Download Deformation Rotational Motion Rotation of Rigid Objects

Deformation
Rotational Motion
Rotation of Rigid Objects
Rotational Kinematics
Torque
Lana Sheridan
De Anza College
Feb 26, 2015
Overview
• deforming systems
• rotation
• rotational kinematics
• relating rotational and translational quantities
• torque
Example 9.14 - Exploding Rocket
A rocket is fired vertically upward. At the instant it reaches an
altitude of 1000 m and a speed of vi = 300 m/s, it explodes into
three fragments having equal mass. One fragment moves upward
with a speed of v1 = 450 m/s following the explosion. The second
fragment has a speed of v2 = 240 m/s and is moving east right
after the explosion. What is the velocity of the third fragment
immediately after the explosion?
Example 9.14 - Exploding Rocket
pi = pf ⇒ M vi =
M
(v1 + v2 + v3 )
3
Let j point in the the upward vertical direction, and i point east.
v3 = 3vi − v1 − v2
= 3 × 300j − 450j − 240i
= (−240i + 450j) m/s
Or, 510 m/s at an angle of 62◦ above the horizontal to the west.
Deformable Systems
Some systems will change the distribution of their mass during
their motion.
Deformable Systems
Some systems will change the distribution of their mass during
their motion.
1
Image from http://northdallasgazette.com
Deformable Systems
In a system that is deformed, the positions of the masses of
particles may change relative to the center of mass, but we can
still study the system by considering what happens to the center of
mass.
leave the floor? (d) Does it make sense to say that this
momentum came from the floor? Explain. (e) With
what kinetic energy does the person leave the floor?
(f) Does it make sense to say that this energy came
from the floor? Explain.
Deformable Systems Problem
Page 288, #59
59. Figure P9.59a shows an overhead view of the initial
S configuration of two pucks of mass m on frictionless
ice. The pucks are tied together with a string of length
, and negligible mass. At time t 5 0, a constant force of
magnitude F begins to pull to the right on the center
point of the string. At time t, the moving pucks strike
each other and stick together. At this time, the force
has moved through a distance d, and the pucks have
attained a speed v (Fig. P9.59b). (a) What is v in terms
of F, d, ,, and m? (b) How much of the energy transferred into the system by work done by the force has
been transformed to internal energy?
(a) It has an
exhaust speed
dizer is requi
design could
amount of fue
same task? (c
(b) is 2.50 tim
why the requ
factor of 2.50.
64. A rocket has
330 kg of fu
it starts from
engine at tim
ative speed v
2.50 kg/s. Th
Mf /k 5 330 k
ing the burn
time is given b
m
d CM
,
F
v
F
d
m
t!0
a
(b) Make a gr
tion of time fo
that the accel
t!t
b
(d) Graph th
the person? (c) With what momentum does the person
leave the floor? (d) Does it make sense to say that this
Deformable
Systems
momentum
came Problem
from the floor? Explain. (e) With
what kinetic energy does the person leave the floor?
Does
it make sense to say that this energy came
Page (f)
288,
#59
from the floor? Explain.
59. Figure P9.59a shows an overhead view of the initial
S configuration of two pucks of mass m on frictionless
ice. The pucks are tied together with a string of length
, and negligible mass. At time t 5 0, a constant force of
magnitude F begins to pull to the right on the center
point of the string. At time t, the moving pucks strike
each other and stick together. At this time, the force
has moved through a distance d, and the pucks have
attained a speed v (Fig. P9.59b). (a) What is v in terms
of F, d, ,, and m? (b) How much of the energy transferred into the system by work done by the force has
been transformed to internal energy?
m
d CM
v
en
(a
ex
di
de
am
sa
(b
wh
fa
64. A
33
it
en
at
2.
M
in
tim
each other and stick together. At this time, the force
Deformable
Systems
Problem
has moved
through
a distance d, and the pucks have
attained a speed v (Fig. P9.59b). (a) What is v in terms
of F, d, ,, and m? (b) How much of the energy transferred into the system by work done by the force has
Pagebeen
288,transformed
#59
to internal energy?
3
i
e
a
2
M
i
t
m
d CM
,
F
v
F
d
m
t!0
(
t
t
t!t
a
b
Figure P9.59
Section 9.9 Rocket Propulsion
60. A model rocket engine has an average thrust of 5.26 N.
(
(
Deformable Systems
(a) Speed of pucks, v , after collision?
Deformable Systems
(a) Speed of pucks, v , after collision?
F = M aCM
Deformable Systems
(a) Speed of pucks, v , after collision?
F = M aCM
F is constant ⇒ aCM is constant.
Deformable Systems
(a) Speed of pucks, v , after collision?
F = M aCM
F is constant ⇒ aCM is constant.
aCM =
F
2m
vf2 = vi2 + 2aCM x
F
`
2
vf = 0 + 2
d−
2m
2
r
2Fd − F `
vf =
2m
Deformable Systems
(b) Increase in internal energy?
Deformable Systems
(b) Increase in internal energy?
∆Eint = W − Kf
Deformable Systems
(b) Increase in internal energy?
∆Eint = W − Kf
1
= Fd − (2m)v 2
2
1
2Fd − F `
= Fd − (2m)
2
2m
`
= Fd − Fd + F
2
1
=
F`
2
Rotation of Rigid Objects
Now we understand that while we can treat a collection of particles
as a single point particle at the center of mass, we do not have to
do that.
This will allow us to describe another important kind of motion:
rotation.
Rotation of Rigid Objects
Now we understand that while we can treat a collection of particles
as a single point particle at the center of mass, we do not have to
do that.
This will allow us to describe another important kind of motion:
rotation.
Begin with rotational kinematics.
here
fordisc
rotational
motion.
As the
rotates, a particle
at
10.1 illustrates
P Figure
moves through
an arc lengthan ove
s onrotates
a circularabout
path ofaradius
disc
fixedr.axis p
The
angular
position
of
P isthe
u. disc
through the center of
ticle at P is at a fixed distance r
radius r. (InPfact, every element o
s
convenientr to represent
the pos
u
r
the distance from the origin to P
O
P
Reference
Reference
ence lineOfixed in space
as show
line
line
changes in time while r remai
cle from the reference line, wh
b
a
length
s as in Figure 10.1b. The
s
s = r θ ; θrelationship
=
Figure
r 10.1 A compact disc
As the disc rotates, a particle at
for the disc, a fixed reference
Rotation
of Rigid
Objects
line is chosen.
A particle
at P
is located at a distance r from
To begin, consider a rotating disc.
the rotation axis through O.
rotating about a fixed axis
Pr moves
through
arcfor
length
is constant
in an
time
a rigid object.
through O perpendicular to the
s on a circular path of radius r.
plane of the figure.
The angular position of P is u.
P
Pitfall Prevention 10.1
Remember the Radian In rota-
here
fordisc
rotational
motion.
As the
rotates, a particle
at
10.1 illustrates
P Figure
moves through
an arc lengthan ove
s onrotates
a circularabout
path ofaradius
disc
fixedr.axis p
The
angular
position
of
P isthe
u. disc
through the center of
ticle at P is at a fixed distance r
radius r. (InPfact, every element o
s
convenientr to represent
the pos
u
r
the distance from the origin to P
O
P
Reference
Reference
ence lineOfixed in space
as show
line
line
changes in time while r remai
cle from the reference line, wh
b
a
length
s as in Figure 10.1b. The
s
s = r θ ; θrelationship
=
Figure
r 10.1 A compact disc
As the disc rotates, a particle at
for the disc, a fixed reference
Rotation
of Rigid
Objects
line is chosen.
A particle
at P
is located at a distance r from
To begin, consider a rotating disc.
the rotation axis through O.
rotating about a fixed axis
Pr moves
through
arcfor
length
is constant
in an
time
a rigid object.
through O perpendicular to the
s on a circular path of radius r.
plane
of the
figure.
The
angular
position
of
P
is
u.
Units for θ: radians. Often written as
“rad”.
But
notice, that a
dimensional analysis gives [m]
[m] = 1, unitless! The radian is an
artificial unit. In fact, angles given inPitfall
radians
are dimensionless.
Prevention
10.1
P
Remember the Radian In rota-
Rotation of
Don’t fall into the trap of using
angles
measured in degrees in
Rigid
Objects
rotational equations.
position of a rigi
the object, such
angular position
the object and th
Such identificati
lational motion
which is the orig
motion that the
As the particl
tion " in a time
sweeps out an an
placement of the
How does the angle advance in time?
y
", t f
r
!,ti
uf
ui
O
Figure 10.2
x
A particle on a rotating rigid object moves from ! to
" along the
circle.
∆θarc
=ofθfa −
θi In the
time interval Dt 5 tf 2 ti , the radial
The rate at whic
spins rapidly, th
slowly, this displ
rates can be qua
Angular Speed
Rate at which the angle advances is a speed: the angular speed, ω.
Average angular speed:
ωavg =
θf − θi
∆θ
=
tf − ti
∆t
Instantaneous angular speed:
ω=
dθ
dt
Angular Acceleration
Rate at which the angular speed changes: the angular acceleration,
α.
Average angular acceleration:
αavg =
ωf − ωi
∆ω
=
tf − ti
∆t
Instantaneous angular acceleration:
α=
dω
dt
Rotation of Rigid Objects and Vector Quantities
We can also define these quantities as vectors!
This might seem a bit strange because the direction of the motion
any point (other than the axis) on a rotating disc is always
changing.
Rotation of Rigid Objects and Vector Quantities
We can also define these quantities as vectors!
This might seem a bit strange because the direction of the motion
any point (other than the axis) on a rotating disc is always
changing.
However, the angle can be positive or negative depending on
whether it is clockwise or counterclockwise from the reference
point.
This is the same way that a 1-dimensional displacement x can be
positive or negative based on whether it is on the left or right of
the origin.
Rotation of Rigid Objects and Vector Quantities
We can also define these quantities as vectors!
This might seem a bit strange because the direction of the motion
any point (other than the axis) on a rotating disc is always
changing.
However, the angle can be positive or negative depending on
whether it is clockwise or counterclockwise from the reference
point.
This is the same way that a 1-dimensional displacement x can be
positive or negative based on whether it is on the left or right of
the origin.
In that case we could write x, and the direction of the vector was
indicated by the sign of the quantity x.
Rotation of Rigid Objects and Vector Quantities
By convention, we define the counterclockwise direction to be
Chapter
Rotation
of athe
Rigid
Object
About a Fixed Axis
The vector
itself10is drawn
along
axis
of rotation.
296
positive.
S
direction of v
for the particle is out o
is counterclockwise and into the pla
wise. To illustrate this convention, it
strated in Figure 10.3. When the fou
direction
of rotation, the extended r
S
v
direction
of S
a follows from its definit
S
if the angular speed is increasing in
v
Figure
10.3 The in
right-hand
speed
is decreasing
time. rule
S
v
for determining the direction of the
angular velocity vector.
Then we can write:
ω
dω
10.2 α Analysis
Model: R
=
dt
Constant Angular
where n̂ is a unit vector perpendicular to the plane of rotation.
θ=
S
v
s
n̂ ;
r
ω=
θ
dθ
;
dt
In our study of translational motion,
Comparison of Linear and Rotational quantities
Linear Quantities
Rotational Quantities
θ
x
v=
dx
dt
ω=
dθ
dt
a=
dv
dt
α=
dω
dt
Rotational Kinematics
If α is constant, we have basically the same kinematics equations
as before, but the relations are between the new quantities.
ωf = ωi + αt
1
θf = θi + ωi t + αt 2
2
ω2f = ω2i + 2α · ∆θ
1
ωavg = (ωi + ωf )
2
1
θf = θi + (ωi + ωf )t
2
Kinematics Comparison
Linear Quantities
Rotational Quantities
vf = vi + at
ωf = ωi + αt
xf = xi + vi t + 21 at 2
θf = θi + ωi t + 12 αt 2
vf2 = vi2 + 2a · ∆x
ω2f = ω2i + 2α · ∆θ
vavg = 12 (vi + vf )
ωavg = 21 (ωi + ωf )
xf = xi + 21 (vi + vf )t
θf = θi + 21 (ωi + ωf )t
and for the velocity,
vf 5 vi 1 at 5 2.00
Relating Rotational Quantities to Translation
of m/s 1 (3
Points
There is no translational analog to part (B) because transl
Consider a point on the rotating object. How does its speed relate
to the angular speed?
y
10.3 Angular a
S
In this section, we derive
acceleration of a rotating
P
of a point in the object.
s
r
rotates about a fixed axi
u
circle whose center is on
x
O
Because point P in Figu
is always tangent to the ci
nitude of the tangential v
v 5 ds/dt, where s is the d
Figure 10.4 As a rigid object
that s 5 r
We know s = r θ, so rotates
since about
the object’s
speed
speedRecalling
along the
the fixed axis
(the is its path.
v
path s,
z axis) through O, the point P
S
v that is
has a tangential
dsvelocitydθ
v = to the
= rcircular path
always tangent
dt
of radius r. dt
Relating Rotational Quantities to Translation of
Points
Since ω =
dθ
dt ,
that gives us and expression for (tangential) speed
v = rω
And differentiating both sides with respect to t again:
at = r α
Relating Rotational Quantities to Translation of
Points
Since ω =
dθ
dt ,
that gives us and expression for (tangential) speed
v = rω
And differentiating both sides with respect to t again:
at = r α
Notice that the above equation gives the rate of change of speed,
which is the tangential acceleration.
Centripetal Acceleration
Remember:
at =
dv
dt
where v is the speed, not velocity.
So,
at = r α
But of course, in order for a mass at that point, radius r , to
continue moving in a circle, there must be a centripetal component
of acceleration also.
v2
ac =
= ω2 r
r
Centripetal Acceleration
Remember:
at =
dv
dt
where v is the speed, not velocity.
So,
at = r α
But of course, in order for a mass at that point, radius r , to
continue moving in a circle, there must be a centripetal component
of acceleration also.
v2
ac =
= ω2 r
r
For a rigid object, the force that supplies this acceleration will be
some internal forces between the mass at the rotating point and
the other masses in the object. Those are the forces that hold the
object together.
Linear and Rotational Kinematics
To get an idea of another way to confirm the relation between the
linear and rotational kinematics equations:
vf
vf
r
ωf
= vi + at
vi
a
=
+ t
r
r
= ωi + αt
Linear and Rotational Kinematics
To get an idea of another way to confirm the relation between the
linear and rotational kinematics equations:
vf
vf
r
ωf
= vi + at
vi
a
=
+ t
r
r
= ωi + αt
The scalar versions of the equations can all be recovered this way.
bout
3.00 s.
pe Example
of
1
4. A bar on a hinge starts from rest and rotates with an
angular acceleration a 5 10 1 6t, where a is in rad/s2
nguand t is in seconds. Determine the angle in radians
rage
through which the bar turns in the first 4.00 s.
ond.
ring Page
Section
10.2#5
Analysis Model: Rigid Object
325,
eed?
Under Constant Angular Acceleration
5. A wheel starts from rest and rotates with constant
ition
0t 1
W angular acceleration to reach an angular speed of
12.0 rad/s in 3.00 s. Find (a) the magnitude of the angueterlar acceleration of the wheel and (b) the angle in radians through which it rotates in this time interval.
6. A centrifuge in a medical laboratory rotates at an angular speed of 3 600 rev/min. When switched off, it rotates
through 50.0 revolutions before coming to rest. Find
the constant angular acceleration of the centrifuge.
7. An electric motor rotating a workshop grinding wheel
2
M at 1.00 3 10 rev/min is switched off. Assume the wheel
has a constant negative angular acceleration of magni-
w
ef
a
fr
e
o
Secti
15. A
Example 1
Known: ωi , ωf , t
(a) Angular accel., α?
Example 1
Known: ωi , ωf , t
(a) Angular accel., α?
ωf
= ωi + αt
Example 1
Known: ωi , ωf , t
(a) Angular accel., α?
ωf
= ωi + αt
α =
ωf − ωi
t
α =
12.0 rad/s − 0
3.00 s
α = 4.00 rad s−2
Example 1
Known: ωi , ωf , t, α
(b) Angle in radians, ∆θ?
Example 1
Known: ωi , ωf , t, α
(b) Angle in radians, ∆θ?
Either use
1
1
∆θ = ωi t + αt 2 = 0 + (4)(3)2 = 18 radians
2
2
or use
∆θ =
1
(ωi + ωf )t
2
Example 1
Known: ωi , ωf , t, α
(b) Angle in radians, ∆θ?
Either use
1
1
∆θ = ωi t + αt 2 = 0 + (4)(3)2 = 18 radians
2
2
or use
∆θ =
1
(ωi + ωf )t
2
∆θ =
1
(0 + 12.0)(3.00)
2
= 18.0 radians
the constant angular acceleration of the centrifuge.
7. An 2electric motor rotating a workshop grinding wheel
Example
2
M at 1.00 3 10 rev/min is switched off. Assume the wheel
has a constant negative angular acceleration of magnitude 2.00 rad/s2. (a) How long does it take the grinding
stop? (b) Through how many radians has the
Pagewheel
325,to#8
wheel turned during the time interval found in part (a)?
8. A machine part rotates at an angular speed of
Q/C 0.060 rad/s; its speed is then increased to 2.2 rad/s
at an angular acceleration of 0.70 rad/s2. (a) Find the
angle through which the part rotates before reaching
this final speed. (b) If both the initial and final angular speeds are doubled and the angular acceleration
remains the same, by what factor is the angular displacement changed? Why?
9. A dentist’s drill starts from rest. After 3.20 s of constant angular acceleration, it turns at a rate of 2.51 3
104 rev/min. (a) Find the drill’s angular acceleration.
(b) Determine the angle (in radians) through which
the drill rotates during this period.
Secti
15. A
A
4
tu
16. M
o
tu
e
17. A
W d
th
o
la
o
to
fo
18. F
h
Example 2
Known: ωi , ωf , α
(a) ∆θ?
Example 2
Known: ωi , ωf , α
(a) ∆θ?
Example 2
Known: ωi , ωf , α
(a) ∆θ?
ω2f = ω2i + 2α∆θ
Example 2
Known: ωi , ωf , α
(a) ∆θ?
ω2f = ω2i + 2α∆θ
∆θ =
=
ω2f − ω2i
2α
(2.2)2 − (0.060)2
2 × 0.70
= 3.5 rad
Example 2
(b) If both ωi and ωf are doubled, α kept constant, what happens
to ∆θ?
∆θ 0 =
(2 × ωf )2 − (2 × ωi )2
2α
= 4
ω2f − ω2i
2α
= 4 ∆θ
Torque
Torque is a measure of force-causing-rotation.
It is not a force, but it is related. It depends on a force vector and
its point of application relative to an axis of rotation.
Torque is given by:
τ=r×F
That is: the cross product between
• a vector r, the displacement of the point of application of the
force from the axis of rotation, and
• an the force vector F
Torque
Torque is a measure of force-causing-rotation.
It is not a force, but it is related. It depends on a force vector and
its point of application relative to an axis of rotation.
Torque is given by:
τ=r×F
That is: the cross product between
• a vector r, the displacement of the point of application of the
force from the axis of rotation, and
• an the force vector F
Units: N m
Newton-meters. These are not Joules!
10.4 Torqu
The component F sin f
tends to rotate the wrench
about an axis through O.
Torque
F sin f
S
F
r
f
O
S
d
Figure 10.7
r
F cos f
f
Line of
action
S
The force F has a
greater rotating tendency about an
axis through
O as
F increases
τ=r×
F=
rF sin φ and
n̂
as the moment arm d increases.
In our study of tra
we studied the cau
What is the cause
Imagine trying
ular to the door s
hinges. You will ac
force near the doo
When a force is
to rotate about th
axis is measured b
but we will consid
Chapter 11.
Consider the wr
perpendicular to t
where φ is the angle between r and F, and n̂ is the unit vector
perpendicular to r and F, as determined by the right-hand rule.
Torque
Diagram also
The component F sin f
tends to rotate the wrench
illustrates
two
points
of O.
view
about
an axis
through
10.4 Torque
about torque:
F sin f
S
F
r
f
O
S
d
r
F cos f
f
Line of
action
S
Figure 10.7 The force F has a
or
greater rotating
tendency
τ = r (F
sin φ)about
n̂ an
axis through O as F increases and
as the moment arm d increases.
In our study of translation
we studied the cause of c
What is the cause of chang
Imagine trying to rotat
ular to the door surface n
hinges. You will achieve a
force near the doorknob t
When a force is exerted
to rotate about that axis.
axis is measured by a quan
but we will consider only
Chapter 11.
Consider the wrench in
perpendicular to the page
τ = F (r sin φ) n̂
In the diagram, the distance d = r sin φ and is called the “moment
arm” or “lever arm” of the torque.
Torque
Torque:
when a force is applied
above the center of mass.
CM
The system rotates clockwise
when a force is applied
a above the center of mass.
CM
The system rotates counteraCM
clockwise
when a force is applied
below the center of mass.
Torque:
a system rotates counterThe
clockwise when a force is applied
below
CM the center of mass.
The system rotates counterclockwise when a force is applied
below
the center of mass.
b
CM
No torque:
The
system moves in the
bCM
direction of the force without
rotating when a force is applied
at the center of mass.
b
The system moves in the
direction of the force without
rotating when a force is applied
The
system
moves
in the
at the
center
of mass.
CM
direction
of the force without
rotating when a force is applied
at the center of mass.
c
CM
the rotating
x axis and
liesFig.
somewhe
out
(see
9.13c)
this
procedure.
is applied at the center of ma
center(see
of mass
of the
outThe
rotating
Fig. 9.13c)
the xprocedure.
axis and lies somewhe
this
For example, if x 1 5 0, x 2
The center
mass
of the
center
of massoflies
closer
to
the
axis and
lies somewher
thexcenter
of mass
lies midw
can extend
ForWe
example,
if x 1 this
5 0,conc
x2 5
dimensions.
The
coordin
center
of mass
liesxcloser
to
the example,
center of mass
lies0,midw
For
if x 1 5
x2 5
m 2xconce
m 1xlies
We can
extend
1 1this
2 1
center
of ;
mass
closer
to
x CM
dimensions.
The
x
coordina
m 1 lies
1 mmidw
the center of mass
2 1
We can extend this conce
m 2x 2 1 m
dimensions.mThe
x coordina
1x 1 1
x CMx ;
where
x coordinate
i is the m
1 1 m2 1 m
the sum runs
all n par
m 1x over
1 1 m 2x 2 1 m
x CM ;defined by the equ
similarly
m1 1 m2 1 m
where xi is the x coordinate
the sum runs over all nyCM
part
;
similarly
by the equo
where
xi isdefined
the x coordinate
of mass
the The
sumcenter
runs over
all ncan
partb
yCMequ
;
The components
ofthe
this
vec
similarly
defined by
9.30. Therefore,
The center of massycan
b
CM ;
S
^i 1
r CM 5ofx CM
yC
The components
this
vect
9.30.
TheTherefore,
center of mass can be
Question
B
A
A torque is supplied by applying a force at point A. To produce the
same torque, the force applied at point B must be:
(A) greater
(B) less
(C) the same
1
Image from Harbor Freight Tools, www.harborfreight.com
the rotation axis to the line of action of
nary line extending out both ends
of the
S
line extending from the tail of F in Fig.
om the right triangle in Figure 10.7 that
that d 5 r sin f. The quantity d is called
rotation axis.
Net Torque
S
Moment
Object that can rotate about
anarmaxis at O:
f F that tends to cause rotation of the
n f, the component perpendicular to a
oint of application of the force. The horie of action passes through O, has no tenpassing through O. From the definition
s as F increases and as d increases, which
we push at the doorknob rather than at a
apply our push as closely perpendicular
o 908. Pushing sideways on the doorknob
S
F1
d1
O
d2
ject as in Figure 10.8,
each tends to proS
In this example, F2 tends to rotate the
counterclockwise. We use the convention
S
F2
a force is positive if the turning tendency
S
tive if the turning
tendency
is
clockwise.
Figure 10.8 The force F 1 tends
S
sulting from
F1, which has a moment arm
to rotate the object counterclockS
are
but
the
twoO, and
torques
ue from F2There
is negative
andtwo
equalforces
to 2F 2d 2.acting,
wise about
an axis
through
S
F 2 tends to rotate it clockwise.
gh O is
τ2 point in opposite directions.
5 F 1d1 2 F 2d 2
rce. Forces can cause a change in translaτ1 would
produce a counterclockwise rotation
cond law. Forces
can also cause a change
ess of the τforces
in
causing
this change
a clockwise rotation
2 would produce
orces and the moment arms of the forces,
has units of force times length—newton
e reported in these units. Do not confuse
produced, τ1 and
om the right triangle in Figure 10.7 that
that d 5 r sin f. The quantity d is called
Net Torque
Moment arm
S
f F that tends to cause rotation of the
n f, the component perpendicular to a
oint of application of the force. The horie of action passes through O, has no tenpassing through O. From the definition
s as F increases and as d increases, which
we push at the doorknob rather than at a
apply our push as closely perpendicular
o 908. Pushing sideways on the doorknob
ject as in Figure 10.8,
each tends to proS
In this example, F2 tends to rotate the
counterclockwise. We use the convention
a force is positive if the turning tendency
tive if the turning
tendency is clockwise.
S
sulting from
F1, which has a moment arm
SThe
net
torque
is2Fthe
ue from F2 is negative and
equal to
2d 2.
gh O is
5 F 1d1 2 F 2d 2
S
F1
d1
O
d2
S
F2
Figure 10.8
S
The force F 1 tends
to rotate the object counterclocksum
of the torques acting on the object:
wise about an axis through O, and
S
F 2 tends to rotate it clockwise.
X
rce. Forces can cause a change in translaτnet =
τi
cond law. Forces can also cause a change
ess of the forces in causing this change
i
orces and the moment arms of the forces,
has units of force times length—newton
this units.
case,Dowith
n̂ pointing out of the slide:
e reportedIn
in these
not confuse
its but are very different concepts.
τnet
oosen a stubborn screw from a piece
of
d you find a screwdriver for which the
= τ1 + τ2 = (F1 d1 − F2 d2 )n̂
Summary
• applying kinematics
• torque
• moment of inertia
3rd Collected Homework!
due Monday, Feb 29.
Next test Friday, Mar 4.
(Uncollected) Homework Serway & Jewett,
• Read ahead in Chapter 10.
• Ch 10, onward from page 288. Probs: 3, 7, 11, 15, 17, 19, 21,
25