Transcript Physics 131: Lecture 14 Notes

REVIEW-2
Phys-151




Work & Energy
Momentum Conservation
Center of Mass
Torque, Angular Momentum,
Moment of Inertia
Physics 151: Lecture 12, Pg 1
See text: 7-1
Definition of Work:
Ingredients: Force ( F ), displacement (  r )
Work, W, of a constant force F
acting through a displacement  r
is:
F

Fr
r
.
W = F  r = F  r cos  = Fr  r
“Dot Product”
Total Energy is Conserved
Physics 151: Lecture 12, Pg 2
Example
Work Kinetic-Energy Theorem

How much will the spring compress to bring the object to a
stop if the object is moving initially at a constant velocity (vo)
on frictionless surface as shown below ?
to vo
F
m
spring at an equilibrium position
x
V=0
t
m
spring compressed
Physics 151: Lecture 12, Pg 3
Lecture 13 - ACT 3
Work of Springs

I have a spring with k = 20 N/m. Its compressed
30 cm by a 200 gram mass. How much work is
done by the spring in this process ?
A) 0.6 J
B) 6 J
C) 0.9 J
D) –0.9 J
Physics 151: Lecture 12, Pg 4
Lecture 13, ACT 5
Work & Energy

Two blocks having mass m1 and m2 where m1 > m2.
They are sliding on a frictionless floor and have the
same kinetic energy when they encounter a long rough
stretch (i.e. m > 0) which slows them down to a stop.
Which one will go farther before stopping ?
(a) m1 (b) m2
(c) they will go the same distance
m1
m2
Physics 151: Lecture 12, Pg 5
Another Example
A 3.0-kg block is dragged over a rough horizontal
surface by a constant force of 16 N acting at an
angle of 37° above the horizontal as shown. The
speed of the block increases from 4.0 m/s to 6.0
m/s in a displacement of 5.0 m. What work was
done by the friction force during this displacement?
a.
b.
c.
d.
e.
–34 J
–64 J
–30 J
–94 J
+64 J
Physics 151: Lecture 12, Pg 6
Work & Power:
Example 3 :

What is the power required for a car (m=1000 kg) to climb a hill (5%) at
v=30m/s assuming the coefficient of friction m = 0.03 ?
Ptot = Phorizontal + Pvertical
5%
100
5
v=const. - > a = 0
Phorizontal = F v = m mg vhorizontal
Phorizontal ~ 0.03 (1000kg) (10m/s2 ) (30m/s) ~ 10 kW
Pvertical = F v = mg vvertical = (1000kg) (10m/s2 ) (30m/s)(5/100)
Pvertical ~ 15 kW
Ptot ~ 10 kW + 15 kW = 25 kW
Physics 151: Lecture 12, Pg 7
Example - 2

A 12-kg block on a horizontal frictionless surface is
attached to a light spring (force constant = 0.80 kN/m). The
block is initially at rest at its equilibrium position when a
force (magnitude P = 80 N) acting parallel to the surface is
applied to the block, as shown. What is the speed of the
block when it is 13 cm from its equilibrium position?
k
v1= 0
F
m
v2=?
0.78 m/s
x
Physics 151: Lecture 12, Pg 8
Lecture 16, ACT 1
The Roller Coaster

I have built a Roller Coaster . A motor tugs the
cars to the top and then they are let go and are in
the hands of gravity. To make the following loop,
how high do I have to let the release the car ??
Car has mass m
h?
R
A) 2R
B) 5R
C) 5/2 R
D) none of the above
Physics 151: Lecture 12, Pg 9
MOMENTUM
F
p = mv
dp
dt
Momentum Conservation
FEXT


dP

dt

dP
0
dt
FEXT  0
A collision is said to be elastic when energy as
well as momentum is conserved before and
after the collision.
A collision is said to be inelastic when energy is
not conserved before and after the collision, but
momentum is conserved.
Kbefore = Kafter
Kbefore  Kafter
Physics 151: Lecture 12, Pg 10
ACT 4

The value of the momentum of a system is the same
at a later time as at an earlier time if there are no
a. collisions between particles within the system.
b. inelastic collisions between particles within the
system.
c. changes of momentum of individual particles
within the system.
d. internal forces acting between particles within the
system.
e. external forces acting on particles of the system.
Physics 151: Lecture 12, Pg 11
Lecture 17, ACT 3
Momentum Conservation


Two balls of equal mass are thrown horizontally with
the same initial velocity. They hit identical stationary
boxes resting on a frictionless horizontal surface.
The ball hitting box 1 bounces back, while the ball
hitting box 2 gets stuck.
Which box ends up moving fastest ?
(a) Box 1
1
(b) Box 2
(c) same
2
Physics 151: Lecture 12, Pg 12
Inelastic collision in 1-D: Example 1

A block of mass M is initially at rest on a frictionless
horizontal surface. A bullet of mass m is fired at the
block with a muzzle velocity (speed) v. The bullet
lodges in the block, and the block ends up with a
speed V. In terms of m, M, and V :
What is the initial speed of the bullet v ?
What is the initial energy of the system ?
What is the final energy of the system ?
Is energy conserved ?
x
v
V
before
See example 12-6
after
Physics 151: Lecture 12, Pg 13
Lecture 17
ACT 4
3.0-kg mass sliding on a frictionless surface has a
velocity of 5.0 m/s east when it undergoes a onedimensional inelastic collision with a 2.0-kg mass that
has an initial velocity of 2.0 m/s west. After the collision
the 3.0-kg mass has a velocity of 1.0 m/s east. How much
kinetic energy does the two-mass system lose during the
collision?
a. 22 J
b. 24 J
c. 26 J
d. 20 J
e. 28
Physics 151: Lecture 12, Pg 14
See text: 9.4
Elastic Collision in 1-D

After some moderately tedious algebra, (see text book
Chapter 9, section3) we can derive the following equations
for the final velocities,
m1  m2
2m2
v1 A  (
)v1 B  (
) v2 B
m1  m2
m1  m2
v2 A  (
In general:
1)
2m1
m  m1
)v1 B  ( 2
) v2 B
m1  m2
m1  m2
m1v1,b + m2v2,b = m1v1,a + m2v2,a
2) v1,b - v2,b = - (v1,a - v2,a)
Physics 151: Lecture 12, Pg 15
See text: 9.4
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
Animation
Physics 151: Lecture 12, Pg 16
See text: Ex. 9.11
2-D Elastic Collisions
Billiards

Consider the case where one ball is initially at rest.
pa
pb
vcm
Pa
F
before
See Figure 12-
the final direction of the
red ball will depend on
where the balls hit.
after
Animation
Physics 151: Lecture 12, Pg 17
ACT
A 5.0-g particle moving 60 m/s collides with a 2.0-g
particle initially at rest. After the collision each of the
particles has a velocity that is directed 30° from the
original direction of motion of the 5.0-g particle. What is
the speed of the 2.0-g particle after the collision?
a. 72 m/s
b. 87 m/s
c. 79 m/s
d. 94 m/s
e. 67 m/s
Physics 151: Lecture 12, Pg 18
See text: 9-2
Force and Impulse

The diagram shows the force vs time for a typical collision.
The impulse, I, of the force is a vector defined as the
integral of the force during the collision.
F
t
I   F dt
Impulse I = area under this curve !
t
t
Impulse has units of Ns.
See Figure 12-
ti
tf
Physics 151: Lecture 12, Pg 19
See text: 9-2
Force and Impulse

F
Using
dP
dt
the impulse becomes:
t
I   F dt  
t
dP
dt
dt
F
t
  dP  Pf  Pi  P
I  P
impulse = change in momentum !
See Figure 12-
t
t
Physics 151: Lecture 12, Pg 20
Lecture 19, ACT 2
Force & Impulse

What is the average force that wall exerts on ball (0.40kg)
if duration of wall-ball contact is 0.01 s ?
a) 20 N
b) 200 N
c) 2,000 N
d) 20,000 N
vi = 30m/s
before collision
vf = 20m/s
after collision
Physics 151: Lecture 12, Pg 21
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:
N
 mi ri
RCM  i 1
M
m2
m1
r1
RCM
r2
y
x
(In this case, N = 2)
Physics 151: Lecture 12, Pg 22
See text: 9.6
Center of Mass Motion: Review

We have the following law for CM motion:
FEXT  MACM

This has several interesting implications:

It tell us that the CM of an extended object behaves like a
simple point mass under the influence of external forces:
We can use it to relate F and A like we are used to doing.
It tells us that if FEXT = 0, the total momentum of the system
does not change.

Physics 151: Lecture 12, Pg 23
Example: Astronauts & Rope

:-)

:-(

A male astronaut and a female astronaut are at rest in outer
space and 20 meters apart. The male has 1.5 times the mass of
the female. The female is right by the ship and the male is out in
space a bit. The male wants to get back to the ship but his jet
pack is broken. Conveniently, there is a rope connected
between the two. So the guy starts pulling in the rope.
Does he get back to the ship?
Does he at least get to meet the woman?
m
M = 1.5m
Physics 151: Lecture 12, Pg 24
Lecture 19, ACT
Center of Mass Motion


A woman weighs exactly as much as her 20 foot long boat.
Initially she stands in the center of the motionless boat, a distance of 20
feet from shore. Next she walks toward the shore until she gets to the
end of the boat.
What is her new distance from the shore.
(There is no horizontal force on the boat by the water).
20 ft
(a) 10 ft
before
20 ft
(b) 15 ft
(c) 16.7 ft
? ft
after
Physics 151: Lecture 12, Pg 25
Recall
Kinematic of Circular Motion:
y
v
x = R cos() = R cos(t)
y = R sin() = R sin(t)
 = tan-1 (y/x)
R
t
(x,y)
s
x
 = t
s=vt
s = R = Rt
 is angular velocity
For uniform
circular motion:
v = R
Animation
v2
a
R
Physics 151: Lecture 12, Pg 26
See text: 10.3
Summary
(with comparison to 1-D kinematics)
Angular
  constant
  0  t
1
  0   0 t  t 2
2

Linear
a  constant
v  v 0  at
x  x 0  v 0t 
1 2
at
2
And for a point at a distance R from the rotation axis:
x = Rv = R
a = R
Physics 151: Lecture 12, Pg 27
See text: 10.1
Example:

The turntable of a record player has an angular
velocity of 8.0 rad/s when it is turned off. The
turntable comes to rest 2.5 s after being turned
off. Through how many radians does the turntable
rotate after being turned off ? Assume constant
angular acceleration.
a.
b.
c.
d.
e.
12 rad
8.0 rad
10 rad
16 rad
6.8 rad
Physics 151: Lecture 12, Pg 28
See text: 10.5
Calculating Moment of Inertia...

For a single object, I clearly depends on the rotation axis !!
I = 2mL2
I = mL2
m
m
m
m
I = 2mL2
L
See example 10.4 (similar)
Physics 151: Lecture 12, Pg 29
Moments of Inertia

Some examples of I for solid objects:
Solid disk or cylinder of mass M and
radius R, about a perpendicular axis
through its center.
dr
L
r
R
I


2
r
 dm
1
I  MR 2
2
Physics 151: Lecture 12, Pg 30
See text: 10.5
Moments of Inertia...

Some examples of I for solid objects:
2
I  MR 2
5
Solid sphere of mass M and radius R,
R
about an axis through its center.
I
R
2
MR 2
3
Thin spherical shell of mass M and radius
R, about an axis through its center.

See Table 10.2, Moments of Inertia
Physics 151: Lecture 12, Pg 31
See text: 10.4
Kinetic Energy of Rotation and
Moment of Inertia

K
1
I 2
2
2
where I   mi ri
i
Parallel Axis Theorem
IPARALLEL = ICM + MR2
Physics 151: Lecture 12, Pg 32
See text: 10.6 and 10.7
Rotational Dynamics:
What makes it spin?
tot = I
= r X F
MAGNITUDE:  = r F sin 
DIRECTION:
Physics 151: Lecture 12, Pg 33
See text: 10.8
How Much WORK is Done ?



W = 
Analogue of W = F •r
W will be negative if  and  have opposite sign !
Work & Kinetic Energy:
K 


1
I  2f   2i  Wnet
2
Physics 151: Lecture 12, Pg 34
Lecture 21, ACT 2

A uniform rod of mass M = 1.2kg and length L = 0.80 m,
lying on a frictionless horizontal plane, is free to pivot
about a vertical axis through one end, as shown. If a
force (F = 5.0 N,  = 40°) acts as shown, what is the
resulting angular acceleration about the pivot point ?
a.
b.
c.
d.
e.
16 rad/s2
12 rad/s2
14 rad/s2
10 rad/s2
33 rad/s2
Physics 151: Lecture 12, Pg 35
See text: 11.3
p=mv

Angular Momentum:
Definitions & Derivations
e have shown that for a system of particles
FEXT 
dp
dt
Momentum is conserved if
FEXT  0

The rotational analogue of force F is torque   r  F

Define the rotational analogue of momentum p to be
angular momentum L  r  p
Animation
Physics 151: Lecture 12, Pg 36
See text: 11.3
Definitions & Derivations...

First consider the rate of change of L:
dL d
 r  p 
dt dt
d
 dr  p   r  dp 
r

p

 


 


dt
dt
dt 
 v  mv 
0
So
dL
dp
r
dt
dt
Physics 151: Lecture 12, Pg 37
Lecture 21, ACT 2

Strings are wrapped around the circumference of two
solid disks and pulled with identical forces for the
same distance. Disk 1 has a bigger radius, but both
are made of identical material (i.e. their density r =
M/V is the same). Both disks rotate freely around
axes though their centers, and start at rest.
Which disk has the
biggest angular velocity
after the pull ?
(a) disk 1
2
1
F
F
(b) disk 2
(c) same
Physics 151: Lecture 12, Pg 38
Example 2

A rope is wrapped around the circumference of a solid
disk (R=0.2m) of mass M=10kg and an object of mass
m=10 kg is attached to the end of the rope 10m above
the ground, as shown in the figure.
a) How long will it take
for the object to hit
the ground ?
1.7 s
a) What will be the
velocity of the object
when it hits the
ground ?
11m/s

M
T
m
h =10 m
a) What is the tension
on the cord ?
32 N
Physics 151: Lecture 12, Pg 39
See text: 10.8
Example: Rotating Road

A uniform rod of length L=0.5m and mass m=1 kg is
free to rotate on a frictionless pin passing through one
end as in the Figure. The rod is released from rest in
the horizontal position. What is
a) angular speed when it reaches the lowest point ?
b) initial angular acceleration ?
c) initial linear acceleration of its free end ?
L
m
See example 10.14
a)
 = 7.67 rad/s
b)
=
c)
a = 15 m/s2
30 rad/s2
Physics 151: Lecture 12, Pg 40
See text: 10.1
Example:

A mass m = 4.0 kg is connected, as shown, by a
light cord to a mass M = 6.0 kg, which slides on a
smooth horizontal surface. The pulley rotates
about a frictionless axle and has a radius R =
0.12 m and a moment of inertia I = 0.090 kg m2.
The cord does not slip on the pulley. What is the
magnitude of the acceleration of m?
a.
b.
c.
d.
e.
2.4 m/s2
2.8 m/s2
3.2 m/s2
4.2 m/s2
1.7 m/s2
Physics 151: Lecture 12, Pg 41
Example : Rolling Motion

A cylinder is about to roll down an inclined plane.
What is its speed at the bottom of the plane ?
Cylinder has radius R
M
h

M
v?
Physics 151: Lecture 12, Pg 42
How high do we have to start the ball ?
1
h
2
h = 2.7 R = (2R + 1/2R) + 2/10 R
-> The rolling motion added an extra 2/10 R to the height)
Physics 151: Lecture 12, Pg 43