Chapter 7 - SteadyServerPages

Report
Chapter 7
• Rotational
Motion
Slide 7-3
Curvilinear coordinates
 = 
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1 2
∆ =   + 
2
2 = 2 + 2∆
1 2
2
2

=

∆ =   + 

 + 2∆
2 © 2015 Pearson Education, Inc.
 = 

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∆
=
∆
This slope is 
© 2015 Pearson Education, Inc.
Checking Understanding
Two coins rotate on a turntable. Coin B is
twice as far from the axis as coin A.
A.
The angular velocity of A is twice
that of B.
B.
The angular velocity of A equals
that of B.
C. The angular velocity of A is half
that of B.
Slide 7-13
Answer
Two coins rotate on a turntable. Coin B is
twice as far from the axis as coin A.
A.
The angular velocity of A is twice
that of B.
B. The angular velocity of A
equals that of B.
C. The angular velocity of A is half
that of B.
All points on the turntable rotate through the same angle in the same time.
All points have the same period.
Slide 7-14
Angular acceleration α measures
how rapidly the angular velocity is
changing:
t
=

t
Tangential acceleration
Slide 7-17
Slide 7-18
Checking Understanding
Two coins rotate on a turntable. Coin B is
twice as far from the axis as coin A.
A.
The speed of A is twice that of B.
B.
The speed of A equals that of B.
C. The speed of A is half that of B.
Slide 7-15
Answer
Two coins rotate on a turntable. Coin B is
twice as far from the axis as coin A.
A.
The speed of A is twice that of B.
B.
The speed of A equals that of B.
C. The speed of A is half that of B.
v  r
Twice the radius means twice the speed
Slide 7-16
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Center of mass follows
original trajectory


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Slide 7-19
Example Problem
A high-speed drill rotating CCW takes 2.5 s to speed up to 2400
rpm.
A. What is the drill’s angular acceleration?
B. How many revolutions does it make as it reaches top speed?
1 2
∆ =   + 
2
2 = 2 + 2∆
2∆
=
2

2
= ∆
2
Slide 7-21
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Slide 7-22
The speed is changing
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Center of Gravity
=
Slide 7-29
Calculating the Center-of-Gravity Position
Slide 7-30
1 1 + 2 2 5kg ∙ m 1
cg =
=
= m
1 + 2
15kg
3
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 =  sin 
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Angle between lever
arm  and force 
Checking Understanding
Which point could be the center of gravity of this L-shaped
piece?
Slide 7-32
Answer
Which point could be the center of gravity of this L-shaped
piece?
(a)
Slide 7-33
Interpreting Torque
Torque is due to the component of the force perpendicular to the
radial line.
  rF  rF sin 
Slide 7-25
Signs and Strengths of the Torque
Slide 7-27
The four forces below are equal in magnitude. Which
force would be most effective in opening the door?
1
2
3
4
Either 1 or 3
98%
2%
$$
4
3
0%
$$
1
or
$$
er
Ei
th
3
0%
$$
2
$$
1
0%
$$
A.
B.
C.
D.
E.
Example torque Problem
 =  sin 
Revolutionaries attempt to pull down a statue of the Great Leader
by pulling on a rope tied to the top of his head. The statue is 17 m
tall, and they pull with a force of 4200 N at an angle of 65° to the
horizontal. What is the torque they exert on the statue? If they are
standing to the right of the statue, is the torque positive or
negative?
 = 17m ∙ 4200N ∙ sin 
°
 ≠ 65
 = 17m ∙ 4200N ∙ sin 25°
17 m
Negative torque, but why?
Rotating it in the CW direction
r = 17m
65°
pivot
F = 4200 N
Slide 7-28
Which force vector on point P would
keep the wheel from spinning?
98%
0%
E
2%
D
0%
C
A
C
D
E
A
A.
B.
C.
D.
 =  sin 

 =  sin 
 = 1.6m 3.2kg
m
°
−9.8 2 sin 65
s
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Which torques are equal?
98%
A. B = C = D = E only
B. A = B and C = D = E
C. None are equal
D. B = E and C = D
2%
=D
C
nd
Ea
=
B
No
ne
C
d
an
B
=
A
ar
e
=
D
=
eq
ua
l
y
Eo
nl
=
D
=
C
=
B
0%
E
0%
What is the Net Torque is exerted by the
gymnast about an axis through the rings?
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=0
1
1
2
2
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Reading Quiz
2. Which factor does the torque on an object not depend on?
A.
B.
C.
D.
The magnitude of the applied force.
The object’s angular velocity.
The angle at which the force is applied.
The distance from the axis to the point at which the
force is applied.
Slide 7-7
Answer
2. Which factor does the torque on an object not depend on?
A.
B.
C.
D.
The magnitude of the applied force.
The object’s angular velocity.
The angle at which the force is applied.
The distance from the axis to the point at which the
force is applied.
Slide 7-8
Example Problem
An object consists of the three balls shown, connected by
massless rods. Find the x- and y-positions of the object’s center
of gravity.
1 1 + 2 2 + 3 3
cg =
1 + 2 + 3
1 1 + 2 2 + 3 3
cg =
1 + 2 + 3
Slide 7-31
An object consists of the three balls shown, connected by
massless rods. Find the x- and y-positions of the object’s center
of gravity.
1 0 + 2 0 + 3 1m
2kg ∙ m 1
cg =
=
= m
1 + 2 + 3
4kg
2
1 1m + 2 0 + 3 0
1kg ∙ m 1
cg =
=
= m
1 + 2 + 3
4kg
4
The center of mass for these 3 bodies
Slide 7-31
Newton’s Second Law for Rotation
  /I
I = moment of inertia. Objects with larger moments of inertia are
harder to get rotating.
I   mi ri
2
Slide 7-34
Rotational and Linear Dynamics Compared
Slide 7-36
Which moment of inertia is greatest?
25%
25%
25%
25%

D
C
B
A
B
C
D
A
A.
B.
C.
D.

20%
20%
20%
E
B
A
B
C
D
E
A
A.
B.
C.
D.
E.
20%
D
20%
C
Which force vector applied to point P
will stop this rolling ball?

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Which red vector is your best bet for
getting this bolt as tight as possible?
25%
25%
25%
D
C
B
A
B
C
D
A
A.
B.
C.
D.
25%
Reading Quiz
1. Moment of inertia is
A.
B.
C.
D.
the rotational equivalent of mass.
the point at which all forces appear to act.
the time at which inertia occurs.
an alternative term for moment arm.
Slide 7-5
Answer
1. Moment of inertia is
A.
B.
C.
D.
the rotational equivalent of mass.
the point at which all forces appear to act.
the time at which inertia occurs.
an alternative term for moment arm.
Slide 7-6
≠0
 = 
What happens to these masses when
you let go?
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What happens to this pulley system?
25% 25% 25%
ac
ce
le
fo
ra
rc
te
e
st
of
..
gr
a
vi
Th
ty
em
on
as
th
sm
e
...
ov
es
up
w
ar
d
at
a.
..
Th
e
Th
e
10
N
fo
rc
e
It
do
es
n
ot
m
ov
e
25%
A. It does not move
B. The 10N force
accelerates the mass
upward
C. The force of gravity on
the mass results in a net
force upward
D. The mass moves upward
at a constant speed
Starting from rest, how long does
it take to hit the ground?
1 = 1 + 1 = 1 1
1
2 = 2 + 2 = 2 2
Newton’s Third
1

1 = −2
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2

Starting from rest, how long does
it take to hit the ground?
2 = 1 = 
1  + 1 = 1 
2  − 1 = 2 
1
Get rid of 1 by solving for it and
substituting into the other equation
m
 = −3.27 2
s
1 2
∆ =   + 
2
=
2∆

1

2
 = .78s
© 2015 Pearson Education, Inc.

2
Reading Quiz
4. A net torque applied to an object causes
A.
a linear acceleration of the object.
B.
the object to rotate at a constant rate.
C. the angular velocity of the object to change.
D. the moment of inertia of the object to change.
Slide 7-11
Answer
4. A net torque applied to an object causes
A.
a linear acceleration of the object.
B.
the object to rotate at a constant rate.
C. the angular velocity of the object to change.
D. the moment of inertia of the object to change.
Slide 7-12
Draw the normal force for
the wheel against the break
Draw the frictional force
from the break


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Is this beam balanced?
A.
B.
C.
D.
Yes
No, it will spin CW
No, it will spin CCW
Not enough information
W
CC
CW
in
gh
ill
No
te
no
u
tw
No
,i
tw
ill
sp
sp
in
in
Ye
s
No
,i
fo
rm
at
io
n
25% 25% 25% 25%
1
 =   2
2
 = 1 + 2 =  
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1
 =   2
2
4 kg
.02m ∙ 20N − .02m ∙ 30N = .0008 kg ∙ m2 
© 2015 Pearson Education, Inc.
Additional Example Problem
A baseball bat has a mass of 0.82 kg and is 0.86 m long. It’s
held vertically and then allowed to fall. What is the bat’s angular
acceleration when it has reached 20° from the vertical? (Model
the bat as a uniform cylinder).
Slide 7-43

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