Transcript Holt Physics Chapter 7

Holt Physics Chapter 7
Rotational Motion
Measuring Rotational Motion
Spinning objects have rotational
motion
Axis of rotation is the line about
which rotation occurs
A point that rotates around an axis
undergoes circular motion
s=arc length, r=radius
=angle of rotation (measured in
degrees or radians) corresponding to s

r
s
  (rad) = s/r
If s = r, then  = 1 radian
If s = 2r (circumference),
then  (rad) = 2r/r= 2 radians
Therefore 360 degrees = 2 radians,
or 6.28
See figure 7-3, page 245

r
s
Equation to convert between
radians and degrees
 (rad) = [/180º] X  (deg)
Sign convention:
clockwise is negative,
counterclockwise is positive!
Angular Displacement ()
describes how much an object
has rotated
 = s/r

r
s
Book Example 1: While riding on a
carousel that is rotating clockwise, a
child travels through an arc length of
11.5m. If the child’s angular
displacement is 165 degrees, what is
the radius of the carousel?
  = -165 degrees but…we need radians!
 rad= [/180º] Xdeg = (/180º)(-165º)
= -2.88 radians
 s = -11.5 m
r=?
  = s/r so,
r=s /
r = -11.5/-2.88 = 3.99m
Example 2. A child riding on a merry-go-round
sits at a distance of 0.345 m from the center.
Calculate his angular displacement in radians,
as he travels in a clock-wise direction through
an arc length of 1.41 m.
Given: r = 0.345 m
s = -1.41 m
Find: 
s - 1.41 m

 - 4.09 rad
Solution: Δθ 
r 0.345 m
The negative sign in our answer indicates direction.
Homework
Practice 7A, page 247
Angular Speed (ave) is equal to the
change in angular displacement(in
radians!) per unit time.
ave= / t , in radians/sec
Remember, One revolution =
_______ radians,
so the number of revolutions
=/2rad

r
s
Book Example 3: A child at an ice
cream parlor spins on a stool. The
child turns counterclockwise with an
average angular speed of 4.0 rad/s.
In what time interval will the child’s
feet have an angular displacement of
8.0  radians.
 =8 rad
 ave= 4.0 rad/s t=?
 ave= / t, so t=/ ave
 t= 8 rad / 4.0 rad/s = 2.0  s
= 6.3 sec
Example 4. A man ties a ball to the end of
a string and swings it around his head. The
ball’s average angular speed is 4.50 rad/s.
A) In what time interval will the ball move
through an angular displacement of 22.7 rad?
B) How many complete revolutions does this
represent?
Given: avg = 4.50 rad/s
Find: A) t
 = 22.7 rad
B) # of revolutions
22.7 ra d
θ

 5.04 s
Solution: Δt 
ωavg
4.50 ra d/s
θ
22.7 ra d

 3.61 rev.
# of rev. 
2rad
2ra d
Angular Acceleration (ave) is
the change in angular speed
() per unit time(s).
ave= /t = (2-1)/(t2-t1)
Unit is rad/s2
Book Example 1:
A car’s tire rotates at an initial angular speed of
21.5 rad/s. The driver accelerates, and after
3.5 sec the tire’s angular speed is 28.0 rad/s.
What is the tire’s average angular acceleration
during the 3.5 s time interval?
1=21.5 rad/s 2=28.0 rad/s
t=3.5 s
ave= ?
ave= 2-1/t =
ave =(28.0rad/s-21.5rad/s)/3.5s
ave=1.9 rad/s2
Example 2. The angular velocity of a rotating
tire increases from 0.96 rev/s to 1.434 rev/s
with an average angular acceleration of
6.0 rad/s2 . Find the time required for given
angular acceleration.
First change revolutions/sec to radians/sec by
multiplying by 2.
(0.96rev/s)X 2 = 6.0 rad/sec
(1.434rev/s)X 2 = 9.01 rad/sec
Given: 1 = 6.0 rad/s 2=9.01 rad/savg=6.0rad/s2
Find: t = ?
Solution:
 avg
ω 2 - ω1

t
t 
ω2 - ω1
 avg
t = 9.01 rad/s – 6.0 rad/s = 0.50s
6.0 rad/s2
Homework
Practice 7B (1,2, and 4) page
248 and 7C (1 and 3), pg 250
Angular Kinematics
All points on a rotating rigid
object have the same angular
acceleration and angular speed.
Angular and linear quantities
correspond if angular speed ()
is instantaneous and angular
acceleration () is constant.
See table 7-2, pg. 251 or your
formula sheet
Equations for Motion with Constant Acceleration
Linear
Rotational
   i  at
ωf  ωi  αt
f
1
x   i t  at 2
2
f 2   i 2  2ax
1
θ  ωit  αt 2
2
ω 2  ω 2  2θ
Δx  12 (νi  νf)Δt
Δθ  12 (ωi  ωf) Δt
f
i
Book Example 1 (page 251)The wheel on
an upside-down bicycle moves through
11.0 rad in 2.0 sec. What is the wheel’s
angular acceleration if its initial angular
speed is 2.0 rad/s?
=11.0 rad
i= 2.00rad/s
t=2.0 s
 = ?
Use one of the angular kinematic
equations from Table 7-2 to solve
for  … which one?
it  ½ (t)2
Solve for 
 =2(-it)/(t)2
  =2[11.0rad-(2.00rad/s)(2.0s)]/(2.0s)2
 = 3.5 rad/s2
Example 2. A CD initially at rest begins to
spin, accelerating with a constant angular
acceleration about its axis through its center,
achieving an angular velocity of 3.98 rev/s
when its angular displacement is 15.0 rad.
what is the value of the CD’s angular
acceleration?
Given: i = 0.0 rad/s
f = 3.98 rev/s X 2 = 25.0 rad/s
 = 15.0 rad
Find: 
Original Formula:
ωf  ωi  2θ
The i = 0.0 rad/s, so
2
2
Given: i = 0.0 rad/s f = 25.0 rad/s
 = 15.0 rad
Find: 
Formula:
ωf  2θ
2
Now, we isolate 
Given: i = 0.0 rad/s f = 25.0 rad/s
 = 15.0 rad
Find: 
ωf
Working Formula: α 
2θ
2
Given: i = 0.0 rad/s f = 25.0 rad/s
 = 15.0 rad
Find: 
Solution:
ωf
(25.0 rad/s)
α

 20.8 rad/s 2
2θ 2(15.0 rad)
2
2
Homework
Problems 7D, 1-5 on page 252
Tangential Speed
Although any point on a fixed
object has the same angular
speed, their tangential speeds vary
based upon their distance from the
center or axis of rotation.
Tangential speed (vt)–
instantaneous linear speed of an
object directed along the tangent
to the objects circular path. (see
fig. 7-6, pg. 253)
Formula
vt = r
Make sure  is in radians/s
r is in meters
vt is in m/s
Book Example 1 (pg. 254): The radius
of a CD in a computer is 0.0600m. If
a microbe riding on the disc’s rim had a
tangential speed of 1.88m/s, what is
the disc’s angular speed?
r=0.0600m
vt=1.88m/s
 =?
Use the tangential speed equation
(vt = r )to solve for angular speed.
 =vt/r
 =1.88m/s/0.0600m
 =31.3 rad/s
Example 2. A CD with a radius of 20.0 cm rotates at
a constant angular speed of 1.91 rev/s. Find the
tangential speed of a point on the rim of the disk.
Given: r =20.0cm= 0.200 m
Find: t
Solution:
 = 2X1.91rev/s= 12.0 rad/s
ν t  rω  (0.200 m)(12.0 rad/s)  2.40 m/s
Tangential Acceleration
Tangential Acceleration (at)instantaneous linear acceleration
of an object directed along the
tangent to the objects circular
path.
Formula
at= r
Make sure that at is in m/s2
r is in meters
  is in radians/s2
Example 1(at): A spinning ride at a
carnival has an angular acceleration of
0.50rad/s2. How far from the center is a
rider who has a tangential acceleration of
3.3 m/s2?
=0.50rad/s2
at= 3.3 m/s2
 r=?
Use the tangential acceleration equation
(at= r) and rearrange to solve for r.
r=at/
r = (3.3 m/s2 )/(0.50rad/s2)
r=6.6m
Example 2. The angular velocity of a rotating CD
with a radius 20.0 cm increases from 2.0 rad/s to
6.0 rad/s in 0.50 s. What Is the tangential
acceleration of a point on the rim of the disk during
this time interval?
Hint: Start by finding the angular acceleration
Given: r = 0.200 m i = 2.0 rad/s f= 6.0 rad/s
t = 0.50 s Find: 
ωf - ωi 6.0 rad/s - 2.0 rad/s
2




8.0
rad/s
Solution:
t
0.50 s
Given: r = 0.200 m
 = 8.0 rad/s2
t = 0.50 s
Find: at
2
2
a

rα

(0.200
m)(8.0
rad/s
)

1.6
m/s
Solution: t
Problems 7E (1,2 and 4),
page 255 and 7F (all) pg 257
Centripetal Acceleration
Centripetal acceleration is the
acceleration directed toward
the center of a circular path.
Caused by the change in
direction
Formula ac = vt2/r
And also……ac = r2
Example 1 pg. 258: A test car moves at a
constant speed around a circular track. If
the car is 48.2 m from the track’s center
and has a centripetal acceleration of 8.05
m/s2. What is its tangential speed?
r=48.2m
ac = 8.05 m/s2
vt=?
Use the first centripetal acceleration
equation (ac = vt2/r) to solve for vt.
 vt2= acXr
vt= acXr
vt= (8.05m/s2X48.2m)
vt= 19.7m/s
Example 2. An object with a mass of 0.345 kg,
Moving in a circular path with a radius of 0.25 m,
Experiences a centripetal acceleration of 8.0 m/s2.
Find the object’s angular speed.
Example 2. An object with a mass of 0.345 kg,
Moving in a circular path with a radius of 0.25 m,
Experiences a centripetal acceleration of 8.0 m/s2.
Find the object’s angular speed.
Given: m = 0.345 kg r = 0.25 m
ac = 8.0 m/s2
Find: 
Original Formula:
a c  rω
2
Example 2. An object with a mass of 0.345 kg,
Moving in a circular path with a radius of 0.25 m,
Experiences a centripetal acceleration of 8.0 m/s2.
Find the object’s angular speed.
Given: m = 0.345 kg r = 0.25 m
ac = 8.0 m/s2
Find: 
Solution: ω 
ac
8.0 m
 /s
2

 32 /s  5.7 rad/s
r
0.25 m

2
Using the Pythagorean Theorem
Tangential and centripetal
acceleration are perpendicular, so
you can use Pythagorean theorem
and trig to find total
acceleration and direction, if
asked for.
See figure 7-9, on page 259
at
ac
atotal
Homework:
7G pg. 258 and Section Review
pg. 259
Suppose a ball (mass m)
moves with a constant
speed V while attached to a
string….
“Is there an outward force, pulling the ball out?”
NO!
aTc
at (and vt)
“What about the “centrifugal force” I feel when my car
goes around a curve at high speed?”
There is no such thing* as centrifugal force!
You feel the centripetal force of the door
pushing you towards the center of the circle of
the turn!
Your confused brain interprets the effect of Newton’s
first law (inertia) as a force pushing you outward.
*Engineers talk about centrifugal force when they attach a coordinate
system to an accelerated object. In order to make Newton’s Laws work,
they have to invent a pseudoforce, which they call “centrifugal force.”
Centripetal Force
Centripetal force is necessary
for circular motion.
Without it, the object would go
in a linear path tangent to the
circular motion at that point.
Then it would follow the motion
of a free-falling body (parabolic
path toward the ground).
Causes of Circular Motion
Centripetal acceleration (ac) is
caused by a FORCE directed toward
the center of the circular path.
This force is called centripetal
force (Fc)
Formula Fc=mac Fc=mvt2/r or,
Fc=mr2
The unit is the Newton (kg·m/s2)
Book Example Pg. 261: A pilot is flying a
small plane at 30.0m/s in a circular path
with a radius of 100.0m. If a force of
635N is needed to maintain the pilot’s
circular motion, what is the pilot’s mass?
r=100.0m
Fc= 635 N
vt=30.0m/s
m=?
Formula
Fc=mvt2/r solve for m…
m=Fc(r/vt2)
m=635NX[100.0m/(30.0m/s)2]
m=70.6kg
Example 2: A raccoon sits 2.2 m from
the center of a merry-go-round with an
angular speed of 2.9 rad/s. If the
magnitude of the force that maintains
the raccoon’s circular motion if 67.0 N,
what is the raccoon’s mass?
r=2.2m Fc= 67.0 N
=2.9 rad/s m=?
Fc=m r 2 solve for m…
Formula
m=Fc/r 2
m=67.0N/[(2.2m)(2.9rad/s)2]
m=3.6kg
Newton’s Law and Gravity
Gravitational force depends on the
mass of the objects involved.
Formula Fg=Gm1m2/r2
G is the constant of universal
2
gravitation = 6.673X10-11Nm /kg2
Gravitational force is localized to the
center of a spherical mass
Book Example pg. 264: Find the distance
between a 0.300kg billiard ball and a
0.400kg billiard ball if the magnitude of
the gravitational force is 8.92X10-11N.
m1=0.300kg m2=0.400kg Fg=8.92X10-11N
r=?
2
2
-11
Fg= Gm1m2/r
G=6.673X10 Nm /kg2
r2=(G/Fg)m1m2

2
-11
r=[(6.673X10 Nm /kg2)/8.92X10-11N](0.30kgX0.400kg)
r= 3.00X10-1m
This is where we get our value for
Free Fall Acceleration
For the earth, Fg=mg, and
Fg= mg= GmEm/rE2
So, g= GmE/rE2
If the earth’s radius is 6.37X106m,
and the earth’s mass is 5.98X1024kg,
what is acceleration due to gravity?
9.83m/s2 …close!!
Assignments…
7H and I
Amusement Park Physics
Review Problems for Ch. 7
1,3,6,7,9,10,12,16,20,21,24,25,
30,38,40