Download PES 1110 Fall 2013, Spendier Lecture 21/Page 1 Today

PES 1110 Fall 2013, Spendier
Lecture 21/Page 1
Today:
- HW 5 due
- HW 6 assigned
- one more general energy problem
- Chapter 9: Center of Mass and Systems of Particles (9.1-9.3)
Example 1:
A system of two paint buckets connected by a lightweight
rope is released from rest with the 12.0 kg bucket 2.00 m
above the floor. During the time that the 12.0 kg bucket drops
to the floor, friction in the pulley removes 10.0 J of energy
from the system. What is the speed of the bucket when it hits
the floor?
PES 1110 Fall 2013, Spendier
Lecture 21/Page 2
Chapter 9:
So far we have approximated our objects as a point with single mass at a given position
(x,y) in 2D.
Most objects, as we know are a collection of particles of mass dmi, each at their own
position (xi,yi) in 2D.
n
Total mass = M =
 dm   dm
i
i
i 1
(integral needed it you consider a continuous distribution of matter)
Center of mass (com)
In order to apply all we have learned to object of finite size, we need to define the center
of mass (com).
The center of mass (com) of a system of particles is the position through all external
forces appear to act.
It is the position we give an extended object.
In practice it is the point around which an object balances:
DEMO 1: point around which an object balances = com
- show balancing eagle
- show wine holder
Find the com in 2D:
DEMO 2: find center of mass by hanging object twice - then show that it indeed
balances.
Note: Center of mass might not be "in" the object.
PES 1110 Fall 2013, Spendier
Lecture 21/Page 3
Center of mass (com): Mathematical definition
The center of mass is the point about which matter is evenly distributed when weighted by
distance
xcom = center of mass position
xi - xcom = vector from center of mass to particle i
Since all the mass must distribute evenly at the center of mass the following condition
must be satisfied
xcom 
1
M
 x dm
i
i
i
n
Total mass = M =
 dm   dm
i
i
i 1
(integral needed it you consider a continuous distribution of matter)
PES 1110 Fall 2013, Spendier
Lecture 21/Page 4
If the particle is distributed in 3D the center of mass must be defined in 3 coordinates:
xcom 
1
M
 x dm
i
i
ycom 
i
call

ri  xi iˆ  yi ˆj  zi kˆ

rcom  xcomiˆ  ycom ˆj  zcom kˆ
rcom 
1
M
 r dm
i
i
i
1
M
 y dm
i
i
i
zcom 
1
M
 z dm
i
i
i
PES 1110 Fall 2013, Spendier
Lecture 21/Page 5
For solid object:
An object contains many atoms. It is best to treat the object as a continues distribution of
matter ==> the sum becomes an integral.
xcom 
1
x dm
M
ycom 
1
M
 y dm
zcom 
1
z dm
M
Example 4
Find the xcom using integration. Mass density of the object is σ kg/m2 and the mass is
uniformly distributed.
Newton's Second Law of Motion for a system of particles (section 9.3)
Earlier I said that forces appear to act through the center of mass. This means that the
center of mass follows the equations of motion.
The center of mass (black dot) of
a baseball bat flipped into the air
follows a parabolic path, but all
other points of the bat follow
more complicated curved paths.
juggling bats...
Projectile motion: Bat is rotating but center of mass follows a parabola. Hence we can
apply our equations of motion for constant acceleration.
PES 1110 Fall 2013, Spendier
Lecture 21/Page 6
Velocity of an object
We assign a velocity to the center of mass position:
vcom 
vcom 
d
d  1
xcom  
dt
dt  M
1
M
dxi
1

 x dm   M  dt
i
dmi
i
i
i
 v dm
i
i
i
vcom velocity of individual particles weighted by mass.
Force of an object
To get an object to accelerate, all of the individual particles must be accelerated.
acom 
1
M
 a dm
i
i
i
so
Fcom  Macom  M
1
M
 a dm   a dm   dF
i
i
i
i
i
i
i
i
The force required to accelerate the center of mass is equal to the sum of the forces
required to accelerate all of the individual particles.
So the observation of the total mass, and the position, velocity, and acceleration of the
center of mass are sufficient to describe all particles in the object.
PES 1110 Fall 2013, Spendier
Lecture 21/Page 7