Transcript ROTATIONAL MOTION and the LAW of GRAVITY

ROTATIONAL MOTION AND
THE LAW OF GRAVITY
Ch 7:
MEASURING ROTATIONAL MOTION


ROTATIONAL MOTION: Motion of a body
that spins about an axis.
Ex: Ferris Wheel
Any point on a Ferris wheel that spins about a
fixed axis undergoes circular motion.
 To analyze the motion is it convenient to set up a
fixed reference line. And assume that at time
t=0, a cart is on the reference line.

Reference
line
After time interval t2 the cart advances to a
new position.
 In this time interval, the line from the center
to the cart moved through angle theta with
respect to the reference line.
 Likewise, the cart moved a distance s,
measured from the circumference of the
circle; s is arc length.

Cart (at time t2)
s
Reference
line
THETA

AND
ARC DISTANCE (S)
If the cart moves through an angle twice as
large, How would the new distance compare
with the distance shown here represented by
s?
Cart (at time t2)
s
Reference
line
THETA
AND
ARC DISTANCE (S)
If the cart moves through an angle twice as
large, How would the new distance compare
with the distance shown here represented by
s?
 THE NEW DISTANCE WOULD BE TWICE
AS LARGE!

Cart (at time t2)
s
Reference
line
ANGLES


CAN BE MEASURED IN
RADIANS
***ALMOST ALL EQUATIONS IN THIS
CHAPTER AND THE NEXT REQUIRE THAT
ANGLES BE MEASURED IN RADIANS***
Radian  an angle whose arc length is equal
to the radius, which is approximately equal
to 57.3 degrees.
ANGLES

IN
RADIANS…
Any angle theta measured in radians is
defined by:
Theta = s (arc length)
r (radius)
Radian is a pure number, with NO
dimensions.
Units cancel, and we use the
abbreviation rad to identify unit.
CIRCUMFERENCE…


When the cart of the Ferris wheel from our
previous example goes through an angle of
360 degrees. The arc length s is equal to the
circumference of the circle, or 2 π r
Thus, theta = s = 2πr = 2 π rad
r
r

Any angle in degrees can be converted to an
angle in radians by multiplying the angle
measured in degrees by:
2π
=
360 degrees
π
180 degrees
 Angular
motion is measured in
units of radians. Because there
are 2 π radians in a full circle,
radians are often expressed as a
multiple of π.


Look at Figure 7-3 in your text on page 245
Determine the radian measure equivalent to
75 degrees.


Look at Figure 7-3 in your text on page 245
Determine the radian measure equivalent to
75 degrees.
_5
12
π
ANGULAR DISPLACEMENT
Describes how much an object has rotated
 Book Definition: the angle through which a
point, line, or body is rotated in a specified
direction and about a specified axis.


EQUATION:
Angular displacement (in radians)= Change in arc length
Distance from axis
OR
θ=
s/r
EXAMPLE:


Earth has an equatorial radius of
approximately 6380 km and rotates 360
degrees every 24 hours.
What is the angular displacement (in
degrees) of a person standing at the equator
for 1 hour?
EXAMPLE:


Earth has an equatorial radius of
approximately 6380 km and rotates 360
degrees every 24 hours.
What is the angular displacement (in degrees)
of a person standing at the equator for 1
hour?
 θ = 360/24 = 15 degrees
EXAMPLE:


Earth has an equatorial radius of
approximately 6380 km and rotates 360
degrees every 24 hours.
Convert the angular displacement (in
degrees) to radians.
EXAMPLE:


Earth has an equatorial radius of approximately
6380 km and rotates 360 degrees every 24
hours.
Convert the angular displacement (in degrees)
to radians.
 θ (rad) = π /180 (θ deg) = π /180 (15 deg)
 θ (rad) = 0.26 rad
EXAMPLE:


Earth has an equatorial radius of
approximately 6380 km and rotates 360
degrees every 24 hours.
What is the arc length traveled by this
person?
EXAMPLE:


Earth has an equatorial radius of
approximately 6380 km and rotates 360
degrees every 24 hours.
What is the angular displacement (in
degrees) of a person standing at the equator
for 1 hour?
 θ = s/r
(r) θ = s
(6380) (0.26) = 1658.8 km ~1700km
ANGULAR SPEED…
Describes rate of rotation
 The average angular speed, θ avg, is the
ratio between the angular displacement,  θ,
to the time interval, t, that the object takes
to undergo that displacement.

θ
avg
=  θ / t
Units are given in rad/s
EXAMPLE

An Indy car can complete 120 laps in 1.5
hours. Even though the track is an oval
rather than a circle, you can still find the
angular speed. Calculate the average
angular speed of the Indy car.
EXAMPLE

An Indy car can complete 120 laps in 1.5 hours.
Even though the track is an oval rather than a
circle, you can still find the angular speed.
Calculate the average angular speed of the Indy car.
1: Convert hours to
seconds.
2: Convert total angular
displacement from
degrees to radians
3: Solve for average
angular velocity.
1.5 hrs = 1.5 * 60 * 60 = 5400 s
 θ = 360 deg * 120
= 43200 degrees
= 43200 (π /180)
= 753.98 radians
avg
=  θ / t
avg
= 753.98 rad / 5400 s
avg
= 0.1396 = 0.14 rad/s
ANGULAR ACCELERATION



Occurs when angular speed changes
It is the time rate of change of angular speed,
expressed in radians per second per second
(think about linear acceleration units…m/s/s or
m/s2)
Equation:

avg
=θ
speed
2-
θ
t2 - t 1
1
=
t
θ
= change in angular
time interval
EXAMPLE…

A top that is spinning at 15 rev/s spins for 55
s before coming to a stop. What is the
average acceleration of the top while it is
slowing?
EXAMPLE…

A top that is spinning at 15 rev/s spins for 55
s before coming to a stop. What is the
average acceleration of the top while it is
slowing?
avg
= θ2 - θ
avg
1
= θ = change in angular speed
t2 - t1
t time interval
= 15 (360) = 0 - 5400 =98.18 (π /180) =
55
55
- 1.7 rad/s/s
HOMEWORK TONIGHT…

PAGE 247 # 1 and 3

Page 248 # 1

Page 250 # 1 and 2

READ 7-1 and 7-2
Q.O.T.D.

•
•
12-1-2009
Convert the following from degrees to
radians:
a. 17 deg
b. 50 deg
c. 170 deg
d. 270 deg
What is the ratio that defines the average
angular speed?
When a wheel rotates about a fixed axis, do
all points on the wheel have the same
angular speed?
QOTD- ANSWERS
1 a. 0.297 rad
c. 2.967 rad
b. 0.873 rad
d. 4.712 rad
2- wavg = change in angular displacement
time interval.
delta theata/delta time (rad/s)
3- YES. Otherwise, the wheel would change shape!
ROTATIONAL MOTION
TANGENTIAL AND CENTRIPETAL
ACCELERATION
Ch 7: Section 2
TANGENTIAL SPEED


The instantaneous linear speed of an object
along the tangent to the object’s circular
path.
Objects in circular motion have a tangential
speed.
TANGENTIAL SPEED


A point on an object rotating about a fixed
axis has tangential speed related to the
object’s angular speed.
When the object’s angular acceleration
changes, the tangential acceleration of a
point on the object changes.
TANGENTIAL SPEED
Which cloud
would have to
move with a
faster
tangential
speed?
Compare their
angular speeds.
TANGENTIAL SPEED
Which cloud
would have to
move with a
faster
tangential
speed?
Compare their
angular speeds.
The cloud
furthest from
the center/axis
has the larger
tangential
speed.
Their angular
speeds are
exactly the
same.
See fig 7-6
Page 253
HOW DO
SPEED?


YOU CALCULATE
TANGENTIAL
Remember that…
Angular displacement = change in arc length/radius
or delta theta = delta s/r
Average angular speed = angular displacement/time or wavg =
delta theta/delta t
Vt = r ω
Tangential Speed = radius x angular speed
TANGENTIAL SPEED



A golfer has a max angular speed of 6.3
rad/s for her swing. She can choose
between two drivers, one placing the club
head 1.9 m from her axis of rotation, and
the other placing it 1.7 m from the axis.
1. Calculate the tangential speed of the club
head for each driver.
2. All other factors being equal, which driver
is likely to hit the ball farther?
TANGENTIAL SPEED


A golfer has a max angular speed of 6.3
rad/s for her swing. She can choose
between two drivers, one placing the club
head 1.9 m from her axis of rotation, and
the other placing it 1.7 m from the axis.
1. Calculate the tangential speed of the club
head for each driver.
Vt = rω = 1.9 (6.3) = 11.97 ~ 12 m/s
 Vt = rω= 1.7 (6.3) = 10.71 ~ 11 m/s

TANGENTIAL SPEED


A golfer has a max angular speed of 6.3
rad/s for her swing. She can choose
between two drivers, one placing the club
head 1.9 m from her axis of rotation, and
the other placing it 1.7 m from the axis.
2. All other factors being equal, which driver
is likely to hit the ball farther?

The longer driver will hit the ball further because
its club head has a higher tangential speed.
TANGENTIAL SPEED EXAMPLE…

A softball pitcher throws a ball with a
tangential speed of 6.93 m/s. If the
pitcher’s are is 0.66m long, what is the
angular speed of the ball before the pitcher
releases it?
TANGENTIAL SPEED EXAMPLE…

A softball pitcher throws a ball with a
tangential speed of 6.93 m/s. If the
pitcher’s are is 0.66m long, what is the
angular speed of the ball before the pitcher
releases it?
Vt = r ω
 6.93 = 0.66 (ω)
 6.93/0.66 = ω
 W = 10.5 rad/s

TANGENTIAL SPEED


A point on an object rotating about a fixed
axis has tangential speed related to the
object’s angular speed.
When the object’s angular acceleration
changes, the tangential acceleration of a
point on the object changes.
TANGENTIAL ACCELERATION
Tangential Acceleration is tangent to a CIRCULAR
path
 It is the instantaneous linear acceleration of an
objected directed along the tangent to the object’s
circular path.
 EQUATION for Tangential Acceleration:

at= r α
Tangential acceleration = radius x angular acceleration
EXAMPLE…

A yo-yo has a tangential acceleration of
0.98m/s/s when it is released. The string is
wound around a central shaft of radius 0.35
cm. What is the angular acceleration of the
yo-yo?
EXAMPLE…

A yo-yo has a tangential acceleration of
0.98m/s/s when it is released. The string is
wound around a central shaft of radius 0.35
cm. What is the angular acceleration of the
yo-yo?
0.35cm = 0.0035 m
at= r α
0.98 = 0.0035 (α)
0.98/.0035 = α
280 rad2 = α
CENTRIPETAL ACCELERATION

The acceleration directed toward the center
of a circular path.

An acceleration due to a change in directioneven when an object is moving at a constant
speed)
Given by…. Ac = vt2/r
Can also be found using the angular speed…
Ac = ω2 x r
UNIFORM CIRCULAR MOTION

Uniform circular motion occurs when an
acceleration of constant magnitude is
perpendicular to the tangential velocity.
TANGENTIAL
VS.
CENTRIPETAL ACCELERATION
Centripetal and tangential acceleration are
NOT the same.
 The tangential component of acceleration is
due to changing speed


The centripetal component of acceleration is
due to changing direction.
REVIEW QUESTIONS…



Describe the path of a moving body whose
acceleration is constant in magnitude at all
times and is perpendicular to the velocity.
An object moves in a circular path with a
constant speed, v. Is the objects velocity
constant? Is its acceleration constant?
Explain.
Give an example of a situation in which an
automobile driver can have a centripetal
acceleration but no tangential acceleration.

Describe the path of a moving body whose acceleration is
constant in magnitude at all times and is perpendicular to
the velocity.
A CIRCLE.

An object moves in a circular path with a constant speed, v.
Is the objects velocity constant? Is its acceleration constant?
Explain.
no. direction is changing.
no. the direction of ac is changing, at = 0

Give an example of a situation in which an automobile driver
can have a centripetal acceleration but no tangential
acceleration.
A car driving in a circle at a Constant Speed.
KEY TERMS TO KNOW FOR YOUR
TEST FRIDAY…
Angular Acceleration
 Angular Displacement
 Angular Speed
 Centripetal Acceleration
 Gravitational Force
 Radian
 Rotational Motion
 Tangential Acceleration
 Tangential Speed

DEMO



LINKS
http://www.upscale.utoronto.ca/GeneralInter
est/Harrison/Flash/ClassMechanics/VertCircul
ar/VertCircular.html
http://www.upscale.utoronto.ca/GeneralInter
est/Harrison/Flash/ClassMechanics/RTZCoord
System/RTZCoordSystem.html
http://w3.shorecrest.org/~Lisa_Peck/Physics
/syllabus/mechanics/circularmotion/hewitt/S
ource_Files/08_CentripetalForce_VID.mov
HOMEWORK!
Read 7-2 and 7-3
 Write out Key Term definitions in your own
words on a sheet of notebook paper.

Page 255 #1
 Page 256 #1-2
 Page 258 #1 and 4

Q.O.T.D.

If an object has a tangential acceleration of
10.0 m/s/s, the angular speed will do which
of the following?

•
12-2-2009
Decrease, increase, or stay the same
What is the tangential speed of a ball swung
at a constant angular speed of 5.0 rad/s on a
rope that is 5.0 m long?
QOTD- ANSWERS
1. increases
2. 25 m/s
CH 7 SECTION 3:
CAUSES OF CIRCULAR MOTION

Centripetal Force (Fc) = force that maintains
circular motion
Fc = mvt2/r
or
 Fc = mrω2

(Where m is mass of the object in motion)

Force maintains circular motion and is
measured in SI units of Newtons (N).
EXAMPLE

A pilot is flying a small plane at 30.0 m/s in
a circular path with a radius of 100.0 m. If a
force of 635 N is needed to maintain the
pilots circular motion, what is the pilot’s
mass?
EXAMPLE

A pilot is flying a small plane at 30.0 m/s in
a circular path with a radius of 100.0 m. If a
force of 635 N is needed to maintain the
pilots circular motion, what is the pilot’s
mass?
Fc = mvt2/r
635 = m (30 x 30)/(100)
635 = m (900/100)
635 = m (9)
635/9 = m
70.55 kg = m
A FORCE DIRECTED TOWARD THE CENTER
NECESSARY FOR CIRCULAR MOTION!

If this force vanishes, an object does not
continue to move in its circular path.
Instead it moves along a straight-line path
tangent to the circle.
SEE FIG. 7-11
 If swinging a ball on a string…then the string
breaks…

IS
NEWTON’S LAW OF
UNIVERSAL GRAVITATION



Gravitational force is a field force that always
exists between two masses regardless of
size.
Gravitational force acts such that objects are
always attracted to one another.
Gravitational force  mutual force of
attraction between two particles.
NEWTON’S LAW OF
UNIVERSAL GRAVITATION


Gravitational force depends on the distance
between the two masses.
If masses m1 and m2 are separated by some
distance, r, the magnitude of gravitational
force is given by:
Fg = G x
(m1m2)
r2
NEWTON’S LAW OF
UNIVERSAL GRAVITATION



G is a universal constant, constant of
universal gravitation
G = 6.673 x 10-11 Nm2/kg2
The law of universal gravitation is an
example of an inverse square law; i.e. the
force between two masses decreases as the
masses move further apart.
EXAMPLE…

Find the distance between a 0.3 kg billiard
ball and a 0.4 kg billiard ball if the
magnitude of the gravitational force is 8.92 x
10-11 N.
EXAMPLE…

Find the distance between a 0.3 kg billiard ball and a
0.4 kg billiard ball if the magnitude of the
gravitational force is 8.92 x 10-11 N.
m1 = 0.3 kg
m2 = 0.4 kg
Fg = 8.92 x 10-11 N
Fg = G x (m1m2)
r2
r2 = (G/ Fg )(m1m2) = (6.673 x 10-11/ 8.92 x 10-11 )(0.3 X 0.4)
√ r2 = √ 8.97 x 10-2 m2
r = 0.30 m
r=?
REVIEW/PRACTICE PROBLEMS…

Understand Practice Problems (shown in
BOLD) on pages:









246(7A)
248 (7B)
249(7C)
251 (7D)
254 (7E)
256 (7F)
258 (7G)
261 (7H)
264 (7I)