Transcript Physics 1422 - Introduction

Physics 203 – College Physics I
Department of Physics – The Citadel
Physics 203
College Physics I
Fall 2012
S. A. Yost
Chapter 8 Part 2
Rotational Motion
Physics 203 – College Physics I
Department of Physics – The Citadel
Announcements
Problem set HW08 is due Thursday.
It covers Chapter 8 except angular momentum,
rolling, and some torque problems.
Today: Rotational Motion, Ch. 10 , mostly sec7 – 8.
Thursday: chapter 9, sec. 1, 2, and 4 .
Please read these sections before class. A problem
set HW09 on them and the remainder of Ch. 8 will
be posted today and due next Tuesday.
Deadline for making up Exam 2: Wed. 5 PM.
Physics 203 – College Physics I
Department of Physics – The Citadel
Rotational Analogs of Linear Motion
Linear Motion
Rotational Motion
Kinematics:
x , v, a
q, w, a
Dynamics:
m, F, Kt
I, t, Kr
F = ma
t = Ia
Kt = ½ mv 2
Kr = ½ Iw 2
W = Fx
W=tq
P=Fv
P = tw
Physics 203 – College Physics I
Department of Physics – The Citadel
Torque
F
R
q
The torque is defined to
be the perpendicular
component of the force
times the distance from
the pivot to where it acts:
t = R F^
Counter-clockwise torque
is considered to be
positive, as for angles.
where
Then
F^ = F sin q.
t = R F sin q .
Physics 203 – College Physics I
Department of Physics – The Citadel
Torque
F
R
q
t = R F^
t = R F sin q
The torque can also be
expressed in terms of the
magnitude of the force and
the distance from the axis
to the line of the force.
t = R^ F
The distance R ^ is called
the lever arm of the torque.
Physics 203 – College Physics I
Department of Physics – The Citadel
Tightening a Nut with a Wrench
Which use of the
wrench is most
effective for
tightening the nut?
A
B
Which is least
effective?
Which of A and D is
more effective?
Choose E if they are
the same.
C
D
The lever arms are the same.
Physics 203 – College Physics I
Department of Physics – The Citadel
Question
A force is applied to the rim of two wheels.
Assuming the only significant mass is in the rim,
what force F2 will give the wheels identical angular
accelerations?
(A) 0.25 N
(D) 2.0 N
(B) 0.5 N
(E) 4.0 N
(C) 1.0 N
I = mR2
Physics 203 – College Physics I
Department of Physics – The Citadel
Question
t = RF = I a = mR2 a.
t2 R2F2
t1 = R1F1 =
mR22 a
mR12 a
F2
R2
→ F = R
1
1
F2 = 2 F1 = 2 N.
I = mR2
Physics 203 – College Physics I
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Example
Mass falling on rope
wrapped around a
massive pulley.
m
R
Assume the pulley is a
uniform disk as shown.
What is the acceleration of
the hanging mass?
M
a
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
Isolate the hanging mass:
Newton’s Law:
M a = Fnet = Mg – FT
where FT is the tension in
the rope.
FT
M
mg
a
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
Isolate the pulley: t = I a
with I = ½ m R2, t = RFT ,
a = a/R.
m
a
Then RFT = (½ mR2)(a/R).
Therefore, FT = ½ ma.
FT
R
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
Combine results:
m
M a = Fnet = Mg – FT
= Mg – ½ ma.
FT
Then (M + m/2) a = Mg.
Result:
a=
g
M
1 + m/2M
a
R
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
Note that the tension does
a
not need to be the same
on two sides of a
massive pulley.
F2
R
R
Net torque =
R(F1 – F2) = Ia.
F1
m
Physics 203 – College Physics I
Department of Physics – The Citadel
Rigid Body Motion
The relation t = I a holds for rigid body rotation
in any inertial frame.
This always holds in the CM frame of the rigid
body, even if it is accelerating.
The energy of a rigid body can be expressed as
a sum K = K cm + K rot with
K cm = ½ mvcm2, K rot = ½ Iw 2.
“Newton’s Law”
F = m acm (ch. 7), t = I a.
Physics 203 – College Physics I
Department of Physics – The Citadel
Dumbbell
→
→
A force F is applied for time t to a
dumbbell in one of two ways
shown.
m
F
m
A
Which gives the greater speed to
the center of mass?
→
(a) A
F
(b) B
(c) the same
→
→
Dp = Ft
m
m
B
Physics 203 – College Physics I
Department of Physics – The Citadel
Dumbbell
→
→
A force F is applied for time t to a
dumbbell in one of two ways
shown.
m
F
m
A
Which gives the greater energy to
the dumbbell?
→
(a) A
(b) B
F
(c) the same
m
m
B
Physics 203 – College Physics I
Department of Physics – The Citadel
Dumbbell
→
The total kinetic energy is
Case A:
m
F
m
A
K = Ktrans + Krot = ½ mvcm2 + ½ Iw2
Case B: no rotation:
K = ½ mvcm2
→
F
There is more energy in case A.
m
m
B
Physics 203 – College Physics I
Department of Physics – The Citadel
Rolling
When an object rolls, its
circumference moves a distance 2pr
every period, so w and v are related:
v = 2pr/T = rw
2pr
2pr
2pr
2pr
Physics 203 – College Physics I
Rolling
A solid wheel and a
hollow wheel roll
down a ramp,
starting from rest
at the same point.
Which gets to the
bottom faster?
Department of Physics – The Citadel
Physics 203 – College Physics I
Department of Physics – The Citadel
Rolling
If an object with mass
m and moment of
inertia I rolls down
an inclined plane of
height h and length
L, how fast is it
rolling when it gets
to the bottom?
m,I
h
L
Physics 203 – College Physics I
Department of Physics – The Citadel
Rolling
Energy conservation:
Ui = Ktrans + Krot
mgh = ½ mv2 + ½ Iw2.
Rolling: w = v/R.
mgh = ½ (m + I/R2) v2.
v=
√
2gh
1 + I/(mR2)
h
L
m,I
Physics 203 – College Physics I
Rolling
The solid wheel gets
to the bottom first,
because the object
with the smaller
moment of inertia
relative to its mass
and size moves
faster.
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Physics 203 – College Physics I
Department of Physics – The Citadel
Rotational Analog of Momentum
Linear Motion:
(one dimension)
Rotational Motion:
(fixed axis)
Momentum:
p = mv
Angular momentum:
L = Iw
Impulse:
Dp = Ft
DL = t t
Physics 203 – College Physics I
Department of Physics – The Citadel
Angular Momentum
Units of angular momentum:
L = I w = [kg m2][s-1] = kg . m2/s
DL = t t = [mN][s] = Nms = J.s
When there is no external torque on a system,
angular momentum is conserved.
In particular, this applies to collisions between
rigid bodies.
Physics 203 – College Physics I
Figure Skater
A figure skater increases
her rotational rate from 1.0
rev/s to 2.5 rev/s in 1.5 s.
Her initial moment of
inertia was 4.6 kg ∙ m2.
(a) What was her final
moment of inertia?
Department of Physics – The Citadel
Physics 203 – College Physics I
Department of Physics – The Citadel
Figure Skater
A figure skater increases
her rotational rate from 1.0
rev/s to 2.5 rev/s in 1.5 s.
Her initial moment of
inertia was 4.6 kg ∙ m2.
I1w1 = I2 w2
w2 = 2.5 w1
I2 = I1 / 2.5 = 1.84 kg ∙ m2
≈ 1.8 kg ∙ m2
Physics 203 – College Physics I
Figure Skater
A figure skater increases
her rotational rate from 1.0
rev/s to 2.5 rev/s in 1.5 s.
Her initial moment of
inertia was 4.6 kg ∙ m2.
(b) What average power did
she apply to pull in her
arms?
Department of Physics – The Citadel
Physics 203 – College Physics I
Department of Physics – The Citadel
Figure Skater
P = W/t,
W = DK = ½ Iw22 – ½ Iw12.
w1 = 1.0 rev/s (2p rad/rev) = 2.0 p rad/s
w2 = 2.5 rev/s (2p rad/rev) = 5.0 p rad/s
I1 = 4.6 kg ∙ m2,
I 2 = 1.84 kg ∙ m2
W = 277 J – 90.8 J ≈ 186 J,
P = 124 W.
t = 1.5 s.
Physics 203 – College Physics I
Department of Physics – The Citadel
Analog of Inelastic Collision
A bar of length 2R is dropped
onto a rotating disk of
radius R.
Suppose M = 2m. If the disk
initially rotates at 120 rpm,
how fast does it rotate if
the stick drops onto it and
rotates together with the
disk?
m
M
R
w0
Physics 203 – College Physics I
Department of Physics – The Citadel
Analog of Inelastic Collision
Angular momentum is conserved: I1w1 = I2 w2 .
I1 =½ MR2 = mR2 and I2 = Ibar + Idisk with
Ibar = mL2/12 = mR2/3,
Idisk = ½ MR2 = mR2
I2 = 4mR2/3
w2= ( I1 / I2) w1 = ¾ w1
= ¾ (120 rpm) = 90 rpm.
m
R
M
wf
Physics 203 – College Physics I
Department of Physics – The Citadel
Angular Quantities as Vectors
If the axis is not fixed, we have to specify a
direction for angular displacements and
velocities. The convention use a vector
pointing along the axis.
→
q
→
→
→
w = d q /dt
w
For fixed axis →rotations,
→
the vectors q and w are
parallel, but they won’t be
if the axis direction changes.
q
w
Right-Hand Rule