Physics 2211: Lecture 21 Todays Agenda

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Physics 2211 Lecture 21Todays Agenda

• Systems of Particles DISTRIBUTED SYSTEMS
• Center of mass
• Velocity and acceleration of the center of mass
  
• Moment of inertia

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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!

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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

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System of Particles Center of Mass

• If the system is made up of only two particles

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System of Particles Center of Mass

• If the system is made up of only two particles

where M m1 m2
r2 - r1

m2
m1
RCM
r2
r1
y
x
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System of Particles Center of Mass

• If the system is made up of only two particles

where M m1 m2
If m1 3m2
r2 - r1

m2
m1
RCM
r2
the CM is now closer to the heavy mass.
r1
y
x
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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.

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System of Particles Center of Mass

• We can consider the components of RCM separately

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Example Calculation

• Consider the following mass distribution

2m
(12,12)
m
m
(0,0)
(24,0)
RCM (12,6)
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System of Particles Center of Mass

• For a continuous solid, we have to do an integral.

dm
r
y
where dm is an infinitesimal mass element.
x
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System of Particles Center of Mass

• We find that the Center of Mass is at the
  center of the object.

y
RCM
x
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System of Particles Center of Mass

• We find that the Center of Mass is at the
  center of the object.

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).
RCM
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System of Particles 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




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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

m2
R2 - R1

R2
so if we have two objects
RCM
R1
m1
y
x
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Center of Mass

• The disk shown below (A) clearly has its CM at
  the center.
• Suppose the disk is cut in half and the pieces
  arranged as shown in (B)
• Where is the CM of (B) as compared to (A)?

(1) higher (2) lower
(3) same
XCM
(A)
(B)
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• The CM of each half-disk will be closer to the
  fat end than to the thin end (think of where it
  would balance).

X
X
(A)
(B)
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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 objectsomewhere else, it
  willorient itself so that theCM is directly
  below the pivot.
• This fact can be used to findthe CM of
  odd-shaped objects.

pivot
CM
pivot
pivot
CM

CM
mg
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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.

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Velocity and Accelerationof the Center of Mass

• If its particles are moving, the CM of a system
  can also move.
• Suppose we know the position ri of every
  particle in the system as a function of time.

So
And

• The velocity and acceleration of the CM is just
  the weighted average velocity and acceleration of
  all the particles.

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Center of Mass and Newtons 2nd Law

• Newtons 2nd Law applies to CM motion
• This has interesting implications
• It tells 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.

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Rotation Kinetic Energy

• Consider the simple rotating system shown below.
  (Assume the masses are attached to the rotation
  axis by massless rigid rods).
• The kinetic energy of this system will be the sum
  of the kinetic energy of each piece

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Rotation Kinetic Energy...

• So but vi ?ri

I has units of kg m2.
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Rotation Kinetic Energy...

• The kinetic energy of a rotating system looks
  similar to that of a point particle Point
  Particle Rotating System

v is linear velocity m is the mass.
? is angular velocity I is the moment of
inertia about the rotation axis.
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Moment of Inertia

• So where

• Notice that the moment of inertia I depends on
  the distribution of mass in the system.
• The further the mass is from the rotation axis,
  the bigger the moment of inertia.
• For a given object, the moment of inertia will
  depend on where we choose the rotation axis
  (unlike the center of mass).
• We will see that in rotational dynamics, the
  moment of inertia I appears in the same way that
  mass m does when we study linear dynamics!

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Calculating Moment of Inertia

• We have shown that for N discrete point masses
  distributed about a fixed axis, the moment of
  inertia is

where r is the distance from the mass to the
axis of rotation.
Example Calculate the moment of inertia of four
point masses (m) on the corners of a square whose
sides have length L, about a perpendicular axis
through the center of the square
m
m
L
m
m
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Calculating Moment of Inertia...

• The squared distance from each point mass to the
  axis is

Using the Pythagorean Theorem
so
L/2
m
m
r
L
m
m
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Calculating Moment of Inertia...

• Now calculate I for the same object about an axis
  through the center, parallel to the plane (as
  shown)

r
L
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Calculating Moment of Inertia...

• Finally, calculate I for the same object about an
  axis along one side (as shown)

r
m
m
L
m
m
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Calculating Moment of Inertia...

• For a single object, I clearly depends on the
  rotation axis!!

I 2mL2
I mL2
I 2mL2
m
m
L
m
m
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Moment of Inertia

• A triangular shape is made from identical balls
  and identical rigid, massless rods as shown. The
  moment of inertia about the a, b, and c axes is
  Ia, Ib, and Ic respectively.
• Which of the following is correct

(1) Ia gt Ib gt Ic (2) Ia gt Ic gt Ib (3)
Ib gt Ia gt Ic
a
b
c
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Moment of Inertia

• Label masses and lengths

m
a
L
b
So (2) is correct Ia gt Ic gt Ib
L
c
m
m
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Calculating Moment of Inertia...

• For a discrete collection of point masses we
  found
• For a continuous solid object we have to add up
  the mr2 contribution for every infinitesimal mass
  element dm.
• We have to do anintegral to find I

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Moments of Inertia

• Some examples of I for solid objects

Thin hoop (or cylinder) of mass M and radius R,
about an axis through its center, perpendicular
to the plane of the hoop.
R
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Moments of Inertia...
Sphere and disk

• Some examples of I for solid objects

Solid sphere of mass M and radius R, about an
axis through its center.
R
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Moment of Inertia

• Two spheres have the same radius and equal
  masses. One is made of solid aluminum, and the
  other is made from a hollow shell of gold.
• Which one has the biggest moment of inertia about
  an axis through its center?

(1) solid aluminum (2) hollow gold (3) same
hollow
solid
same mass radius
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Moment of Inertia

• Moment of inertia depends on mass (same for both)
  and distance from axis squared, which is bigger
  for the shell since its mass is located farther
  from the center.
• The spherical shell (gold) will have a bigger
  moment of inertia.

ISOLID lt ISHELL
hollow
solid
same mass radius
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Moments of Inertia...

• Some examples of I for solid objects

Thin rod of mass M and length L, about a
perpendicular axis through its center.
L
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Parallel Axis Theorem

• Suppose the moment of inertia of a solid object
  of mass M about an axis through the center of
  mass, ICM, is known.
• The moment of inertia about an axis parallel to
  this axis but a distance D away is given by
• IPARALLEL ICM MD2
• So if we know ICM , it is easy to calculate the
  moment of inertia about a parallel axis.

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Parallel Axis Theorem Example

• Consider a thin uniform rod of mass M and length
  D. Figure out the moment of inertia about an axis
  through the end of the rod.
• IPARALLEL ICM MD2

DL/2
M
CM
x
L
ICM
IEND
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Recap of todays lecture

• Distributed systems Center of mass Moment of
  Inertia
• Kinetic energy of a rotating system
• Moment of inertia
• Discrete particles
• Continuous solid objects