Rotational Kinetic Energy and Moment of Inertia CH9

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Rotational Kinetic Energy and Moment of Inertia
CH-9
The latest results perhaps indicate the
Blandford-Znajek effect in action whereby
rotational energy can escape from the black hole
as it is braked by the magnetic fields threading
the hole and accretion disk.
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• Rotational Kinetic Energy
• A Rotating Object has the ability to do work,
  thus it possesses
• ENERGY
• This energy is
• Rotational Energy
• For a system of particles rotating around an
  axis, Rotational Energy
• The Total KE of all particles on a that body

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Moment of Inertia
On a rotating object each particle at a distance
r will have a different linear velocity (v) but
all particles have the same angular velocity
(?). Related by v ?r Kinetic Energy of ith
particle is KEi ½ mivi2 ½ mi(?ri)2 Total
for all particles in object KE ½ m1?2r12 ½
m2?2r22 ½ m3?2r32 . ?I ½ mi(?ri)2 Simplified
we get KE ½ (?I miri2) ?2 ?I miri2 is a
quality of the object and is called I, the
Moment of Inertia Therefore KErot ½ I?2 Where
did I come from?? Moment of Inertia is an
inherent characteristic of an object, analogous
to MASS and INERTIA. This is why I is used.
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Moment of Inertia

• I ?I miri2 Moment of Inertia depends on
  MASS distribution.
• I is the rotational equivalent of MASS
• Mass resistance to linear motion
• I resistance to ROTATIONAL motion
• For example Bar 1 Bar 2
• Bar 1 vs Bar 2
• WHICH BAR WILL YOUR CLASSMATE TIRE OUT FASTER
  TWISTING?????

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Moment of Inertia of Rigid Bodies

• To calculate the Moment of Inertia of a rigid
  body, divide the body into infinitesimally small
  point masses (?m)
• Moment of Inertia I lim?M 0 ??mi ri2
• lim?M 0 ? ri2 (?mi)
• ? ri2 dm
• (This is a similar process to xdm for center of
  mass on p306)
• Example 1 Thin Ring/Hoop rotated about an axis
  perpendicular
• to the hoop and through the center.
• I ? ri2 dm
• R2 dm (Where dm represents small masses
• added up to equal total mass M)
• I MR2

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Moment of Inertia of Rigid Bodies
Example 2 Thin rod rotated about an axis
perpendicular to the rod and through the center

• I ? ri2 dm
• ? x2 dm
• (Not integrable 2 variables)
• Ratio of mass element to M is equal to length
  segment and total length
• dm/M dx/L (linear density)
• So, dm M/Ldx and we can get rid of dm
• I ? x2 (M/L)dx (from above)
• M/L ? x2 dx ( from -L/2 to L/2)
• M/L (1/3 x3)
• M/L(1/24) L3 (1/24) L3)
• I 1/12 ML2

x
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Moment of Inertia of Rigid Bodies
Example 3 Solid, Uniform Disk rotated about an
axis perpendicular to the disk and through the
center

• Start with I ? ri2 dm
• (NOTE dm must be the same distance from the
  axis. Little chunks of dm are thin rings!!!!)
• Now we add up all the little cylinders
• Volume of dm is just a rectangle (Think of taking
  a receipt and forming a cylinder. Volume is area
  times thickness of paper)
• Length 2pr, h h, thickness dr
• Volume 2pr(h)dr
• dm/dV ? (density is mass per unit volume)
• dm ?dV (dV Volume that changes as r
  changes)
• dm ?(2pr(h)dr) (Now we have volume in terms of
  r, a variable we can integrate)

I ? ri2 dm ? ri2 ?(2pr(h)dr) ?2ph ? ri3
dr ?2ph (r4/4) from 0 to R TAMO I ½
MR2 (Note Independent of HEIGHT)
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Moment of Inertia of Rigid Bodies
I 2/5 MR2
I 1/2 M(R12 R22)
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Parallel Axis Threorem
If you know the moment of inertia for a rigid
body about any axis through the CoM, you can
compute the moment of inertia about any axis
parallel to the one through the CoM Parallel Axis
Theorem I ICM Md2 where M mass of object
and d distance from CM to new axis (at
P) Example 1 Find the I at the end of a
rod IEND 1/12 ML2 Md2 (d L/2) IEND
1/12 ML2 M(L/2)2 IEND 1/3ML2 Is this higher
or lower than ICM?? Higher
View looking down the z-axis
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Parallel Axis Threorem
Example 2 Find the I at the corner of a
rectangular board with dimensions length a,
width b and mass M (Board behaves like a thin
rod... I 1/12 ML2) ICM 1/12 M (a2 b2) (L2
a2 b2) ICorner ICM Md2 Icorner 1/12 M
(a2 b2) M(a/2)2 (b/2)2 1/12 M (a2
b2) ¼ M (a2 b2) Icorner 1/3 M(a2 b2)
Is this higher or lower than ICM??
HIGHER What other object has a similar
I? Moment of Inertia about the end of a thin
rod
a2 b2
b
ICM
a
ICorner
New axis a b
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Parallel Axis Threorem
Example 3 Find the I about the center of a
hoop with a bar through the diameter and masses
located in the midway between center and
edge. Mass of hoop mass of masses mass of bar
M Diameter of hoop L Ihoop MR2 Ibar 1/12
ML2 (about middle) Imasses MR2 MR2 ITotal
Ihoop Ibar Imasses M(L/2)2 1/12 ML2
2M(L/4)2 11/24 ML2
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Parallel Axis Threorem

• Applications of Moment of Inertia
• Tight Rope walker
• Guy balancing chairs, etc on nose.
• Jordans jump shot.
• Others
• Homework
• For Mon Quiz on Info through Thursdays HW
• The rest of Ch-9 problems
• 34, 36, 38, 41, 44, 51, 52, 55, 69, 78,
  85, 89,92