Transcript Holt Physics Chapter 8

Try to lift the meter stick about an axis at the end
by lifting on the far end. KEEP THE “AXIS” ( ) ON
THE TABLE!
F
F
F
Is it easy to lift?
Now change the position of your force to the
middle.
Now change the position to about 20 cm from the
axis.
How is the force required changing?
Try to lift the meter stick about an axis at the end
by lifting on the far end. KEEP THE “AXIS” ( ) ON
THE TABLE!
F
F
F
Is it easy to lift?
Now change the position of your force to the
middle.
Now change the position to about 20 cm from the
axis.
How is the force required changing?
Holt Physics Chapter 8
Rotational Equilibrium
and Dynamics
Apply two equal and opposite forces acting at the
center of mass of a stationary meter stick.
F1
F2
F1=F2
Does the meter stick move?
Fext = 0.
Apply two equal and opposite forces acting on a
stationary meter stick.
F
F
Does the meter stick move?
The center of mass of the meter stick does not
accelerate, so it does not undergo translational
motion.
However, the meter stick would begin to rotate
about its center of mass.
A torque is produced by a force
acting on an extended (not
point-like) object.
The torque depends on how
strong the force is, and where
it acts on the object.
F
A
You must always specify your reference axis for
calculation of torque. By convention, we indicate
that axis with the letter “A” and a dot.
Torques cause changes in rotational motion.
Torque is a vector. It is not a force,* but is related
to force.
*So never set a force equal to a torque!
Torque
Torque is a quantity that measures
the ability of a force to rotate an
object around some axis.
Torque depends on force and the
lever arm.
Lever arm (moment arm)is the
perpendicular distance from the
axis of rotation to a line drawn
along the direction of the force.
See figure 8-3, page 279
The most torque is produced when
the force is perpendicular to the
object.
Formula
=Fd(sin)
d is the lever arm and  is the
angle between the lever arm and
the force. (If 90º, then sin=1).
See figure 8-5, page 280.
Force is POSITIVE if rotation is
counterclockwise.
If there is more than one force,
add the two resultant torques,
using the appropriate signs.
EX: Wishbone…sum the two
torques
F1
F2
A
net =  = 1+2 =
F1d1 + (-F2d2)
The sign of the net torque
will tell you which direction
the object will rotate.
F1
F2
d1
A
d2
Example 8A, page 281
A basketball is being pushed by two
players during tip-off. One
player(to the right) exerts a
downward force of 11 N at a
distance of 7.0 cm from the axis of
rotation. The second player(to the
left) applies an upward fore of 15 N
at a perpendicular distance of 14cm
from the axis of rotation. Find the
net torque acting on the ball.
Direction and sign??? Units???
F1 = -15 N F2 = -11N
d1 = 0.14m d2 = 0.070m
net=?
net= 1+ 2= F1d1+ F2 d 2
(-15N X 0.14m)+(-11N X 0.070m) =
-2.9Nm
Example with an Angle
An upward 34N force is exerted on
the right side of a meter stick
0.30 m from the axis of rotation at
an angle of 35 degrees. A second
downward force of 67N is exerted
at an angle of 49 degrees to the
meter stick 0.40m to the left of
the axis of rotation.
What is the net torque?
49
0.30m
67N
0.40m
35
34N
Signs???
BOTH are POSITIVE!!
F1 = 34N
F2 = 67N
d1 = 0.30m d2 = 0.40m
1= 35 2= 49
net=?
net= 1+ 2= F1d1 sin1+ F2d2 sin2
(34NX.30m)sin35+(67NX.40m)sin49=
26Nm
Practice 8A, page 282
Equilibrium
Complete equilibrium requires zero
net force and zero net torque.
Translational equilibrium: net force
in x and y direction = 0
Called 1st condition of equilibrium
∑Fx = 0, ∑Fy = 0
Rotational equilibrium: net torque=0
Called 2nd condition of equilibrium
∑ = 0
A 45.0m beam that weighs
60.0N is supported in the
center by a cable. The beam is
in equilibrium and supports
three masses. A 67.0kg mass is
on one end, an 89.0kg mass is
on the other. A fish is hanging
10.0m from the 67.0 kg mass.
What is the mass (kg!) of the
fish and what is the tension
(force!) in the cable.
FT
45m
10m
67kg
?
60N
89kg
Convert to Newtons!
FT
45m
10m
67kg
?
60N
89kg
FT
45m
10m
657N
?
60N
873N
Choose Axis
Choose center to eliminate a
variable!!
FT
45m
10m
657N
?
60N
873N
How far is the fish from the
axis?
FT
45m
10m 12.5m
657N
?
60N
873N
Calculate Torques and sum to
zero
FT
45m
10m 12.5m
657N
?
60N
873N
Assign sign to torques
FT
45m
10m 12.5m
657N
+
?
+
60N
0
873N
-
∑=0=
657N(22.5m)+Wf(12.5m)–873N(22.5m)=0
Wf=4860Nm/(12.5m) = 388.8N
mf=388.8N/9.81m/s2=39.6kg
FT
45m
10m 12.5m
657N
39.6kg 60N
873N
∑Fy = 0, ∑Fdown = ∑Fup , change all
to forces
Fish is 39.6kgX9.81m/s2= 389N
FT
45m
10m12.5m
-657N
-60N
-389N
39.6kg
-873N
∑Fy = 0, ∑Fdown = ∑Fup
FT= -∑Fdown
=-(-657N- 389N -60N-873N)
=1980N
FT
45m
10m12.5m
-657N
-389N -60N
-873N
How to draw a bridge with
pillars…
F, pillar 1
F, pillar 2
Fw, car
Fw,
bridge
Begin Worksheet
Example 8B, page 287-8
A uniform 5.00 m long horizontal beam
that weighs 315N is attached to a
wall by a pin connection that allows
the beam to rotate. Its far end is
supported by a cable that makes an
angle of 53 degrees with the
horizontal, and a 545 N person is
standing 1.5 m from the pin. Find
the force in the cable, FT, and the
force exerted on the beam by the
wall, R, if the beam is in equilibrium.
L=5.00 m Fg,beam=315 N =53
Fg,person=545 N d=1.50m
FT=? R=?
Record distances, put weight of
object at center of mass and
position all forces.
R
1.50 m
5.00 m
53
315 N
545 N
FT
The unknowns are R ( Rx, Ry), and FT
Because we have equilibrium,
∑Fx = 0, ∑Fy = 0
Rx - FTcos = 0
Ry + FTsin - Fg,p – Fg,b = 0
Rx
Ry
FTx=FTcos
R
1.50 m
5.00 m
53
315 N
545 N
FT FTy=FTsin
Rx - FTcos = 0
Ry + FTsin - Fg,p – Fg,b = 0
Because there are too many
unknowns, pause (don’t panic) and go
to the second condition of
equilibrium.Rx
Ry
R
1.50 m
5.00 m
53
315 N
545 N
FT
Choose an axis and sum the
torques…remembering signs for
direction of rotation!! Why choose the pin
for A?
Rx
Ry
0
A
+
R
53
1.50 m
5.00 m
-
-
545 N
315 N
This eliminates R as
a variable!!
FT
=FT L(sin) – Fg,bL/2 –Fg,p d=0
Now substitute and solve for FT.
Rx
Ry
A
R
1.50 m
5.00 m
53
315 N
545 N
FT
FT 5.00m(sin53) –
(315N)(5.00m)/2 –(545N)(1.5m)=0
FT= 1606Nm/4.0m
FT= 4.0X102 N
Rx
Ry
A
R
1.50 m
5.00 m
53
315 N
545 N
FT
We’re not done yet!!
Rx
Ry
A
R
1.50 m
5.00 m
53
315 N
545 N
FT
Now we can substitute force in
the wire (FT) into the Rx and Ry
equations to find Rx and Ry, and
then solve for R…
Rx
Ry
A
R
1.50 m
5.00 m
53
315 N
545 N
FT
Rx - FTcos = 0 Rx = FTcos
Rx = 400N X cos53 = 240N
 Ry+FTsin-Fg,p–Fg,b=0
Ry= -FTsin53+
Fg,p+Fg,b
 Ry = -3.2X102N + 860 N = 540 N
 R = (Rx2 + Ry2) = (240N2 + 540N2) =590 N
Rx
Ry
A
R
1.50 m
5.00 m
53
315 N
545 N
FT
Homework
Finish Worksheet!
Moment of Inertia
The moment of inertia is the
resistance of an object to
changes in rotational motion about
some axis.
Similar to mass…mass ( simple
inertia) is the measure of resistance
to translational motion…
Moment of Inertia depends on the
object’s mass and the distribution of
mass around the axis of rotation.
The farther the center of mass
from the axis of rotation, the more
difficult it is to rotate the object,
and therefore, the higher the
moment of inertia.
Use Table 8-1, page 285
Newton’s
nd
2
Law
F=ma can be translated to
rotational motion.
net = I = Iat/r
I = moment of inertia
  = angular acceleration
 net = net torque
at=tangential acceleration
r=radius
Example page 291
A student tosses dart using only the
rotation of her forearm to accelerate
the dart. The forearm rotates in a
vertical plane about an axis at the
elbow joint. The forearm and dart
have a combined moment of inertia of
0.075 kgm2 about the axis, and the
length of the forearm is 0.26 m. If
the dart has a tangential acceleration
of 45m/s2 just before it is released,
what is the net torque on the arm
and dart?
I = 0.075 kgm2 at=45m/s2
r=0.26m
=?
=I, where =at/r
=I at/r
= 0.075 kgm2 X45m/s2/0.26m=
13Nm
Example 2
A 25 g CD (radius =7.0 cm)
is rotating at 100 rev/min.
If it stops in 8.5 sec, what
is the angular acceleration
of the CD? How much
torque is required to stop
the CD?
r=0.07m, m=0.025kg, t=8.5 sec,
i = 100rev/min=100x260=10.5rad/s
 f= 0
 = /t = (0-10.5rad/s)/8.5s
r=0.07m, m=0.025kg, t=8.5 sec,
i = 100rev/min=100x260=10.5rad/s
 f= 0
 = /t = (0-10.5rad/s)/8.5s
 = -1.2 rad/s2
=I…What is I? Look in table
on page 285
Rotating disk is 1/2mr2
 = -1.2 rad/s2
=I = 1/2mr2=
=1/2(.025kg)(.07)2 -1.2rad/s2
-5
=-7.35X10 Nm
8C, page 291
Be ready to use table 8-1
on page 285 to calculate I
and use old chapter 7
formulas for quantities like
, , s, at and .
Momentum and Rotation
Linear momentum can be
translated to angular momentum
L = I
 L = angular momentum
 I = moment of inertia – look in
the table again (page 285!!)
  = angular speed (you may
need to use ch. 7 formulas
again)
Conservation of Angular
Momentum
As in linear momentum,
angular momentum is also
conserved.
Li = Lf
Example page 293
A 65 kg student is spinning on a
merry-go-round that has a mass of
5.25X102 kg and a radius of 2.00 m.
She walks from the edge of the
merry-go-round toward the center.
If the angular speed of the merrygo-round is initially 0.20 rad/sec,
what is its angular speed when the
student reaches a point 0.50m from
the center?
mm = 525 kg
ms =65 kg
ri,s= rm =2.00m
rf,s = 0.50 m
i =0.20 rad/s f =?
Use conservation of momentum
Li = Lf
Lm.i + Ls,i = Lm,f + Ls,f
Need moments of inertia!!
Because L = I= Ivt/r
The merry-go-round is a _____
The Student is a _____
Merry-go-round (I = ½ MR2 )
Student (I = MR2 )
Lm.i + Ls,i = Lm,f + Ls,f
½MmRm2i+MsRs,i2i =
½MmRm2f+MsRs,f2f
½MmRm2i+MsRs,i2i =
½MmRm2f+MsRs,f2f
½525kg(2.00m)2(0.20rad/s)
+65kg(2.00m)2(0.20rad/s) =
½525kg(2.00)2f +65kg(0.50m)2f
Plug it into the calculator and solve for f
f =0.2435rad/s =0.24 rad/s
Keep ch. 7 formulas handy too! Esp. vt=r
Example 2
A comet has a speed of
7.056 X 104 m/s at a
distance of 4.95X1010 m.
At what distance from the
sun would the comet have a
speed of 5.0278X104 m/s?
Li=Lf
Point
Mass
So, I =
2
MR
Mass is constant
ω= vt/r
Ri = 4.95X1010 m Vi=7.056 X 104 m/s
Rf= ?
Vf= 5.0278X104 m/s
MRi2 vi/ri = MRf2 vf/rf
Ri vi= Rf vf
Rf =Ri vi/ vf
Rf = 4.95X1010 m X 7.056 X 104 m/s
5.0278X104 m/s
Rf= 6.95X1010m
8 D, page 294
Happy Friday!
All labs in?
Today we will cover Conservation
of Mechanical Energy.
We will skip simple machines and
start review problems on Monday
for a Wednesday Test.
Kinetic Energy
Rotational Kinetic energy (KErot) is the
kinetic energy associated with their
angular speed.
Formula KErot = ½ I2 = ½ I(vt/r)2
Conservation of Kinetic Energy also applies…
KEtrans + KErot + PEi = KEtrans + KErot + PEf
 ½ mvi
2
+ ½ Ii
2
+ mghi = ½ mvf
2
+ ½ If
2
+ mghf
Be sure to keep track of initial and final
conditions as well as angular vs.
translational speeds and moments of
inertia.
Example page 296
A solid ball with a mass of 4.10 kg
and a radius of 0.050m starts from
rest at a height of 2.00 m and
rolls down a 30 slope. What is
the translational speed of the ball
when it leaves the incline?

v
2.00m
30 
hi = 2.00 m m = 4.10 kg
R = 0.050 m vi = 0.0 m/s
 = 30.0 hf = 0 m vf = ?

2
IWhat
= 2/5MR
will I be???
v
2.00m
30 
hi = 2.00 m m = 4.10 kg
R = 0.050 m
vi = 0.0 m/s
 = 30.0 hf = 0 m
vf = ?
½ mvi 2 + ½ Ii 2 + mghi = ½ mvf
+ ½ If 2 + mghf
2
WE have two variables
So we need to find one
In terms of the other…

REMEMBER!!!
 =vt/r
v
2.00m
30 
hi = 2.00 m
m = 4.10 kg
R = 0.050 m
vi = 0.0 m/s
 = 30.0 hf = 2.00 m
vf = ?
½ mvi 2 + ½ Ii 2 + mghi = ½ mvt,f
2 +mgh
(vt,f
/r)
+½I 
f
f
2
Substitute vt/r into
The equation for
f

v
2.00m
30 
PLUG IN THE NUMBERS
4.10kg(9.81m/s2) (2.00m) = ½ 4.10kgvf 2
+½I (vf/0.050m) 2
I = 2/5MR2
4.10kg(9.81m/s2) (2.00m) = ½ 4.10kgvf 2
+½(2/5MR2) (vf/0.050m) 2
4.10kg(9.81m/s 2) (2.00m) =½ 4.10kg vf
+½(2/5)(4.10kg)(0.050m) 2 (vf/0.050m) 2
80.442kgm2/s2 = 2.05kg vf 2 +
0.82kg vf 2
2.87kg vf 2 = 80.442kgm2/s2
vf 2 =28 m2/s2
vf = 5.29 m/s
2
Example 2
A 3.5 kg spherical potato
(radius .070m) is kicked up a
30 degree slope at a speed
of 5.4 m/s. What distance
along the slope did the
potato roll before it
stopped?
mp= 3.5kg, vi,p= 5.4m/s θ=30°
hf=? Hypotenuse (slope dist.)= ?
hf=2.08m
D=2.08/sin30=4.2m
30
height
2 2 +mgh = ½ mv 2 + ½ I 2 + mgh
½ mvi 2 + ½ I(vit,i/r)
i
f
f
f
*Substitute vt/r for ω
2 2 = mgh
½ mvi 2 + ½ I(vit,i/r)
f
*Solve for hf
hf=½ mvi 2 + ½ I(vt /r)2 =(51.03 + 20.412)/34.335
mg
Kinetic Energy
 Practice 8E, page 297