ch8 - University of Florida · 3/12/2014 3 University of Florida PHY 2053 Page 13 Exam 2 Fall 2011:...
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Transcript of ch8 - University of Florida · 3/12/2014 3 University of Florida PHY 2053 Page 13 Exam 2 Fall 2011:...
3/12/2014
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Smooth Rolling
Smooth rolling, if a wheel does not slip at the point of contact
cm
cm
s R
v R
a R
θ
ωα
=
==
θso, because arc length s is the same as linear distance moved forward,
s
s
rolling without slipping: translation + rotation
cmR vω =
vcm
pure translation
+
pure rotation
0
2 cmv
rolling
As seen by an observer at rest:
= vcm
Rolling and Kinetic Energy
ω
2cm2
22
22cm
rotTtot
)(2
1
)(2
12
1
2
1
vR
Im
ImR
Imv
KKK
+=
+⇒
+=
+=
ω
ω
If the object rolls without slipping then vcm = Rω.
4
Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp. Both objects start from rest and from the same height. Which object reaches the bottom of the ramp first?
h
θ
The object with the largest linear velocity (v) at the bottom of the ramp will win the race.
Rolling Example
5
+=
+=
+=++=+
+=+
=
2
22
2222
2
2
1
2
1
2
1
2
1
2
100
RI
m
mghv
vR
Immgh
R
vImvImvmgh
KUKU
EE
ffii
fi
ω
Apply conservation of mechanical energy:
Example continued:
Solving for v:
6
2sphere
2disk
5
22
1
mRI
mRI
=
=
For the disk:
For the sphere:
ghv3
4disk =
ghv7
10sphere =
Since Vsphere> Vdisk the sphere wins the race.
ghv 2box =Compare these to a box sliding down the frictionless ramp.
Example continued:
The moments of inertia are:
3/12/2014
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7
How do objects in the previous example roll?
FBD:
Both the normal force and the weight act through the center of mass so Στ = 0. This means that the object cannot rotate when only these two forces are applied.
N
w
y
x
Acceleration
8
FBD:
Add friction:
0cos
sin cm
=−=
=−=
==
∑∑∑
θ
θ
ατ
wNF
maFwF
IrF
y
sx
s
y
x
N
w
Fs
θAlso need acm = αR and
The above system of equations can be solved for v at the bottom of the ramp. The result is the same as when using energy methods. (See text example 8.13.)
xavv ∆+= 220
2
It is static friction that makes an object roll.
angular momentum
REMINDERS
Linear Motion Rotations
coordinate xvelocity v
acceleration a
mass mforce F
1st Newton’s Law F=0: v=const2nd Newton’s Law F = ma
Kinetic Energy K = ½mv2
Momentum p = mv
θθθθ angleωωωω angular velocity αααα angular acceleration
ΙΙΙΙ rotational inertiaττττ torqueττττ=0: ωωωω=constττττ = Iαααα
K = ½Iωωωω2 Kinetic Energy??? Angular Momentum
10
Angular Momentum and AMC
vp
pF
mtt
=∆∆=
→∆ 0net lim
ωL
L
Itt
=∆∆=
→∆ 0net limτ
Units of p are kg m/s Units of L are kg m2/s
When no net external forces act, the momentum of a system remains constant (pi = pf)
When no net external torques act, the angular momentum of a system remains constant (Li = Lf).
x-axis
r
P
m
p θθθθ
y-axis p⊥⊥⊥⊥ Angular momentum of a moving particle:
prrprpL ⊥⊥ === θsin
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Example (text problem 8.69): A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a frequency of 0.500 rev/sec. What is the angular momentum? Assume a uniform disk.
rad/sec 14.3rev 1
rad 2
sec
rev500.0 =
= πω
/sm kg 079.02
1 22 =
== ωω MRIL
Example
12
Example (text problem 8.75): A skater is initially spinning at a rate of 10.0 rad/sec with I=2.50 kg m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces I to 1.60 kg m2?
The skater is on ice, so we can ignore external torques.
( ) rad/sec 6.15rad/sec 0.10m kg 60.1
m kg 50.22
2
=
=
=
=
=
if
if
ffii
fi
I
I
II
LL
ωω
ωω
AMC Example
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University of Florida PHY 2053 Page 13
Exam 2 Fall 2011: Problem 17
R FT
Mg
y-axis
• A cloth tape is wound around the outside of a non-uniform solid cylinder (mass M, radius R) and fastened to the ceiling as shown in the figure. The cylinder is held with the tape vertical and then release from rest. If the acceleration of the center-of-mass of the cylinder is 3g/5, what is its moment of inertia about its symmetry axis?
Answer: % Right: 48%
232 MR
R
yT MaFMg =−
R
IaIRF y
T === ατ)( yT agMF −=
2322
53
53
22
MRMRg
ggMR
a
ag
a
FRI
y
y
y
T =
−=
−==
University of Florida PHY 2053 Page 14
Exam 2 Fall 2011: Problem 31
• A uniform solid disk with mass M and radius R is mounted on a vertical shaft with negligible rotational inertia and is initially rotating with angular speed ωωωω. A non-rotating uniform solid disk with mass M and radius R/2 is suddenly dropped onto the same shaft as shown in the figure. The two disks stick together and rotate at the same angluar speed. What is the new angular speed of the two disk system? Answer: 4ωωωω/5% Right: 47%
R M
M
ffffiii MRMRILMRIL ωωωω )( 222
1212
1212
1 +=====
ωωωω 54
22
2
22
21
21
)2/(=
+=
+=
RR
R
RR
Rf
University of Florida PHY 2053 Page 15
Exam 2 Fall 2010: Problem 16
Initial
P
R
M
m
• A mouse of mass M/6 lies on the rim of a solid uniform disk of mass M that can rotate freely about its center like a merry-go-round. Initially the mouse and disk rotate together with an angular velocity of ωωωω. If the mouse walks to a new position that is at the center of the disk what is the new angular velocity of the mouse-disk system?Answer: 4ωωωω/3% Right: 58%
Final
P
R
M
m ωω )( 2mRIIL diskiii +==fdiskfff IIL ωω ==
ωω )( 2mRII disknewdisk +=fi LL =
ωωωωω 34
221
22 2111 =
+=
+=
+=
M
m
MR
mR
I
mR
disknew
2291
22
22ωMR
I
L
I
L
EEE
i
i
f
f
if
=−=
−=∆Note that energy is not conserved in
this problem!