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Holt Physics Chapter 7
Rotational Motion
Measuring Rotational Motion
Spinning objects have rotationalmotion
Axis of rotation is the line about which rotation occurs
A point that moves around an axis undergoes circular motion (revolution)
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 = 2r (circumference),
then (rad) = 2r/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
Example 1. 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:
Find:
Solution:
r = 0.345 m s = -1.41 m
rad 4.09- m 0.345
m 1.41-
r
sΔθ
The negative sign in our answer indicates direction.
Book Example pg. 246: 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] Xdeg = (/180º)(-165º) = -2.88 radians
s = -11.5 m r=?
= s/r so, r=s /
r = -11.5/-2.88 = 3.99m
Angular Speed
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,
Therefore, if you want to know the number of revolutions:
# of Revolutions = rad /2rad
Or rad = # of rev. X 2rad
r
s
Book Example pg. 248: 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:
Find:
Solution:
avg = 4.50 rad/s = 22.7 rad
A) t B) # of revolutions
s 5.04 d/sar 4.50
dar 22.7
rev. of #
ω
θΔt
avg
rev. 3.61 dar2
dar 22.7
rad2
θ
Homework 7A page 247 and Practice 7B page 248
Angular Acceleration
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 2rad.
(0.96rev/s)X 2rad = 6.0 rad/sec
(1.434rev/s)X 2rad = 9.01 rad/sec
Given:
Find:
Solution:
1 = 6.0 rad/s 2=9.01 rad/savg=6.0rad/s2
t = ?
t = 9.01 rad/s – 6.0 rad/s =
6.0 rad/s2
0.50s
ω - ω
tavg
12
t
ω - ω 12
avg
Homework
7C 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
taif
2
2
1tatx i
xaif 222
Linear Rotational
tαωωf i
2tα2
1tωθ i
θ2ωω 22if
νf)Δt (ν Δx i21 Δt )ω (ω Δθ fi2
1
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 t=2.0 s
i= 2.00rad/s = ?
Use one of the angular kinematic equations from Table 7-2 to solve for … which one?
it ½ (t)2
Solve for
=2(-it)/(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:
Find:
Original Formula:
i = 0.0 rad/s
f = 3.98 rev/s X 2 = 25.0 rad/s
= 15.0 rad
θ2ωω2
i
2f
The i = 0.0 rad/s, so
Given:
Find:
Formula:
i = 0.0 rad/s f = 25.0 rad/s
= 15.0 rad
θ2ω 2f
Now, we isolate
Given:
Find:
Working Formula:
i = 0.0 rad/s f = 25.0 rad/s
= 15.0 rad
θ2
ω α
2f
Given:
Find:
Solution:
i = 0.0 rad/s f = 25.0 rad/s
= 15.0 rad
222
rad/s 20.8 rad) 0.15(2
rad/s) (25.0
θ2
ω α
f
Homework
Problems 7D, page 252
Review and Test
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:
Find:
Solution:
r =20.0cm= 0.200 m = 2X1.91rev/s= 12.0 rad/s
t
m/s 2.40 rad/s) m)(12.0 (0.200 rω ν t
Tangential Acceleration (at)-instantaneous linear acceleration of an object directed along the tangent to the objects circular path.
Formulaat= r
Make sure that at is in m/s2
r is in meters
is in radians/s2
Tangential Acceleration
Book Example pg.256: 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:
Find:
Solution:
r = 0.200 m i = 2.0 rad/s f= 6.0 rad/s
t = 0.50 s
2rad/s 8.0 s 0.50
rad/s 2.0 - rad/s 6.0
t
ω - ω if
Given:
Find:
Solution:
r = 0.200 m = 8.0 rad/s2 t = 0.50 s
at
22
t m/s 1.6 )rad/s m)(8.0 (0.200 rα a
7E, page 255, 7F page 256
Centripetal Acceleration
Centripetal acceleration is the acceleration directed toward the center of a circular path.Caused by the change in directionFormula ac = vt
2/rAnd also……ac = r2
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 = vt
2/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:
Find:
Original Formula:
m = 0.345 kg r = 0.25 m ac = 8.0 m/s2
2
c rω a
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:
Find:
Solution:
m = 0.345 kg r = 0.25 m ac = 8.0 m/s2
rad/s 5.7 /s32m 0.25
/sm 8.0
r
a ω 2
2
c
Homework:
7G pg. 258
Lock down drill!!!
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
Using the Pythagorean Theorem
ac
at
atotal
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!Tac
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 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).
Centripetal Force
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=mr2
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=2.2mFc= 67.0 N =2.9 rad/s m=?
Formula Fc=m r 2 solve for m…
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 gravitation = 6.673X10-11Nm
2/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=0.300kgm2=0.400kg Fg=8.92X10-11Nr=?
Fg= Gm1m2/r2 G=6.673X10-11Nm2/kg2
r2=(G/Fg)m1m2
r=[(6.673X10-11Nm2/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/rE
2
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!!
Homework7H pg. 261, 7I pg. 265 and complete Amusement Park Physics Lab
Review Problems for Ch. 7
1,3,6,7,9,10,12,16,20,21,24,25,30,38,40,43
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