7/1/2008 - universitas...
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7/1/2008
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Ontang Anting
Moment of Inertia
Energy
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Adhi Harmoko S
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Passenger undergo “uniform circular motion” (circular path at constant speed)Therefore, there must be a:
centripetal acceleration, ac.Therefore there must be a
centripetal force, Fc.› it is directed toward the circle’s center› it depends on speed and size of circle:
ac = v2/r
Fc = m x ac = m x v2/r
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When Ontang Anting is turning:› acceleration is toward the center of circle (inward)› so “fictitious force” is outward.
This is “fictitious force” is called “centrifugal force”Centrifugal force is an experience of inertia: It is NOT a real force.
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Dufan duduk di bangku paling luar pada permainan ontang anting, and Dufi duduk di tengah tengah antara dufan dan titik tengah. Ontang anting bergerak membuat satu putaran penuh setiap dua detik.› Kecepatan angular Dufi sama dengan
(a) Dufan (b) Dua kali Dufan(c) Setengah Dufan
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Kecepatan angular ω pada sembarang titik pada benda pejal yang berputar pada pusat yang tetap adalah sama.› Dufan & Dufi berputar sekali (2π radians) selama dua detik
ω
(Kecepatan linier dufan dan dufi v akan berbeda karena v = ωr).
DufanDufi VV21
=
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Rotation about a fixed axis:› Consider a disk rotating aboutan axis through its center:
First, recall what we learned aboutUniform Circular Motion:
(Analogous to )
ω
θ
dtdθω =
dt
dxv =
8
Now suppose ω can change as a function of time:We define the angular acceleration:
Consider the case when αis constant.› We can integrate this to
find ω and θ as a function of time:
α
2
2
dtd
dtd θωα ==
200
0
2
1
constant
tt
t
αωθθ
αωωα
++=
+==
ω
θ
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Recall also that for a point at a distance R away from the axis of rotation:› x = θR› v = ωRAnd taking the derivative of this we find:
› a = αR
200
0
2
1
constant
tt
t
αωθθ
αωωα
++=
+=
=
ω
θ
α
x
v
R Angular Linier
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constant=α
tαωω += 0
200 2
1tt αωθθ ++=
constant=a
atvv += 0
200 2
1attvxx ++=
And for a point at a distance R from the rotation axis:
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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:
rr1
rr2rr3
rr4
m4
m1
m2
m3
ω
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So: but vi = ωri ∑=i
iivmK 2
2
1
( ) ∑∑ ==i
iii
ii rmrmK 222
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21 ωω
rr1
rr2rr3
rr4
m4
m1
m2
m3
ωvv4
vv1
vv3
vv2
which we write as:
2I21 ω=K
∑=i
ii rm 2I
Define the moment of inertiamoment of inertia
about the rotation axis
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Point ParticlePoint Particle Rotating SystemRotating System
v is “linear” velocitym is the mass
ω is angular velocityI is the moment of inertiaabout the rotation axis
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The kinetic energy of a rotating system looks similar to that of a point particle:
∑=i
ii rm 2I
.
2I21 ω=K2
21mvK =
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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 Iappears in the same way that mass m does when we study linear dynamics!
2I21 ω=K ∑=
iii rm 2I
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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:
∑=i
ii rm 2I
mm
mm
L
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The squared distance from each point mass to the axis is:
222
222 LLr =⎟
⎠⎞
⎜⎝⎛=
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2222I
22222
1
2 Lm
Lm
Lm
Lm
Lmrm
N
iii =+++== ∑
=
so
Using the Pythagorean Theorem
mm
mm
L r
L/2
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Now calculate I for the same object about an axis through the center, parallel to the plane (as shown):
44
4444I
22222
1
2 Lm
Lm
Lm
Lm
Lmrm
N
iii =+++== ∑
=
mm
mm
L
r
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Finally, calculate I for the same object about an axis along one side (as shown):
mm
mm
L
r
2222
1
2 00I mmmLmLrmN
iii +++== ∑
=
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For a single object, I clearly depends on the rotation axis!!
mm
mm
L
mm
mm
mm
mm
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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 :
r
dm
dmr∫= 2I
∑=i
ii rm 2I
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Some examples of I for solid objects:Thin hoop (or cylinder) of mass Mand radius R, about an axis through its center, perpendicular to the plane of the hoop.
2I MR=
R
2
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I MR=
Thin hoop of mass M and radius R, about an axis through a diameter.R
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Some examples of I for solid objects:
Solid sphere of mass M and radius R, about an axis through its center.
2
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I MR=
2
21
I MR=
Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center.
R
R
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Some examples of I for solid objects:
Thin rod of mass M and length L, about a perpendicular axis through its center.
2
121
I MR=
2
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I MR=
Thin rod of mass M and length L, about a perpendicular axis through its end.
L
L
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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?
(a) solid aluminum (b) hollow gold (c) same
same mass & radius
solid hollow
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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
same mass & radius
solid hollow
ISOLID < ISHELL
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Suppose the moment of inertia of a solid object of mass Mabout 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:
So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.
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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
2
121
I MLCM =We know
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2
31
2121
I MLL
MMLEND =⎟⎠⎞
⎜⎝⎛+=So
which agrees with the result on a previous slide.
L
D=L/2M
xCM
ICMIEND
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Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the direction at some instant:
› aθ = αr
Now use Newton’s 2nd Law in the direction:› Fθ = maθ = mα r
Multiply by r :› rFθ = mr2α
θθ̂̂
θθ̂̂
r
aaθ
α
FF
m
rr̂̂θθ^̂
Fθ
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rFθ = mr2α use
= Ι αDefine torque: τ = rFθ .› τ is the tangential force Fθ times the lever arm r.
Torque has a direction:› + z if it tries to make the systemspin CCW.
› ‐ z if it tries to make the systemspin CW.
ατ I=
r
aaθ
α
FF
m
rr̂̂θθ^̂
Fθ
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So for a collection of many particles arranged in a rigid configuration:
Since the particles are connected rigidly,they all have the same α
rr1
rr2rr3
rr4
m4
m1
m2
m3
ωFF4
FF1
FF3FF2
{ { ii
iii
ii rmFr
i
ατ
θ ∑∑Ι
= 2,
ατ I=∑i
i
ατ I=NET
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This is the rotational analogue of FNET = ma
Torque is the rotational analogue of force:Torque is the rotational analogue of force:
› The amount of “twist” provided by a force.Moment of inertiaMoment of inertia I I is the rotational analogue of mass.is the rotational analogue of mass.› If I is big, more torque is required to achieve a given angular acceleration.
Torque has units of kg m2/s2 = (kg m/s2) m = Nm
ατ I=NET
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Recall the definition of torque:
rp= “distance of closest approach”
Equivalent definitions
Fr
Fr
Frτ
φφ
θ
sin
sin
=
=
=
Frp=τ
rr
φ
rp
FF
φFθ
Fr φ
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So if φ = 0o, then τ = 0
And if φ = 90o, then τ = maximum
Fsinr φτ =
rr
FF
rr
FF
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In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.
(a) (a) case 1 (b) (b) case 2(c) (c) same
L
L
F F
axis
case 1 case 2
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Torque = F x (distance of closest approach)› The applied force is the same.› The distance of closest approach is the same
› Torque is the same!
L
L
F F
case 1 case 2
L
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Consider the work done by a force FF acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement dθ:
› dW = FF.drdr = FR dθ cos(β)
= FR dθ cos(90‐φ)= FR dθ sin(φ)= FR sin(φ) dθ
dW = τ dθ
We can integrate this to find: W = τθAnalogue of W = F •ΔrW will be negative if τ and θ have opposite signs!
R
FF
dr = R dθdθ
axis
φ
β
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Recall the Work/Kinetic Energy Theorem: ΔK = WNET
This is true in general, and hence applies to rotational motion as well as linear motion.
So for an object that rotates about a fixed axis:
( ) NETif WK =−=Δ 22I21 ωω
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A massless string is wrapped 10 times around a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning).› How fast is the disk spinning after the string has unwound?
F
RM
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The work done is W = τ θ› The torque is τ = RF (since φ = 90o)
› The angular displacement θ is2π rad/rev x 10 rev
SoW = (.1 m)(10 N)(20π rad) = 62.8 J
τ θ
F
RM
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Recall that �I� for a disk about its central axis is given by
2NET I
21
K J62.8 W W ω=Δ===
2
2
1I MR=
WMRK =⎟⎠⎞
⎜⎝⎛=Δ 22
2
1
2
1ω
( )( )( )22 1.04.
8.6244
kg
J
MR
W==ω ω = 792.5 rad/s
RM
ω
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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 have the same moment of inertia. 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 (b) disk 2(c) same FF
ω1ω2
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The work done on both disks is the same!› W = FdThe change in kinetic energy of each will therefore also be the same since W = ΔK
But we know
So since I1 = I2
2I21 ω=ΔK
d
FF
ω1ω2
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Adhi Harmoko S