenta are the same in magnitude.
Only say that the magnitudes of their momenta are equal
since they’re aren’t moving in the same direction.
The difficulty in bringing each vehicle to rest--in terms of a
combination of the force and time required--would be the
same, since they each have the same momentum.
Impulse Defined
Impulse is defined as the product force acting on an
object and the time during which the force acts. The
symbol for impulse is J. So, by definition:
J=Ft
Example: A 50 N force is applied to a 100 kg boulder
for 3 s. The impulse of this force is J = (50 N) (3 s)
= 150 N · s.
Note that we didn’t need to know the mass of the
object in the above example.
Impulse Units
J = F t shows why the SI unit for impulse is the Newton · second.
proof: 1 N · s = 1 (kg · m /s2) (s) = 1 kg · m /s
Fnet = m a shows this is
equivalent to a newton.
Therefore, impulse and momentum have the same units, which leads
to a useful theorem.
Impulse - Momentum Theorem
The impulse due to all forces acting on an object (the net force) is
equal to the change in momentum of the object:
Fnet t =  p
We know the units on both sides of the equation are the same
(last slide), but let’s prove the theorem formally:
Fnet t = m a t = m( v / t)t = m v =  p
Stopping Time
Ft = Ft
Imagine a car hitting a wall and coming to rest. The force on the car due
to the wall is large (big F), but that force only acts for a small amount of
time (little t). Now imagine the same car moving at the same speed but
this time hitting a giant haystack and coming to rest. The force on the
car is much smaller now (little F), but it acts for a much longer time (big
t). In each case the impulse involved is the same since the change in
momentum of the car is the same. Any net force, no matter how small,
can bring an object to rest if it has enough time. A pole vaulter can fall
from a great height without getting hurt because the mat applies a smaller
force over a longer period of time than the ground alone would.
Impulse - Momentum Example
A 1.3 kg ball is coming straight at a 75 kg soccer player at 13 m/s who
kicks it in the exact opposite direction at 22 m/s with an average force
of 1200 N. How long are his foot and the ball in contact?
answer: We’ll use Fnet t = p. Since the ball
changes direction, p = m v = m (vf - v0)
= 1.3 [22 - (-13)] = (1.3 kg) (35 m/s)
= 45.5 kg · m /s. Thus, t = 45.5 / 1200
= 0.0379 s, which is just under 40 ms.
During this contact time the ball compresses substantially and then
decompresses. This happens too quickly for us to see, though. This
compression occurs in many cases, such as hitting a baseball or golf
ball.
Fnet (N)
Fnet vs. t graph
Net area = p
t (s)
6
A variable strength net force acts on an object in the positive direction
for 6 s, thereafter in the opposite direction. Since impulse is Fnet t, the
area under the curve is equal to the impulse, which is the change in
momentum. The net change in momentum is the area above the curve
minus the area below the curve. This is just like a v vs. t graph, in
which net displacement is given area under the curve.
• As long as there are no external forces
acting on a system of particles, collisions
between the particles will exhibit
conservation of linear momentum.
• This means that the vector sum of the
momenta before collision is equal to the
vector sum of the momenta of the particles
afterwards.


p  mv
Conservation of linear momentum
p   p  constant
or:
 p  p
i
f
p1,i  p2,i  p1, f  p2, f
Conservation of Momentum in 1-D
Whenever two objects collide (or when they exert forces on each
other without colliding, such as gravity) momentum of the system
(both objects together) is conserved. This mean the total momentum
of the objects is the same before and after the collision.
(Choosing right as the +
before: p = m1 v1 - m2 v2
v2
v1
m1
direction, m2 has - momentum.)
m2
m1 v1 - m2 v2 = - m1 va + m2 vb
after: p = - m1 va + m2 vb
va
m1
m2
vb
Directions after a collision
On the last slide the boxes were drawn going in the opposite direction
after colliding. This isn’t always the case. For example, when a bat hits
a ball, the ball changes direction, but the bat doesn’t. It doesn’t really
matter, though, which way we draw the velocity vectors in “after”
picture. If we solved the conservation of momentum equation (red box)
for vb and got a negative answer, it would mean that m2 was still moving
to the left after the collision. As long as we interpret our answers
correctly, it matters not how the velocity vectors are drawn.
v2
v1
m1
m2
m1 v1 - m2 v2 = - m1 va + m2 vb
va
m1
m2
vb
Simple Examples of Head-On Collisions
(Elastic)(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One at rest initially has twice the mass.
Collision between two objects. One not at rest initially has twice the mass.


p  mv
Simple Examples of Head-On Collisions
(Totally Inelastic Collision, only Momentum Conserved)
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One at rest initially has twice the mass.
Collision between two objects. One not at rest initially has twice the mass.


p  mv
Elastic and inelastic collisions in one dimension
Momentum is conserved in any collision, elastic and
inelastic.
Mechanical Energy is only conserved in elastic collisions.
Perfectly inelastic collision: After colliding, particles stick
together. There is a loss of energy (deformation).
Elastic collision: Particles bounce off each other without loss of
energy.
Inelastic collision: Particles collide with some loss of energy,
but don’t stick together.
Perfectly inelastic collision of two particles
(Particles stick together)
 
pi  p f



m1v1i  m2v2i  (m1  m2 )v f
Notice that p and v are
vectors and, thus have a
direction (+/-)
Ki  Eloss  K f
1
1
1
2
2
2
m1v1i  m2v2i  (m1  m2 )v f  Eloss
2
2
2
There is a loss
in energy Eloss
For elastic collisions in one dimension:
Suppose we know the initial masses and velocities.
Then:
 m1  m2 
 2m2 
v1i  
v2i
v1 f  
 m1  m2 
 m1  m2 
(9.20)
 2m1 
 m2  m1 
v1i  
v2i
 
 m1  m2 
 m1  m2 
(9.21)
v2 f
Note, that these are pretty specialized equations, (elastic collision in one dimension,
known initial velocities, and masses)
Black board example 9.2
Two carts collide elastically on a frictionless track. The first
cart (m1 = 1kg) has a velocity in the positive x-direction of
2 m/s; the other cart (m = 0.5 kg) has velocity in the
negative x-direction of 5 m/s.
(a) Find the speed of both carts after the collision.
(b) What is the speed if the collision is perfectly inelastic?
(c) How much energy is lost in the inelastic collision?
Black board example 9.3 and demo
Determining the speed of a bullet
A bullet (m = 0.01kg) is fired into a block (0.1 kg) sitting at the edge of a table. The
block (with the embedded bullet) flies off the table (h = 1.2 m) and lands on the
floor 2 m away from the edge of the table.
a.) What was the speed of the bullet?
b.) What was the energy loss in the bullet-block collision? (skip)
vb = ?
h = 1.2 m
x=2m
Motion of a System of Particles.
Newton’s second law for a System of Particles
The center of mass of a system of particles (combined mass M)
moves like one equivalent particle of mass M would move under
the influence of an external force.


Fnet  MaCM
Fnet , x  MaCM , x
Fnet , y  MaCM , y
Fnet , z  MaCM , z
Center of mass
Center of mass for many particles:

rCM 

 mi ri
i
M
Black board example 9.6
Where is the center of mass
of this arrangement of
particles.
(m3 = 2 kg; m1 = m2 = 1 kg)?
Velocity of the center of mass:

vCM 

 mi vi
i
M
Acceleration of the center of mass:

aCM 

 miai
i
M
A rocket is shot up in the air and explodes.
Describe the motion of the center of mass before and after
the explosion.
A method for finding the center of mass of any object.
- Hang object from two
or more points.
- Draw extension of
suspension line.
- Center of mass is at
intercept of these lines.
Impulse (change in momentum)

 

A change in momentum is called “impulse”: J  p  p f  pi
During a collision, a force F acts on
an object, thus causing a change in
momentum of the object:
For a constant (average) force:
tf


p  J   F (t )dt
ti
 
p  J  Favg  t
Think of hitting a soccer ball: A force F acting over a time t
causes a change p in the momentum (velocity) of the ball.
Black board example 9.6
A soccer player hits a ball (mass m =
440 g) coming at him with a
velocity of 20 m/s. After it was
hit, the ball travels in the
opposite direction with a velocity
of 30 m/s.
(a) What impulse acts on the ball
while it is in contact with the
foot?
(b) The impact time is 0.1s. What
average force is the acting on the
ball?
(c) How much work was done by the
foot? (Assume and elastic
collision.) (skip)
Sample Problem 1
35 g
7 kg
700 m/s
v=0
A rifle fires a bullet into a giant slab of butter on a frictionless surface.
The bullet penetrates the butter, but while passing through it, the bullet
pushes the butter to the left, and the butter pushes the bullet just as hard
to the right, slowing the bullet down. If the butter skids off at 4 cm/s
after the bullet passes through it, what is the final speed of the bullet?
(The mass of the rifle matters not.)
35 g
v=?
4 cm/s
7 kg
continued on next slide
Sample Problem 1
(cont.)
Let’s choose left to be the + direction & use conservation of
momentum, converting all units to meters and kilograms.
35 g
p before = 7 (0) + (0.035)(700)
7 kg
= 24.5 kg · m /s
v=0
35 g
4 cm/s
v=?
p before = p after
7 kg
700 m/s
p after = 7 (0.04) + 0.035 v
= 0.28 + 0.035 v
24.5 = 0.28 + 0.035 v
v = 692 m/s
v came out positive. This means we chose the correct
direction of the bullet in the “after” picture.
Sample Problem 2
35 g
7 kg
700 m/s
v=0
Same as the last problem except this time it’s a block of wood rather than
butter, and the bullet does not pass all the way through it. How fast do
they move together after impact?
v
7. 035 kg
(0.035) (700) = 7.035 v
v = 3.48 m/s
Note: Once again we’re assuming a frictionless surface, otherwise there
would be a frictional force on the wood in addition to that of the bullet,
and the “system” would have to include the table as well.
Proof of Conservation of Momentum
The proof is based on Newton’s 3rd Law. Whenever two objects collide
(or exert forces on each other from a distance), the forces involved are an
action-reaction pair, equal in strength, opposite in direction. This means
the net force on the system (the two objects together) is zero, since these
forces cancel out.
F
F
M m
force on M due to m
force on m due to M
For each object, F = (mass) (a) = (mass) (v / t) = (mass v)/ t = p / t.
Since the force applied and the contact time is the same for each mass,
they each undergo the same change in momentum, but in opposite
directions. The result is that even though the momenta of the individual
objects changes, p for the system is zero. The momentum that one
mass gains, the other loses. Hence, the momentum of the system before
equals the momentum of the system after.
Conservation of Momentum applies only
in the absence of external forces!
In the first two sample problems, we dealt with a frictionless surface.
We couldn’t simply conserve momentum if friction had been present
because, as the proof on the last slide shows, there would be another
force (friction) in addition to the contact forces. Friction wouldn’t
cancel out, and it would be a net force on the system.
The only way to conserve momentum with an external force like
friction is to make it internal by including the tabletop, floor, or the
entire Earth as part of the system. For example, if a rubber ball hits a
brick wall, p for the ball is not conserved, neither is p for the ballwall system, since the wall is connected to the ground and subject to
force by it. However, p for the ball-Earth system is conserved!
Black board example 9.1
(similar to blocks and spring HW problem)
You (100kg) and your skinny friend (50.0 kg)
stand face-to-face on a frictionless, frozen
pond. You push off each other. You move
backwards with a speed of 5.00 m/s.
(a) What is the total momentum of the you-andyour-friend system?
(b) What is your momentum after you pushed
off?
(c) What is your friends speed after you pushed
off?
(d) How much energy (work) did you and your
friend expend?(skip)
Sample Problem 3
An apple is originally at rest and then dropped. After falling a short
time, it’s moving pretty fast, say at a speed V. Obviously, momentum
is not conserved for the apple, since it didn’t have any at first. How can
this be?
answer: Gravity is an external force on the
apple
m
V
F
v
Earth
M
apple, so momentum for it alone is not
conserved. To make gravity “internal,” we
must define a system that includes the
other object responsible for the
gravitational force--Earth. The net force
on the apple-Earth system is zero, and
momentum is conserved for it. During the
fall the Earth attains a very small speed v.
So, by conservation of momentum:
F
mV = M v
Sample Problem 4
A crate of raspberry donut filling collides with a tub of lime Kool Aid
on a frictionless surface. Which way on how fast does the Kool Aid
rebound? answer: Let’s draw v to the right in the after picture.
3 (10) - 6 (15) = -3 (4.5) + 15 v
v = -3.1 m/s
Since v came out negative, we guessed wrong in drawing v to the
right, but that’s OK as long as we interpret our answer correctly.
After the collision the lime Kool Aid is moving 3.1 m/s to the left.
before
3 kg
10 m/s
6 m/s
15 kg
after
4.5 m/s
3 kg
15 kg
v
Conservation of Momentum in 2-D
To handle a collision in 2-D, we conserve momentum in each
dimension separately.
Choosing down & right as positive:
m1
v1
m2
2 v
2
1
a
m1
va
m2
vb
b
before:
px = m1 v1 cos1 - m2 v2 cos2
py = m1 v1 sin1 + m2 v2 sin2
after:
px = -m1 va cosa + m2 vb cos b
py = m1 va sina + m2 vb sin b
Conservation of momentum equations:
m1 v1 cos1 - m2 v2 cos2 = -m1 va cosa + m2 vb cos b
m1 v1 sin1 + m2 v2 sin 2 = m1 va sina + m2 vb sin b
Conserving Momentum w/ Vectors
B
E m1
1
F
O
p1
R
E
A
F
T
E
R
a
m1
pa
2
p1
m2
p before
p2
p2
m2
pa
b
pb
p after
pb
This diagram shows momentum vectors, which are parallel to
their respective velocity vectors. Note p1 + p 2 = p a + p b and
p before = p after as conservation of momentum demands.
Exploding Bomb
Acme
after
before
A bomb, which was originally at rest, explodes and shrapnel flies
every which way, each piece with a different mass and speed. The
momentum vectors are shown in the after picture.
continued on next slide
Exploding Bomb
(cont.)
Since the momentum of the bomb was zero before the
explosion, it must be zero after it as well. Each piece does
have momentum, but the total momentum of the exploded
bomb must be zero afterwards. This means that it must be
possible to place the momentum vectors tip to tail and form a
closed polygon, which means the vector sum is zero.
If the original momentum of
the bomb were not zero,
these vectors would add up
to the original momentum
vector.
Two-dimensional collisions (Two particles)
Conservation of momentum:
pi  p f
m1v1i  m2v2i  m1v1 f  m2v2 f
Split into components:
p x ,i  p x , f
m1v1ix  m2v2ix  m1v1 fx  m2v2 fx
p y ,i  p y , f
m1v1iy  m2v2iy  m1v1 fy  m2v2 fy
If the collision is elastic, we can also use conservation of energy.
Velocity Components in Projectile Motion
(In the absence of air resistance.)
Note that the horizontal component
of the velocity remains the same if air
resistance can be ignored.
Example of Non-Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
If you vector add the total momentum after collision,
you get the total momentum before collision.


p  mv
2-D Sample Problem
152 g
before
40
34 m/s
0.3 kg
5 m/s
A mean, old dart strikes an innocent
mango that was just passing by
minding its own business. Which
way and how fast do they move off
together?
Working in grams and taking left & down as + :
152 (34) sin 40 = 452v sin
152 (34) cos 40 - 300 (5) = 452 v cos
after
452 g

v
Dividing equations : 1.35097 = tan
 = 53.4908
Substituting into either of the first two
equations :
v = 9.14 m/s
Alternate Solution
40
5168
Shown are momentum vectors (in g m/s).
The black vector is the total momentum
before the collision. Because of
conservation of momentum, it is also the
total momentum after the collisions. We
can use trig to find its magnitude and
direction.

p
40
1500
Law of Cosines : p2 = 5168 2 + 1500 2 - 2 5168  1500 cos 40
p = 4132.9736 g m/s
Dividing by total mass : v = (4132.9736 g m/s) / (452 g) = 9.14 m/s
Law of Sines :
sin 
1500
sin 40
=
4132.9736
 = 13.4908
Angle w/ resp. to horiz. = 40 + 13. 4908   53.49
Comments on Alternate Method
• Note that the alternate method gave us the exact
same solution.
• This method can only be used when two objects
collide and stick, or when one object breaks into
two. Otherwise, we’d be dealing with a polygon
with more sides than a triangle.
• In using the Law of Sines (last step), the angle
involved (ß) is the angle inside the triangle. A little
geometry gives us the angle with respect to the
horizontal.
Black board example 9.5
Accident investigation. Two
automobiles of equal mass approach
an intersection. One vehicle is
traveling towards the east with 29 mi/h
(13.0 m/s) and the other is traveling
13.0 m/s
north with unknown speed. The
vehicles collide in the intersection and
stick together, leaving skid marks at an
angle of 55º north of east. The second
driver claims he was driving below the
speed limit of 35 mi/h (15.6 m/s).
a) Is he telling the truth?
??? m/s
b) What is the speed of the “combined vehicles” right after the
collision?
c) How long are the skid marks (mk = 0.5)?
ROTATIONAL INERTIA &
ANGULAR MOMENTUM
• For every type of linear quantity we have a
rotational quantity that does much the same
thing
Linear Quantities
Speed
Force
Mass
Momentum
Distance
Rotational Quantities
Rotational (Angular) Speed
Torque
Rotational Inertia
Angular Momentum
Angle
Rotational Inertia(I)
• AKA (not really but could be) Rotational
Mass
• Resistance to change in rotational motion
– Objects that are rotating about an axis tend to stay
rotating, objects not rotating tend to remain at rest,
unless an outside torque is applied
• A torque is required to change the status of an
object’s rotation
• It’s the rotational equivalent to mass,
– Harder to give an ang. acc. to an object w/ a larger I
Moment of Inertia
Any moving body has inertia.
The more inertia or rotational inertia a body has, the harder it is to
change its linear/rotational motion.
Single point-like mass
m
r
System of masses
Q
I=
m2
m1
r1
mr2
I=
 mi ri 2
= m1 r12 + m2 r22
r2
Q
Moment of Inertia Example
Two merry-go-rounds have the same mass and are spinning with the
same angular velocity. One is solid wood (a disc), and the other is a
metal ring. Which has a bigger moment of inertia relative to its center
of mass?
r
r

m

m
answer: I is independent of the angular speed. Since their masses and
radii are the same, the ring has a greater moment of inertia. This is
because more of its mass is farther from the axis of rotation. Since I
is bigger for the ring, it would more difficult to increase or decrease its
angular speed.
The big idea
• Rotational Inertia depends on mass and
radius
• If either one of these is large, then rotational
inertia is large, and object will be harder to
rotate
• Different types of objects have different
equations for rotational inertia
• But all equations have m and r2 in them.
Rotational Inertia (cont.)
• Some objects have
more rotational inertia
than others
– Objects with mass closer
to axis of rotation are
easier to rotate, b/c it they
have less rotational
inertia
– If the mass is farther
away from the axis, then
object will have more
rotational inertia, and will
therefore be harder to
rotate
Angular Momentum, L
Depends on linear momentum and the distance from a particular point.
If r and v are  then the magnitude of angular momentum w/ resp.
to point Q is given by L = rp = mvr.
In this case L points out of the page. If the mass were moving in the
opposite direction, L would point into the page.
Unit: kgm2 / s
v
r
Q
A torque is needed to change L, just
a force is needed to change p.
m Anything spinning has angular has
angular momentum. The more it
has, the harder it is to stop it from
spinning.
Angular Momentum: General Definition
If r and v are not  then the angle between these two vectors must
be taken into account. The general definition of angular momentum is
given by a vector cross product:
L = r p
This formula works regardless of the angle. As you know from our
study of cross products, the magnitude of the angular momentum
of m relative to point Q is: L = r p sin = m v r sin. In this
case, by the right-hand rule, L points out of the page. If the mass
were moving in the opposite direction, L would point into the
page.
v

r
Q
m
Comparison: Linear & Angular Momentum
Linear Momentum, p
Angular Momentum, L
• Tendency for a mass to continue • Tendency for a mass to continue
moving in a straight line.
rotating.
• Parallel to v.
• Perpendicular to both v and r.
• A conserved, vector quantity.
• A conserved, vector quantity.
• Magnitude is inertia (mass)
times speed.
• Magnitude is rotational inertia
times angular speed.
• Net force required to change it.
• Net torque required to change it.
• The greater the mass, the greater • The greater the moment of
the force needed to change
inertia, the greater the torque
momentum.
needed to change angular
momentum.
Angular Acceleration
Angular acceleration occurs when a spinning object spins faster or
slower.
 =   /t
Note how this is very similar to a =  v /t for linear acceleration.
Ex: If a wind turbine spinning at 21 rpm speeds up to 30 rpm over
10 s due to a gust of wind, its average angular acceleration is
9 rpm /10 s. This means every second it’s spinning 9 revolutions per
minute faster than the second before. Let’s convert the units:
9 rpm
9 rev
9 rev/min
9  (2 rad)
2
=
=
=
0.094
rad/s
=
10 s
min  10 s
10 s
(60 s)  10 s
Torque & Angular Acceleration
Newton’s 2nd Law, as you know, is Fnet = m a
The 2nd Law has a rotational analog:
 net = I 
A force is required for a body to undergo acceleration. A
“turning force” (a torque) is required for a body to undergo
angular acceleration.
The bigger a body’s mass, the more force is required to
accelerate it. Similarly, the bigger a body’s rotational inertia,
the more torque is required to accelerate it angularly.
Both m and I are measures of a body’s inertia
(resistance to change in motion).
Linear Momentum & Angular Momentum
Recall, angular momentum’s magnitude is given by
L = mvr
r
v
m
r must be perpendicular
So, if a net torque is applied, angular velocity must
change, which changes angular momentum.
proof:
 net = r sinθFnet = rperpm a
= rperp m  v / t =  L / t
So net torque is the rate of change of angular
momentum, just as net force is the rate of change of
linear momentum.
From the formula v = r , we get
L = mvrperp= m r (r ) = m rperp 2  = I 
Why does a tightrope walker carry a long
pole?
• The pole is usually fairly heavy and by carrying it,
he creates a lot of mass far away from the axis of
rotation
• This increases his rotational inertia
• And therefore makes it harder for him to rotate/tip
over
• http://www.youtube.com/watch?v=w8Tfa5fHr3s
Sports Connection
• Running
– When you run you
bend your legs to
reduce your
rotational inertia
• Gymnastics/Diving
– Pull body into tight
ball to achieve fast
rotation
Angular Momentum
• “inertia of rotation”
• Ang. Momentum= Rotational Inertia X Rotational
Speed
– L=Iω
Conservation of Angular Momentum
• If no outside torque is being applied, then total
angular momentum in a system must stay the
same
• This means, if you decrease radius, you
increase rotational speed
• Increase radius, then rotational speed
decreases
I – represents rotational inertia
ω -represents angular speed
Sports Connection…
• Ice skating
– Skater starts out in slow spin with arms and
legs out
– http://www.youtube.com/watch?v=AQLtcEAG9v0
– http://www.youtube.com/watch?v=NtEnEeEyw_s
– Skater pulls arms and legs in tight to body
– Skater is then spinning much fast (higher
rotational speed)
• Gymnastics (pummel horse or floor
routine)
– Small radius to achieve fast rotational speed
during moves, increase radius when low
rotational speed is desired (during landing)
Do cats violate physical law?
• Video
• No rotate their tail
one way, so that
their body rotates
the other so that
their feet are
facing the ground
and they land on
their feet.
• This combined
with their
flexibility all them
to almost always
69
• Helicopter tail rotor failure
• Tail rotor failure #2
Universe Connection
• Rotating star shrinks
radius…. What
happens to rotational
speed??
– Goes way up….. Spins
very fast
• Rotating star explodes
outward…. What
happens to rotational
speed??
– Goes way down … spins
much slower