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Physics
Chapter 8:
Rotational Equilibrium and
Dynamics
Rotational Equilibrium and Dynamics
Rotational Dynamics
What is Involved in Rotating an Object
Force
Distance from the Point of Rotation
Rotational Equilibrium and Dynamics
Rotational Dynamics
To swing open a door, you
exert a force.
The doorknob is near the outer
edge of the door. You exert the
force on the doorknob at right
angles to the door, away from
the hinges.
Rotational Equilibrium and Dynamics
Rotational Dynamics
What is Involved in Rotating an Object
Force
Distance from the Point of Rotation
Torque (t)
t  Fr
Rotational Equilibrium and Dynamics
Torque (t)
Product of the Rotational Force and the
Length of a Lever Arm
t  Fr
Rotational Equilibrium and Dynamics
Torque
Magnitude
Fr
Direction
+ When Rotation is Counterclockwise
- When Rotation is Clockwise
t  Fr
Rotational Equilibrium and Dynamics
Torque
Units
Force = Newton
Length = meter
Torque = Newton*Meter (N*m)
t  Fr
Rotational Equilibrium and Dynamics
Torque
Direction is Not Always Perpendicular to the
Radius
t  Fr sin 
Rotational Equilibrium and Dynamics
Problem
The gardening tool shown is used to pull
weeds. If a 1.23N*m torque is required to pull a
given weed, what force did the weed exert on
the tool?
Rotational Equilibrium and Dynamics
Solution
t = 1.23 N*m
r = 0.040m
F = ?
t  Fw (0.040 m)  1.23 N  m  0
1.23 N  m
Fw 
 31 N
0.040 m
Rotational Equilibrium and Dynamics
Problem
• What is the torque on a bolt produced by a
15N force exerted perpendicular to a wrench
that is 25cm long.
Rotational Equilibrium and Dynamics
Solution
F = 15N
r = 25cm
t  Fr sin 
t  (15N )(0.25m)(sin 90.0 )  3.8N  m
o
Rotational Equilibrium and Dynamics
Problem
A person slowly lowers a 3.3kg crab trap over
the side of a dock. What torque does the trap
exert about the person’s shoulder?
Rotational Equilibrium and Dynamics
Solution
t = ?
r = 0.70m
m = 3.3kg
m

t  r mg  (0.70 m)(3.3 kg)  9.81 2   23 N  m
s 

Rotational Equilibrium and Dynamics
Problem
At the local playground, a 16kg child sits on
the end of a horizontal teeter-totter, 1.5 m from
the pivot point. On the other side of the pivot
an adult pushes straight down on the teetertotter with a force of 95N. In which direction
does the teeter-totter rotate if the adult applies
the force at a distance of 3.0 m from the pivot?
Rotational Equilibrium and Dynamics
Solution
rchild = 1.5m
mchild = 16kg
radult = 3.0m
Fadult = 95N
t  0
rc Fc  rA FA
rc (mc g ) (1.5m)(16kg)(9.81m / s 2 )
rA 

 2.48m
FA
95 N
3.0 m > 2.48 m; the adult has enough momentum to push down and push the child up.
Rotational Equilibrium and Dynamics
Problem
At the local playground, a 16kg child sits on
the end of a horizontal teeter-totter, 1.5 m from
the pivot point. On the other side of the pivot
an adult pushes straight down on the teetertotter with a force of 95N. In which direction
does the teeter-totter rotate if the adult applies
the force at a distance of 2.5m from the pivot?
Rotational Equilibrium and Dynamics
Solution
rchild = 1.5m
mchild = 16kg
radult = 2.5m
Fadult = 95N
t  0
rc Fc  rA FA
rc (mc g ) (1.5m)(16kg)(9.81m / s 2 )
rA 

 2.48m
FA
95 N
2.5 m > 2.48 m; the adult has just barely enough momentum to push the child up.
Rotational Equilibrium and Dynamics
Problem
At the local playground, a 16kg child sits on
the end of a horizontal teeter-totter, 1.5 m from
the pivot point. On the other side of the pivot
an adult pushes straight down on the teetertotter with a force of 95N. In which direction
does the teeter-totter rotate if the adult applies
the force at a distance of 2.0m from the pivot?
Rotational Equilibrium and Dynamics
Solution
rchild = 1.5m
mchild = 16kg
radult = 2.0m
Fadult = 95N
t  0
rc Fc  rA FA
rc (mc g ) (1.5m)(16kg)(9.81m / s 2 )
rA 

 2.48m
FA
95 N
2.0 m < 2.48 m; the adult does not have enough momentum to hold the child up, and the child goes down.
Rotational Equilibrium and Dynamics
Center of Mass
The center of mass of an object is the point on the
object that moves in the same way that a point particle
would move.
The path of center of mass of the object is a straight
line.
Rotational Equilibrium and Dynamics
Center of Mass
To locate the center of mass of an
object, suspend the object from any
point.
When the object stops swinging, the
center of mass is along the vertical
line drawn from the suspension point.
Draw the line, and then suspend the
object from another point. Again, the
center of mass must be below this
point.
Rotational Equilibrium and Dynamics
Center of Mass
Draw a second vertical line.
The center of mass is at the
point where the two lines
cross.
Rotational Equilibrium and Dynamics
Center of Mass and Stability
An object is said to be stable if an external force is
required to tip it.
The object is stable as long as the direction of the
torque due to its weight, τw tends to keep it upright.
This occurs as long as the object’s center of mass lies
above its base.
To tip the object over, you must rotate its center of
mass around the axis of rotation until it is no longer
above the base of the object.
To rotate the object, you must lift its center of mass.
The broader the base, the more stable the object is.
Rotational Equilibrium and Dynamics
Center of Mass and Stability
If the center of mass is outside the base of an
object, it is unstable and will roll over without
additional torque.
If the center of mass is above the base of the
object, it is stable.
If the base of the object is very narrow and the
center of mass is high, then the object is stable,
but the slightest force will cause it to tip over.
Rotational Equilibrium and Dynamics
Center of Mass and Stability
Rotational Equilibrium and Dynamics
Moment of Inertia (I)
The Inertia of a Rotating Mass
For a Point Mass…
I  mr
2
Rotational Equilibrium and Dynamics
Moment of Inertia (I)
However, Smi
changes with
different objects
Rotational Equilibrium and Dynamics
Moment of Inertia (I)
Observe how the moment of
inertia depends on the location
of the rotational axis.
Hold a book in the upright
position and put your hands at
the bottom of the book. Feel the
torque needed to rock the book
towards and away from you.
Repeat with your hands at the
middle of the book. Less torque is
needed as the average distance of
the mass from the axis is less.
Rotational Equilibrium and Dynamics
Zero Torque and Static Equilibrium
No Change in Linear Motion
No Acceleration
No Change in Rotational Motion
In Both X and Y Components
Rotational Equilibrium and Dynamics
Zero Torque and Static Equilibrium
F1 + F2 = mg
Rotational Equilibrium and Dynamics
Static Equilibrium
SFx  0
SFy  0
Rotational Equilibrium and Dynamics
Static Equilibrium

SFx  0
and
SFy  0
St  0
Rotational Equilibrium and Dynamics
Conditions for Equilibrium
An object is said to be in static equilibrium if both
its velocity and angular velocity are zero or
constant.
First, it must be in translational equilibrium
The net force exerted on the object must be zero.
Second, it must be in rotational equilibrium
The net torque exerted on the object must be zero.
Rotational Equilibrium and Dynamics
Conditions for
Equilibrium
Rotational Equilibrium and Dynamics
Problem
• A 12.5kg board, 4.00m long, is being held up
on one end by Jessica. She calls for help, and
Bobbi responds. What is the least force that
Bobbi could exert to lift the board to the
horizontal position? What part of the board
should she lift to exert this force?
Rotational Equilibrium and Dynamics
Solution
m = 12.5kg
r = 4.00m
F  mg
If Bobbi lifts the opposite end she will
only lift half of the mass of the board.
1
2
F  (12.5kg)(9.8m / s )  61.2 N
2
Rotational Equilibrium and Dynamics
Problem
• A 12.5kg board, 4.00m long, is being held up
on one end by Jessica. She calls for help, and
Bobbi responds. What is the greatest force
that Bobbi could exert to lift the board to the
horizontal position? What part of the board
should she lift to exert this force?
Rotational Equilibrium and Dynamics
Solution
m = 12.5kg
r = 4.00m
F  mg
If Bobbi lifts the center she will
lift the mass of the board.
F  (12.5kg)(9.8m / s )  122 N
2
Rotational Equilibrium and Dynamics
Problem
• A car’s specifications state that its weight
distribution is 53 percent on the front tires
and 47 percent on the rear tires. The wheel
base is 2.46m. Where is the car’s center of
mass?
Rotational Equilibrium and Dynamics
t front  t rear
F frontrfront  Frear rrear
(0.53Fg )rfront  (0.47 Fg )( 2.46  rfront )
(0.53Fg )rfront  (1.16 Fg )  (0.47 Fg )rfront
(0.47 Fg )rfront  (0.53Fg )rfront  1.16 Fg
r front  1.16m
Rotational Equilibrium and Dynamics
Homework
Pages 305-307
Problems
9 (26Nm)
11 (a, 6300Nm b, 550N)
21 (a, 392N b,Rx=339N, Ry=0N)
23 (11N, 1.6N & 7.1 N )
Rotational Equilibrium and Dynamics
Torque and Angular Acceleration
a

r
F  ma
F
a
m
F

mr
rF
r F
  

2
 r  mr mr
rF t
 2
mr
I
Rotational Equilibrium and Dynamics
Torque and Angular Acceleration
Remember that “I” Changes with the
Object
F

mr
rF
r F
  

2
 r  mr mr
rF t
 2
mr
I
Rotational Equilibrium and Dynamics
Newton’s Second Law of Rotational
Motion
The Torque Produced by this Force is…

t
I
t  I
Rotational Equilibrium and Dynamics
Problem
• A bicycle wheel with a radius of 38cm is
given an angular acceleration of
2.67rad/s2 by applying a force of 0.35N on
the edge of the wheel. What is the
wheel’s moment of inertia?
Rotational Equilibrium and Dynamics
t

I
Solution
  = 2.67rad/s2
r = 38cm
F = 0.35N
t Fr sin 
I 


o
(0.35 N )(0.38m)(sin 90.0 )
2
I
 0.050kg  m
2
2.67rad / s
Rotational Equilibrium and Dynamics
Angular Momentum (L)
Mass ~ Moment of Inertia
Velocity ~ Angular Velocity
L  I
Rotational Equilibrium and Dynamics
Angular Momentum (L)
v
L  I  (mr )   rmv  rp
r
or L  rp sin  if the Momentum is Not
2
Perpendicular to the Tangent
Rotational Equilibrium and Dynamics
Angular Momentum (L)
L  I
t  I
I
t
t

L
t
t

t
Rotational Equilibrium and Dynamics
Conservation of Angular Momentum
L
t
t
L
t
t
Angular Impulse-Angular Momentum
Theorem
L  L f  Li  tt
Rotational Equilibrium and Dynamics
Conservation of Angular Momentum
L  L f  Li  tt
L f  Li  tt
L f  Li (if t  0)
Rotational Equilibrium and Dynamics
Rotational Kinetic Energy
With Rotation
mass ~ Moment of Inertia
velocity ~ Angular Velocity
1 2
KE  mv
2
KErot
1 2
 I
2
Rotational Equilibrium and Dynamics
Problem
When a ceiling fan rotating with an angular
speed of 2.55rad/s is turned off, a frictional
torque of 0.220N*m slows it to a stop in 5.75s.
What is the moment of inertia of the fan?
Rotational Equilibrium and Dynamics
Solution
  = 2.55rad/s
 t = 0.220N*m
t = 5.75s
   t and t  I  , so
t 
    t
I
tt
I

(0.220 N  m)(5.75 s)

2.55 rad
s
 0.496 kg  m 2
Rotational Equilibrium and Dynamics
Problem
A person holds a ladder horizontally at its
center. Treating the ladder as a uniform rod of
length 3.25m and mass 8.40kg, find the torque
the person must exert on the ladder to give it
an angular acceleration of 0.322rad/s2.
Rotational Equilibrium and Dynamics
Solution
  = 0.322rad/s2
r = 3.25m
m = 8.40kg
1
2
I  mL .
12
1
1
rad 
2
2
t  I  mL   (8.40 kg)(3.25 m)  0.322 2   2.38 N  m
12
12
s 

Rotational Equilibrium and Dynamics
Problem
Calculate the angular momentum of the Earth
about its own axis, due to its daily rotation.
Assume that the Earth is a uniform sphere.
L
Rotational Equilibrium and Dynamics
Solution
I = 2/5 mr2
m = 5.97x1024kg
r = 6.38x106m
t = 8.64x104s
L  I
2 2
L  mr 
5

2rad
5



7
.
27
x
10
rad / s
4
t 8.64 x10 s
2
(5.97 x10 24 kg)(6.38 x106 m) 2 (7.27 x10 5 rad / s)  7 x1033 kg * m 2 rad / s
5
Rotational Equilibrium and Dynamics
Problem
A torque of 0.12N*m is applied to an egg beater.
If the egg beater starts at rest, what is its
angular momentum after 0.50s?
Rotational Equilibrium and Dynamics
Solution
 t = 0.12N*m
t = 0.50s
L  tt
L  (0.12 N * m)(0.50s)  0.06kg * m / s
2
Rotational Equilibrium and Dynamics
Problem
A student sits at rest on a piano stool that can rotate
without friction. The moment of inertia of the studentstool system is 4.1kg*m2. A second student tosses a
1.5kg mass with a speed of 2.7m/s to the student on the
stool, who catches it at a distance of 0.40m from the axis
of rotation. What is the resulting angular speed of the
student and the stool?
Rotational Equilibrium and Dynamics
Solution
I = 4.1kg*m2
mp = 1.5kg
vp = 2.7m/s
r = 0.40m
mvr  I
mvr  ( I system  mr )
2
mvr

I system  mr 2
(1.5kg)( 2.7m / s)(0.40m)

 0.37rad / s
2
2
4.1kg * m  (1.5kg)(0.40m)
Rotational Equilibrium and Dynamics
Problem
A person exerts a tangential force of 36.1N on
the rim of a disk-shaped merry-go-round of
radius 2.74m and mass 167kg. If the merry-goround starts at rest, what is its angular speed
after the person has rotated it through an angle
of 60.0°?
Rotational Equilibrium and Dynamics
Solution
FT = 36.1N
  = 60o
m = 167kg
r = 2.74m
1 2

o
Fr  t , I  mr , and 60.0  rad
2
3
1
2
W  t  If
2
2t
2 Fr
F
f 

2
1 2
I
mr
mr
2
(36.1N )( / 3rad )
f  2
 0.575rad / s
(167kg)( 2.74m)
Rotational Equilibrium and Dynamics
Homework
Pages 307-310
Problems
27 (1.12x10-2Nm)
35 (1.2rad/s)
37 (7.0m/s)
69 (a, 3.1m/s2 b, 27N, 9.3N)
Rotational Equilibrium and Dynamics
• Machines
– Perform Work
– Makes Tasks Easier
– Changes Either the Magnitude or the Direction
of an Applied Force as it Transmits Energy to
the Task
Rotational Equilibrium and Dynamics
• Machines
– Work
• Input Work (Wi)
– The Work that You Perform on the Machine
• Output Work (Wo)
– The Work that the Machine Performs
Rotational Equilibrium and Dynamics
• Machines
– Work
• According to the Law of Conservation of Energy, the
Output Work Can Never be Greater than the Input
Work
Wo  Wi
Rotational Equilibrium and Dynamics
• Machines
– Mechanical Advantage
• Input Force (Fin)
– The Force You Exert on the Machine
• Output Force (Fout)
– The Force Exerted by the Machine
• Distance from Input (din)
– The Distance Achieved from Effort Force
• Distance from Output (dout)
– The Distance Achieved from Resistance Force
Rotational Equilibrium and Dynamics
• Machines
– Mechanical Advantage (MA)
• The Ratio of Forces
• The Ratio of the Resistance Force to the Effort Force
is the Mechanical Advantage
Fout
MA 
Fin
Rotational Equilibrium and Dynamics
• Machines
– Mechanical Advantage (MA)
• MA > 1.0
– The Machine Increases the Applied Force
• MA < 1.0
– More Force is Applied than the Machine Exerts
• MA = 1.0
– Machine Exerts the Same Magnitude of Force as what was
Applied (but may change direction)
Rotational Equilibrium and Dynamics
• Machines
– Ideal Mechanical Advantage (IMA)
• Ratio of Distances Moved
• The Ratio of the Distance from Effort to the Distance
from Resistance is the Ideal Mechanical Advantage
d in
IMA 
d out
Rotational Equilibrium and Dynamics
• Machines
– Efficiency
• Some Energy May Be Lost by the Machine
– Friction and Heat
Wout
% Efficiency 
x100
Win
Rotational Equilibrium and Dynamics
• Machines
– Ideal Machine
• Efficiency of 100%
Wo
Efficiency 
x100
Wi
Rotational Equilibrium and Dynamics
• Machines
– Simple Machines
•
•
•
•
•
•
Lever
Pulley
Wheel & Axle
Inclined Plane
Wedge
Screw
Rotational Equilibrium and Dynamics
• Machines
– Compound Machines
• Two or More Simple Machines Linked so the
Resistance Force of the First Machine Becomes the
Effort Force of the Second Machine
Rotational Equilibrium and Dynamics
• Problem
– Tyler raises a 1200N piano a distance of
5.00 m using a set of pulleys. Tyler pulls in
20.0 m of rope. How much effort force
would Tyler apply if this were an ideal
machine?
Rotational Equilibrium and Dynamics
• Solution
– Fr = 1200N
– dr = 5m
– de = 20m
Findin  Fout d out
Fout d out
Fin 
d in
(1200 N )(5.0m)
Fin 
 300 N
20.0m
Rotational Equilibrium and Dynamics
• Problem
– Tyler raises a 1200N piano a distance of
5.00 m using a set of pulleys. Tyler pulls in
20.0 m of rope. What force is used to
balance the friction force if the actual
effort is 340 N?
Rotational Equilibrium and Dynamics
• Solution
– Fr = 1200N
– dr = 5m
– de = 20m
– Fe = 340N
– Fe ideal = 300N
Fin  F f  Fin ideal
F f  Fin  Fin ideal
F f  340 N  300 N  40 N
Rotational Equilibrium and Dynamics
• Problem
– Tyler raises a 1200N piano a distance of
5.00 m using a set of pulleys. Tyler pulls in
20.0 m of rope. What is the work output?
Rotational Equilibrium and Dynamics
• Solution
– Fr = 1200N
– dr = 5m
– de = 20m
– Fe = 340N
– Fe ideal = 300N
Wo  Fout d out
Wo  (1200 N )(5.0m)  6 x10 J
3
Rotational Equilibrium and Dynamics
• Problem
– Tyler raises a 1200N piano a distance of
5.00 m using a set of pulleys. Tyler pulls in
20.0 m of rope. What is the input work?
Rotational Equilibrium and Dynamics
• Solution
– Fr = 1200N
– dr = 5m
– de = 20m
– Fe = 340N
– Fe ideal = 300N
Wi  Fin d in
Wi  (340 N )( 20.0m)  6.8 x10 J
3
Rotational Equilibrium and Dynamics
• Problem
– Tyler raises a 1200N piano a distance of
5.00 m using a set of pulleys. Tyler pulls in
20.0 m of rope. What is the mechanical
advantage?
Rotational Equilibrium and Dynamics
• Solution
– Fr = 1200N
– dr = 5m
– de = 20m
– Fe = 340N
– Fe ideal = 300N
Fout
MA 
Fin
1200 N
MA 
 3.53
340 N
Rotational Equilibrium and Dynamics
• Problem
– Tyler raises a 1200N piano a distance of
5.00 m using a set of pulleys. Tyler pulls in
20.0 m of rope. What is the ideal
mechanical advantage?
Rotational Equilibrium and Dynamics
• Solution
– Fr = 1200N
– dr = 5m
– de = 20m
– Fe = 340N
– Fe ideal = 300N
d in
IMA 
d out
20m
IMA 
4
5m
Rotational Equilibrium and Dynamics
• Problem
– Tyler raises a 1200N piano a distance of
5.00 m using a set of pulleys. Tyler pulls in
20.0 m of rope. What is the efficiency of
the set of pulleys?
Rotational Equilibrium and Dynamics
• Solution
– Fr = 1200N
– dr = 5m
– de = 20m
– Fe = 340N
– Fe ideal = 300N
MA
Efficiency 
x100
IMA
3.53
Efficiency 
x100  88%
4
Rotational Equilibrium and Dynamics
• Homework