m - De Anza College

Transcription

m - De Anza College

Rotation
Rolling Motion
Chapter 10 Review
Lana Sheridan
De Anza College
June 2, 2015
Overview
• rolling motion
• Chapter 10 practice
Energy
In total then,
W = ∆Ktrans + ∆Krot + ∆U + ∆Eint
where
• ∆Ktrans represents the kinetic energy of the CM motion, and
• ∆Krot is the rotational kinetic energy
Rolling Motion
e translation
Combination of
translation and rotation
A combination of translation and rotation.
v CM
CM
v ! v CM " R v ! 2v CM
CM
v CM
v ! v CM
Figure 10
P
v CM
P
v !0
c
The CM is translated at velocity vCM .
rolling obj
a combina
and pure r
Notice that instantaneously P is the pivot point of the rotation.
Henry Leap and Ji
Rolling Motion
The pivot point of the rotation changes as different parts of the
wheel contact the surface.
ranslational speed of
(10.28)
holds whenever a cylpure rolling motion.
mass for pure rolling
(10.29)
at speed v CM, staying
mass of the object. As
R
u
s
s !R u
Figure 10.24
For pure rolling
motion, as the cylinder rotates
ds
through an angle
center
vCM = u=itsRω
moves a linear distance
s 5 R u.
dt
Kinetic Energy of a Rolling Object
The kinetic energy of a rolling object is just the sum of the
rotational KE and translational KE. Can see this by considering
just an instantaneous rotation about the point P:
K
=
1
IP ω2
2
Using the parallel axis theorem: IP = ICM + mR 2 .
K
1
(ICM + mR 2 )ω2
2
1
1 2
=
ICM ω2 + mvCM
2
2
= KCM,rot + KCM,trans
=
Equation 10.30 to obtain
Rolling down an incline
K5
A sphere starts from rest at the top of an incline
andv CM
rolls
down.
Using
5 Rv,
this equation ca
Find the (translational) velocity
of
the
center
of
mass
at
the
K
Total kinetic energy
of a rolling object
bottom of the incline.
1
2
M
R
h
x
u
v
S
vCM
Figure 10.26 A sphere rolling down an incline. Mechanical
energy of the sphere–Earth system
is conserved if no slipping occurs.
The term 2 ICMv represents the
center of mass, and the term 12Mv
have if it were just translating th
kinetic energy of a rolling objec
the center of mass and the trans
statement is consistent with the
that the velocity of a point on th
mass and the tangential velocity
Energy methods can be used
ing motion of an object on a ro
which shows a sphere rolling wit
top of the incline. Accelerated
is present between the sphere a
center of mass. Despite the pre
occurs because the contact poin
(On the other hand, if the sphe
incline–Earth system would dec
friction.)
In reality, rolling friction cau
energy. Rolling friction is due to
For example, automobile tires fl
K5
andv CM
rolls
down.
Using
5 Rv,
this equation ca
of
the
center
of
mass
at
the
K
of a rolling object
1
2
R
h
v
x
S
vCM mass and the tangential velocity
u
Figure 10.26 A sphere rolltop of the incline. Accelerated
ing down an incline. Mechanical
∆K + ∆U
= 0
is present between the sphere a
energy of the sphere–Earth system
is conserved if no slipping occurs.
friction.)
M
K5
andv CM
rolls
down.
Using
5 Rv,
this equation ca
of
the
center
of
mass
at
the
K
of a rolling object
1
2
R
h
v
x
S
u
∆K
+
∆U
= 0
system is present
between the sphere a
energy of the sphere–Earth
1
1
2 is conserved
2 if no slipping occurs.
ICM ω + MvCM − 0 + (0 − Mgh)
= 0
2
2
friction.)
M
K5
andv CM
rolls
down.
Using
5 Rv,
this equation ca
of
the
center
of
mass
at
the
K
of a rolling object
1
2
R
h
v
x
S
u
∆K
+
∆U
= 0
system is present
between the sphere a
energy of the sphere–Earth
1
1
2 is conserved
2 if no slipping occurs.
ICM ω + MvCM − 0 + (0 − Mgh)
= 0
2
2
s
friction.)
2gh
vCM =
1 + (ICM /MR 2 )) energy. Rolling friction is due to
M
Question
Quick Quiz 10.71 A ball rolls without slipping down incline A,
starting from rest. At the same time, a box starts from rest and
slides down incline B, which is identical to incline A except that it
is frictionless. Which arrives at the bottom first?
(A) The ball arrives first.
(B) The box arrives first.
(C) Both arrive at the same time.
(D) It is impossible to determine.
1
Serway & Jewett, page 318.
Question
Quick Quiz 10.71 A ball rolls without slipping down incline A,
starting from rest. At the same time, a box starts from rest and
slides down incline B, which is identical to incline A except that it
is frictionless. Which arrives at the bottom first?
(A) The ball arrives first.
(B) The box arrives first.
←
(C) Both arrive at the same time.
(D) It is impossible to determine.
1
Serway & Jewett, page 318.
ch as a marble and a croquet ball.
phere, a solid cylinder, or a hoop, we would obtain
Example 10.14 - Pulling a Spool
fer. The constant factors that appear in the exprescylindrically
symmetric
of mass
m and
ut the Acenter
of mass
for the spool
specific
object.
In radius
all R sits at rest
on acceleration
a horizontal table
withhave
friction.
pull on
on a light string
alue the
would
if theYou
incline
were
wrapped around the axle (radius r ) of the spool with a constant
horizontal force of magnitude T to the right. As a result, the spool
rolls without slipping a distance L along the table with no rolling
friction.
Find the final translational speed of the center of mass of the spool.
on a horizontal
rapped around
al force of magg a distance L
L
R
S
r
spool.
Figure 10.27 (Example 10.14)
T
ss than g sin u, the value the acceleration would have if the incline were
AM
adius R sits at rest on a horizontal
on a light string wrapped around
constant horizontal force of magolls without slipping a distance L
L
R
S
T
r
nter of mass of the spool.
Can use: W = ∆K .
he motion of the spool when you
distance L, notice that your hand
ent from L.
A spool rests on a horizontal table.
A string is wrapped around the axle
and is pulled to the right by a hand.
continued
n work–energy relationships for introductory physics,” The Physics Teacher, 43:10, 2005.
AM
L
R
S
r
T
Can use: W = ∆K .
ent from L.
r
W =T L+L
continued
R
AM
L
R
S
r
T
Can use: W = ∆K .
ent from L.
r
W =T L+L
continued
R
s
2TL (1 + r /R)
vCM =
m + I/R 2
mine the moment of inertia of
about
axis. and
Examplethis- structure
Moment
of this
Inertia
Section
10.7
Rotational
Page
328,
#44
Kinetic Energy
Axis of
Rotational
rotation
x
KE
Figure P10.43
44. Rigid rods of negligible mass lying along the y axis conW nect three particles (Fig. P10.44). The system rotates
Q/C about the x axis with an
y
angular speed of 2.00 rad/s.
Find (a) the moment of iner- 4.00 kg
y ! 3.00 m
tia about the x axis, (b) the
total rotational kinetic energy
x
evaluated from 12Iv 2, (c) the
O
tangential speed of each
2.00 kg
y ! "2.00 m
particle, and (d) the total
kinetic energy evaluated from 3.00 kg
y ! "4.00 m
1
2
m
v
.
(e)
Compare
the
a2 i i
answers for kinetic energy in
Figure P10.44
parts (a) and (b).
45. The four particles in Figure P10.45 are connected by
W rigid rods of negligible mass. The origin is at the cen-
u
s
a
p
e
p
p
in
s
s
c
t
w
c
s
Example - Moment of Inertia and Rotational KE
(a) Moment of inertia about x-axis?
Ix = 92 kg m2
(b) & (d) Kinetic energy?
Ix = 92 kg m2
K = 184 J
(c) tangential speeds?
Ix = 92 kg m2
K = 184 J
(c) tangential speeds?
v = rω
Assuming pg333,
the board#
is 81
1.00
Example:(b)Rolling,
m long and is supported at this limiting angle, show that the cup must be
18.4 cm from the moving end.
81. A uniform solid sphere of radius r is placed on the
S inside surface of a hemispherical bowl with radius R.
The sphere is released from rest at an angle u to the
vertical and rolls without slipping (Fig. P10.81). Determine the angular speed of the sphere when it reaches
the bottom of the bowl.
r
u
R
Figure P10.81
82. Review. A spool of wire of mass M and radius R is
top
is a
tom
of
(c)
Fin
tal
be
for
85. A t
S wit
tal
(a)
bef
No
De
bef
86. Re
the
as
Example: Rolling, pg333, # 81
Energy conservation:
∆K + ∆U = 0
1 2 1 2
Iω + mv − 0 + (0 − mg (R − r )(1 − cos θ)) = 0
2
2
v = rω
1 2 2 2 1
( mr )ω + m(r ω)2
2 5
2
+ mg (R − r )(cos θ − 1) = 0
7mr 2 ω2 + 10mg (R − r )(cos θ − 1) = 0
r
ω=
10g (R − r )(1 − cos θ)
7r 2
eter R and also of mass M. What is the kinetic energy of
theAcam–shaft
combination
when
Example:
Trebuchet,
pg328,
# it47is rotating with
angular speed v about the shaft’s axis?
y
47. A war-wolf or trebuchet is a device used during the Mid-
x
43
s conotates
00 m
x
.00 m
4.00 m
Q/C dle Ages to throw rocks at castles and now sometimes
used to fling large vegetables and pianos as a sport. A
simple trebuchet is shown in Figure P10.47. Model it
as a stiff rod of negligible mass, 3.00 m long, joining
particles of mass m1 5 0.120 kg and m 2 5 60.0 kg at its
ends. It can turn on a frictionless, horizontal axle perpendicular to the rod and 14.0 cm from the large-mass
particle. The operator releases the trebuchet from rest
in a horizontal orientation. (a) Find the maximum
speed that the small-mass object attains. (b) While the
small-mass object is gaining speed, does it move with
constant acceleration? (c) Does it move with constant
tangential acceleration? (d) Does the trebuchet move
with constant angular acceleration? (e) Does it have
constant momentum? (f) Does the trebuchet–Earth
system have constant mechanical energy?
nstant momentum? (f) Does the trebuchet–
Example:
Trebuchet
stem
have A
constant
mechanical energy?
m1
m2
3.00 m
Figure P10.47
Example: A Trebuchet
(a) Maximum speed of small mass?
Will occur when arm is vertical, because up until that point gravity
is providing a clockwise torque and doing positive work. (After, the
work is negative)
Will occur when arm is vertical, because up until that point gravity
is providing a clockwise torque and doing positive work. (After, the
work is negative)
The rod has negligible mass. For the rest of the system, use energy
conservation:
∆K + ∆U = 0
1 2
Iω − 0 + m1 g (3 − 0.14) + m2 g (0 − 0.14) = 0
2
Solve for ω, v = r ω.
v = 24.5 m s−1
Summary
• rolling motion
• practice problems
Next test Friday, June 5.
(Uncollected) Homework Serway & Jewett,
• From last time: Ch 10, page 328-. Questions: Section Qs 45,
47, 51, 53
• Ch 10, page 328-. Questions: Section Qs 49, 59, 65, 69

m - De Anza College

Transcription

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