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 1
Rotational Motion
Physics 203 – College Physics I
Department of Physics – The Citadel
Announcements
A problem set HW07B was due today.
Today: Rotational Motion, Ch. 10.
sec. 1 – 6
Next: sec. 7 – 8, skipping sec. 9.
Homework set HW08 on sections 1 – 6 will be due
next Thursday.
Sections 7 – 8 will be combined with chapter 9, sec.
1, 2, and 4 in a later set, HW09.
Physics 203 – College Physics I
Department of Physics – The Citadel
Motion of the Center of Mass
If an external force F acts on an extended object or
collection of objects of mass M, the acceleration of the
CM is given by
F = Macm.
You can apply Newton’s 2nd Law as if it were a particle
located at the CM, as far as the collective motion is
concerned.
This says nothing about the relative motion, rotation,
etc., about the CM. That comes up in chapter 8.
Physics 203 – College Physics I
Department of Physics – The Citadel
Motion of Extended Objects
The motion of extended objects or collections of
particles is such that the CM obeys Newton’s 2nd
Law.
Physics 203 – College Physics I
Department of Physics – The Citadel
Motion of the Center of Mass
The CM of a wrench sliding on a frictionless table will move in
a straight line because there is no external force. In this
sense, the wrench may be though of as a particle located
at the CM.
cm motion
Physics 203 – College Physics I
Department of Physics – The Citadel
Motion of the Center of Mass
For example, if a hammer is thrown, its CM follows a
parabolic trajectory under the influence of gravity,
as a point object would.
Physics 203 – College Physics I
Department of Physics – The Citadel
Motion of the Center of Mass
For example, if a hammer is thrown, its CM follows a
parabolic trajectory under the influence of gravity,
as a point object would.
Physics 203 – College Physics I
Department of Physics – The Citadel
Rotational Motion
Everything we have done for linear motion has an
analog for rotational motion.
We will consider only fixed axis rotations.
Rotational displacements can be expressed in various
units: radians are most standard, but degrees and
revolutions are common as well.
Conversions: 1 rev = 2p rad = 360o, 180o = p rad.
Physics 203 – College Physics I
Department of Physics – The Citadel
Rotational Kinematics
The physics of rotations of a rigid body about a fixed
axis is in many ways analogous to the onedimensional motion of a point particle.
v
x
q
w
Physics 203 – College Physics I
Department of Physics – The Citadel
Rotational Kinematics
Greek letters are used for rotational quantities…
Linear Motion:
Rotational Motion:
distance x
angle q
velocity v
angular velocity w
acceleration a
angular acceleration a
Physics 203 – College Physics I
Department of Physics – The Citadel
Constant (Angular) Acceleration
Linear Motion:
Rotational Motion:
v = v0 + at
w = w0 + a t
x = x0 + v0t + ½ at2
q = q0 + w0t + ½ a t2
v2 = v02 + 2a (x – x0)
w2 = w02 + 2a (q – q0)
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
A centrifuge accelerates from rest to 20,000
rpm in 5.0 min.
(a) What is its average angular acceleration?
(b) How many times does it spin during these
five minutes?
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
(a) a = w / t. t = 300 s.
w = 2p f = 2p (20,000 rev / 60 s) = 2094 rad/s
a = (2094 rad/s) / 300 s = 6.98 rad/s2
20,000 rpm
f
10,000 rpm
(b) Revolutions = favg t
= ½ fmax t = 10,000 rpm × 5 min
= 50,000 rev.
0
t
5 min
Or, use q = ½ a t2 to get the same result with more effort.
Physics 203 – College Physics I
Department of Physics – The Citadel
Rotational Energy
Consider a massless
rotating disk with small
masses mi inserted in it.
The total kinetic energy is
K = S ½ mivi2 .
v = rw for an object a
distance r from the axis.
We’d like to express K in
terms of w.
w
m1
r1 r2
r5
m5
m2
r3
r4
m4
m3
Physics 203 – College Physics I
Department of Physics – The Citadel
Rotational Kinetic Energy
In terms of the angular velocity,
K = S ½ mivi2 = ½ S miri2w2
=½
w2 S
K=½I
w
m1
r1 r2
miri2
w2
r5
m5
I is called the
I = S miri2
r3
r4
m4
moment of inertia.
Units: kg m2.
m2
m3
Physics 203 – College Physics I
Department of Physics – The Citadel
Moment of Inertia
For example, the
moment of inertia of a
hoop of radius R and
mass M about an axis
through the center is
I = MR2
since all parts are the
same distance from
the center.
M
R
Physics 203 – College Physics I
Department of Physics – The Citadel
Moments of Inertia
Thin Hoop
Disk
R
I = MR2
Thin
Rod
I =ML2 /12
R
I = ½ MR2
L
Solid
Sphere
I = 2/5 MR2
Physics 203 – College Physics I
Department of Physics – The Citadel
Solid vs Hollow Sphere
If a solid and hollow sphere
have the same mass and
size, which has a bigger
moment of inertia about an
axis through the center?
Solid
2/5 MR2
A: solid
B: hollow
C: the same
The hollow one, since the mass
is distributed further from the
axis.
Hollow
2/3 MR2
Physics 203 – College Physics I
Department of Physics – The Citadel
Parallel Axis Theorem
The moment of inertia about
any axis can be found if its
value ICM is known about
an axis through the CM.
M
CM
Ih = ICM + Mh2
The moment of inertia is
always the smallest about
an axis through the CM.
h
Physics 203 – College Physics I
Department of Physics – The Citadel
Moment of Inertia of Rod
What is the moment of inertia of uniform rod of mass
M and length L about the end?
Iend = ICM + M(L/2)2
ICM = ML2/12
Iend = ML2/12 + ML2/4
CM
= ML2/3
h = L/2
Physics 203 – College Physics I
Department of Physics – The Citadel
Rotational Work
An object can be given
kinetic energy by doing
work: DK = W
To do work, I can apply a
force F at a point given,
relative to the axis, by a
vector R.
The force does work through
a distance d = Rq.
q
d
F
R
F
Only the component
of F perp. to R does
work: W = F┴ R q.
Physics 203 – College Physics I
Department of Physics – The Citadel
Torque
W = F R q.
The product t = F R is
called torque.
If f is the angle between R
and F, t = RF sin f
Linear motion: W = F x
Rotational motion: W = t q.
_
Power: W = Pt, P = tw.
F
F
R
f
Physics 203 – College Physics I
Department of Physics – The Citadel
Torque
t = F R = RF sin f
F
Equivalent expression:
If R is the component of R
perpendicular to F, then
t=FR .
R is the distance of the line
of the force from the axis,
and is called the lever
arm of the force.
f
R
R
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
Rotational Dynamics
Linear Motion:
Rotational Motion:
mass m
moment of inertia I
force F = ma
torque t = I a
kinetic energy
Kt = ½ mv2
kinetic energy
Kr = ½ Iw2
work Wt = Fx
work Wr = tq
power Pt = Fv
power Pr = tw
Physics 203 – College Physics I
Department of Physics – The Citadel
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. 9), 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
Dumbbell Energy
How much more energy
does it have if I push at
the top?
The bottom mass is initially
at rest, so the speed of
any point on the
dumbbell is initially
proportional to its
distance from the bottom
mass.
m
2vcm
vcm
m
v=0
Physics 203 – College Physics I
Department of Physics – The Citadel
Dumbbell Energy
The initial kinetic energy is then
Ktotal = ½ m (2vcm)2 = 2m vcm2.
m
2vcm
This is conserved as the dumbbell
rotates.
vcm
Kcm = ½ (2m)vcm2 = mvcm2
Krot = Kcm = ½ Ktotal
There is twice as much total
energy.
m
v=0