chap. 10: rotational motion - texas a&m...
TRANSCRIPT
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Chap. 10: Rotational
Motion
1
I. Rotational Kinematics II. Rotational Dynamics - Newton’s Law for
Rotation III. Angular Momentum Conservation (Chap.
10)
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Rotational Motion
1. Remember how Newton’s Laws for translational
motion were studied:
1. Kinematics (x = x0 + v0 t + ½ a t2 )
2. Dynamics (F = m a)
3. Momentum Conservation
2. Now, we repeat them
again, but for rotational
motion:
1. Kinematics (q, w, a)
2. Dynamics (t = I a)
3. Angular Momentum
Part III
2
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Rotational Motion
2
c m
2
2
1
2
1 wIMM g H v
Icm = Large v = Small
3
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Rotational Motion
Rotational Motion
Rolling Motion w/o SlippingRolling Motion w/o Slipping
FasterFaster
Instantaneously restInstantaneously rest
Instantaneous axisInstantaneous axis
w
w
Newton’s 2nd Law of Motion
Translational Motion
Rotational Motion
)(integralnet tv
m
Fa
Quick Question: All shapes have the same mass,
but the speeds are different. Why?
Unlike a = F/m, even if F and m are given,
we have to calculate t and I.
4
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Rotational Motion Rotational Motion
Mechanical Energy ConservationMechanical Energy Conservation
K = Km + KMK = Km + KM
Rotational Motion
ParallelParallel--axis Theoremaxis Theorem
Icm = ½ MR02
d
2 MdII cm
Idisk = Icylinder
Rotational Motion
t = r x F
q
F
r
tz
y
xF
r
z
Rotational Motion
Torque due to Gravity?Torque due to Gravity?
)(sin2
q
t
mg xl
F r
r
x
F
lCM
mass m
? F r
t
Rotational Motion
Vector Nature of Vector Nature of
Angular QuantitiesAngular QuantitiesKinematical variables to describe the rotational motion:
Angular position, velocity and acceleration
c.w. or c.c.w. rotation (like +x or –x direction in 1D)
Vector natures!
)(
)(
)(
2rad/s k
dt
dk
rad/s k dt
dk
rad R
l
ˆˆ
ˆˆ
wa
qw
q
x
y
z
R.-H. Rule
>0 or <0
Newton’s 2nd Law for Rigid-body Rotation
at
In et
Krot = (1/2) I w2
qtt s in Fr
6
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Rotational Motion
9
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Rotational Motion
10
Try This.
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Rotational Motion
Angular Momentum
pprL
FFr
v m pIL
v
mI
t
w
w
11
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Rotational Motion
Rotational Dynamics
dt
pd amF
dt
Ld I
at (torque )
m
I
a
ax
No external force
Momentum conservation
No external torque
L conservation
F
F
r
Speeding up
12
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Rotational Motion
Which one has a bigger I?
Fp
Fp
13
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Rotational Motion
Which one has a bigger I?
15
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Rotational Motion
x
y
)()( 21 tF tF xx
fx ,fx ,x ,ix ,i p ppp 2121
Ia: big Ib: small z
17
Angular Momentum Conservation
The gravitational force and normal force don’t do anything about the rotation. The net torque is zero.
0)()( 21 tF tF xx
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Rotational Motion
Ia: big Ib: small z
)(aw
)(bw
18
Angular Momentum Conservation
We observe the rotational speed increases. angular velocity (“omega”) increases. Why?
0ˆ )( Ng
kτττ i
x
y
)()( 21 tF tF xx
fx ,fx ,x ,ix ,i p ppp 2121
0)()( 21 tF tF xx
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Rotational Motion
Ia: big Ib: small z
)(aw
)(bw
19
Angular Momentum Conservation
x
y
)()( 21 tF tF xx
fx ,fx ,x ,ix ,i p ppp 2121
0)()( 21 tF tF xx
0ˆ )( Ng
kτττ i
bbaa
ba
I I
L L
ww
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Rotational Motion
2
cm
2
2
1
2
1 wIMMgH v
Icm = Large v = Small
bbaa
ba
I I
L L
ww
0ˆ )( Ng
kτττ i
Ia: big Ib: small
z
)(aw
)(bw
Angular momentum conservation Energy conservation
20
Conceptual Training
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Rotational Motion
A person stands, hands at the side, on a platform that is
rotating at a rate of 1.60 rev/s. If the person now raises
his arms to a horizontal position, the speed of rotation
decreases to 0.800 rev/s.
(a) Why does the speed of
rotation decrease?
Explain using the two
key words;
external torque
angular momentum
(b) By what factor has the
moment of inertia of the
person changed?
(c) Compare Ki and Kf Iiwi Ifwf
21
Example 1
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Rotational Motion
22
Example 2 P10-43 Simplified Model
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Rotational Motion
23
Angular Momentum Conservation in Life
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Rotational Motion
24
Angular Momentum Conservation?
Yes.
Let’s try to see in the next page.
E. 10.42
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Rotational Motion
m2
m1
m1
wi
wf
Ii=MR2
If=MRf2
Rf
27
Iiwi Ifwf
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Rotational Motion
A uniform disk turns at 7.00 rev/s around a
frictionless spindle. A non-rotating rod, of the
same mass (m) as the disk and length (l) equal
to the disk’s diameter, is dropped onto the disk.
They then both turnaround the spindle with
their centers superposed. There is no slipping
between the rod and the disk. What is the
moment of inertia of the disk+rod system about
the axis?
28
Pre-Example 2 E. 10.45
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Rotational Motion
Example 2
A uniform disk turns at 7.00 rev/s around a
frictionless spindle. A non-rotating rod, of the
same mass (m) as the disk and length (l) equal
to the disk’s diameter, is dropped onto the disk.
They then both turnaround the spindle with
their centers superposed. There is no slipping
between the rod and the disk. What is the
angular velocity (in rev/s) of the disk+rod
system about the axis?
29
Iiwi Ifwf
E. 10.45
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Rotational Motion
A uniform disk turns at 7.00 rev/s around a
frictionless spindle. A non-rotating rod, of the
same mass (m) as the disk and length (l) equal
to the disk’s diameter, is dropped onto the disk.
They then both turnaround the spindle with
their centers superposed. There is no slipping
between the rod and the disk. What is the
angular velocity (in rev/s) of the disk+rod
system about the axis?
Example 2
32
Iiwi Ifwf
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Rotational Motion
Example 3
34
Example 10.10
Example 10.11
Example 10.12
P10-43
P10-45
P10-50
P10-88
P10-91
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Rotational Motion
Example 3 Hint
35
Example 10.10
Example 10.11
Example 10.12
P10-43
P10-45
P10-50
P10-88
P10-91
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Rotational Motion
Example 3 – Deep Impact Suppose a meteor (m = 7.0 x 1010 kg) struck the Earth at the
equator with a speed v = 1.0 x 104 m/s and remained stuck. By what
factor would this affect the rotational frequency of the Earth?
41
r
v
E a rthRr
m w(before)
prLm
Basic concept:
“By what factor..”
R = w(after)/w(before)
Lm =
LE = IEarth w(before)
Lm+E = Imeteor+Earth w(after)
REarth m v sinq
(into page)
(into page)
(out of page)
Emmm LLL
Imeteor+Earth = Imeteor + IEarth
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Table 9.2
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A spinning figure
skater pulls his arms
in as he rotates on the
ice. As he pulls his
arms in, what
happens to his
angular momentum L
and kinetic energy K?
A. L and K both increase.
B. L stays the same; K increases.
C. L increases; K stays the same.
D. L and K both stay the same.
Q10.11
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The four forces shown all have the
same magnitude: F1 = F2 = F3 = F4.
Which force produces the greatest
torque about the point O (marked by
the blue dot)?
A. F1
B. F2
C. F3
D. F4
E. not enough information given to decide
Q10.1 F1
F2
F3
F4
O
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Which of the four forces shown
here produces a torque about O that
is directed out of the plane of the
drawing?
A. F1
B. F2
C. F3
D. F4
E. more than one of these
Q10.2 F1
F2
F3
F4
O
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A plumber pushes
straight down on the
end of a long wrench
as shown. What is the
magnitude of the
torque he applies about
the pipe at lower right?
A. (0.80 m)(900 N)sin 19°
B. (0.80 m)(900 N)cos 19°
C. (0.80 m)(900 N)tan 19°
D. none of the above
Q10.3
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Q10.4
What is the torque that this force applies about the origin?
A. zero
B.
C.
D.
E.
ˆ ˆ24 18 N mi + j •
ˆ ˆ24 18 N mi j •
ˆ ˆ18 24 N mi + j •
ˆ ˆ18 24 N mi j •
A force ˆ ˆ4 3 NF = i j acts on an object at a point
ˆ6 m.r = klocated at the position
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A. m2g = T2 = T1
B. m2g > T2 = T1
C. m2g > T2 > T1
D. m2g = T2 > T1
E. none of the above
Q10.5
A glider of mass m1 on a frictionless horizontal track is connected
to an object of mass m2 by a massless string. The glider accelerates
to the right, the object accelerates downward, and the string rotates
the pulley. What is the relationship among T1 (the tension in the
horizontal part of the string), T2 (the tension in the vertical part of
the string), and the weight m2g of the object?
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A lightweight string is wrapped
several times around the rim of a
small hoop. If the free end of the
string is held in place and the hoop
is released from rest, the string
unwinds and the hoop descends.
How does the tension in the string
(T) compare to the weight of the
hoop (w)?
A. T = w
B. T > w
C. T < w
D. not enough information given to
decide
Q10.6
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A solid bowling ball rolls
down a ramp.
Which of the following forces
exerts a torque on the bowling
ball about its center?
A. the weight of the ball
B. the normal force exerted by the ramp
C. the friction force exerted by the ramp
D. more than one of the above
E. The answer depends on whether the ball rolls without
slipping.
Q10.7
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Rotational MotionAngular
Momentum
Challenge Change in
Angular Momentum Conservation
dt
Ld I
at (torque )
59
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Rotational MotionAngular
Momentum
Challenge: Change in
Angular Momentum (I)
x
y
z
R.-H. Rule
x
y
z
R.-H. Rule
wf Iwi Iwf wi
dt
Ld I
at (torque )
F
60
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Rotational MotionAngular
Momentum
Only the direction of L changes;
The magnitude of L does not change.
Out of page (+x axis)
F
iL
fL
if LLL
dt
Ld I
at (torque )
61
Challenge: Change in
Angular Momentum (II)
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Rotational MotionAngular
Momentum and Torque
dt
Ld I
at (torque)
62
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Rotational MotionAngular
Momentum and Torque
z
y
x
Spinning
L Spinning Axis
Rope
x
z
Rotation (in xy plane) of
Spinning Bicycle Wheel
63
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Rotational Motion
Example 1
64
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Rotational Motion
Recap: Example 1
Calculate the torque on the 2.00-m long
beam due to a 50.0 N force (top) about
(a) point C (= c.m.)
(b) point P
Calculate the torque on the 2.00-m long
beam due to a 60.0 N force about
(a) point C (= c.m.)
(b) point P
Calculate the torque on the 2.00-m long
beam due to a 50.0 N force (bottom) about
(a) point C (= c.m.)
(b) point P
qtt s in Fr
0ˆ )( ? ]W hy[ Ng
kτττ i
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Rotational Motion
L = r p sinq Direction and magnitude of angular momentum?
Magnitude: L = r p sinq
Lm = r (mv) sinq = R0 m v
r
q v
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Rotational Motion
L = r p sinq Direction and magnitude of angular momentum?
Magnitude: L = r p sinq
Lm = r (mv) sinq = R0 m v Lm = ?
q
w
r
v
E a rthRr
r
q v
m
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Rotational Motion
L = r p sinq Direction and magnitude of angular momentum?
Magnitude: L = r p sinq
Lm = r (mv) sinq = R0 m v Lm = REarth m v sinq
q
w
r
v
E a rthRr
r
q v
m
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Rotational Motion
(a) What is the angular momentum of a 3.00-kg uniform cylindrical
grinding wheel of radius 20 cm and height 40 cm when rotating
at 1500 rpm?
(b) How much torque is required to stop it in 10 s?
H 0.4H
w
if
if
tt
LL
td
Ld b
ILa
t
w
)(
)(Hints:
69
Example 2