Download Rotational Motion and the Law of Gravity 1 Rotational quantities

Rotational Motion and the Law of Gravity
Linear motion is described by position, velocity,
and acceleration.
Circular motion repeats itself in circles around
the axis of rotation
 Ex. Planets in orbit, children on a
merry-go-round, spinning a basketball
Rotational quantities
Rotational motion: when an object spins
 Axis of rotation: line about which the
rotation occurs
 Point on an object that rotates has
circular motion around that axis
o Can’t use linear quantities
because we are considering the
rotational motion of the extended
object instead of just the linear
motion of a particle
 Circular motion = change in angular
position
o All points on a rigid rotating
object (except points on the axis)
move through the same angle
during any time interval
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Angular quantities
Angular displacement (θ) describes
how much an object has rotated
 Change in angular position
 Change in arc length/distance
from axis
Θ=s
Or, s = rΘ (only works
r
in radians)
 Measured in radians (rad)
o Unit has no dimension
(placeholder only)
Reference line:
 defines Θ = 0
at some time
(t) = 0.
 analogous to
the origin in
the linear
coordinate
When s = r, Θ = 1 rad
system.

Begins at axis
Conversion factors:
of rotation.
Θ = s = 2πr = 2 π rad

Can be any
r
r
point, but
must be used
Full circle: s = 2πr
consistently
thereafter.
1 revolution (rev) = 2π rad = 360°
Θ (rad) = π Θ (degree)
180
Clockwise = negative
Counterclockwise = positive
Sample problem A
While riding on a carousel that is
rotating clockwise, a child travels
through an arc length of 11.5 m. If the
child’s angular displacement is 165°,
what is the radius of the carousel?
Rotational Motion and the Law of Gravity
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Angular velocity (ω)
Sample problem B
Rate at which an object rotates or revolves
 Also called rotational velocity
 Amount an object turns in a specific
time period
 Ratio of angular displacement to time
interval
ωavg = ΔΘ
Unit: rad/s = s-1
Δt
o Unit may also be
revolutions/time period (RPM)
 Convert revolutions to
radians
 Radian is more convenient
unit to calculate angular
velocity than degrees
 Ratio of two lengths
 Magnitude of angular velocity = angular
speed (not a vector)
 Instantaneous angular velocity
ω = lim
ΔΘ
Δt0 Δt
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 8.0π rad?
Angular acceleration (α) = change in angular velocity
αavg = Δω = ωf - ωi
Unit: rad/s2 = s-2
counterclockwise = +, clockwise = ―
Δt
Δt
Every point on a rigid rotating object has the same angular velocity and angular
acceleration.
 Instantaneous angular acceleration α = lim
Δω
Δt0 Δt
Practice problem C
A car’s tire rotates at an initial angular
speed of 21.5 rad/s. The driver accelerates,
and after 3.5 s, 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?
Kinematic equations for constant angular
acceleration
Rotational motion
ωf = ωi + α∆t
Linear motion
vf = vi + a∆t
∆θ = ωi∆t + ½α(∆t)2
∆x = vi∆t + ½a(∆t)2
ωf2 = ωi2 + 2α(∆θ)
vf2 = vi2 + 2a(∆x)
∆θ = ½(ωi + ωf)∆t
∆x = ½(vi + vf)∆t
ω represents the instantaneous angular
speed of the rotating object instead of the
average angular speed.
Rotational Motion and the Law of Gravity
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Practice problem D
Tangential speed
The wheel on an upside-down bicycle moves
through 11.0 rad in 2.0 s. What is the wheel’s
angular acceleration if its initial angular
speed is 2.0 rad/s?
There is a relationship between angular
speed and angular acceleration and
linear speed and linear acceleration of
a point on the rotating object.
 A wheel rolling along the ground
has both a linear speed and an
angular speed
Horses on a carousel rotating about its
center:
 Same angular speed
 Same angular acceleration
 Different tangential speed
depending upon their distance
from the axis of rotation
Angular speed and linear speed are related
by the circumference of the wheel.
 Circumference is the distance around
the circle and depends upon the
radius of the circle
 Circumference = 2πr
 A point at the edge of a wheel moves
1 circumference in each turn of the
circle
 Linear speed (v) is the circumference
divided by the time it takes to make
1 turn
 Angular speed (ω) = 2π
t
 Linear speed = angular speed x radius
v = ωr
o ω is instantaneous angular speed,
not average angular speed
because the time is very short
o ω must be in rad/s
Outer object covers more distance in the
same time, so its speed is greater!
Point at center (r=0) doesn’t move at all.
Tangential speed: “instantaneous
linear speed” of any point on a rotating
object
 For horses on a carousel = speed
along a line drawn tangent to its
circular path.

Angular speed is the same for all
points on the wheel

Linear speed on a point on a wheel
depends on how far the point is
from the center of rotation

Tangential speed of an object is its
speed along a line drawn tangent to
its circular path.

Time for a demonstration!
Rotational Motion and the Law of Gravity
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Practice problem E
Another practice problem
The radius of a CD in a computer is 0.0600 m.
If a microbe riding on the disc’s rim has a
tangential speed of 1.88 m/s, what is the
disc’s angular speed?
Two children are spinning around on a
merry-go-round. Joe is standing 4 meters
from the axis of rotation and Sue is
standing 2 meters from the axis.
Calculate each child’s linear speed when
the angular speed = 1 rad/s.
Back on the carousel …
Practice problem F
If the carousel speeds up, the horses on it
experience an angular acceleration.
A spinning ride at a carnival has an
angular acceleration of 0.50 rad/s2. How
far from the center is a rider who has a
tangential acceleration of 3.3 m/s2?
There will also be a linear acceleration
tangent to the circular path, and it is called
tangential acceleration.

Tangential =
acceleration
radius
x
angular
acceleration
vt = rω
t
t
at = rα

Instantaneous angular acceleration, and
unit must be in rad/s2.
Rotational Motion and the Law of Gravity
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Centripetal Acceleration
A Ferris wheel moving at constant
speed.
Acceleration?
Yes, change in
direction
Centripetal acceleration: always
directed toward the center of a
circular path; center-seeking.
ac = vt2
r
Units: m/s2
ac = rω2
a = vf – vi
tf - t i
Acceleration and velocity
are both vectors
Acceleration is due to its change in direction,
change in magnitude of velocity, or both.
Practice problem 7A (back in the textbook!)
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 the car’s tangential speed?
Note:
Centripetal acceleration is due to
changes in direction. Tangential
acceleration is due to a change in
speed.
A car on a circular track has
centripetal acceleration because it is
moving in a circle.
When the car speeds up to pass
another car, it has tangential
acceleration because of its change in
speed.
Total acceleration
Any object moving in a circle has both angular acceleration and centripetal acceleration.
When both components exist simultaneously, tangential acceleration is tangent to the
circular path and centripetal acceleration points to the center of the path.
These two components are perpendicular to each other, so total acceleration can be
found by using Pythagorean theorem.
With tangential acceleration, tangential
2
2
speed is changing. Not uniform circular
Atotal = √ac + at
motion.
-1
Direction can be found by using tan .
Rotational Motion and the Law of Gravity
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Centripetal Force
Ball of mass, m, tied to a string of length r. The ball moves
with constant speed, so it has centripetal acceleration.
ac = vt2
r
The ball’s inertia wants to travel in a
straight path, but the string exerts a
force on the ball, making it travel in a
circular path.
Fc = mac
Forces on the ball:

Force exerted by string horizontal and
vertical

Gravitational force
Gravitational force is directed downward.
The vertical component of the string’s
force counteracts the force of gravity
upward.
Fc = mvt2
r
Centripetal force:
name given to any object in uniform
circular motion.
Forces that provide centripetal force:
 Friction – ex. Car wheels and the
road around a curve
 Gravity – keeps the moon in
orbit
Net force is the horizontal force. The net
force is directed toward the center of the
object’s circular path, so it is the
centripetal force.
Practice problem 7B (book)
A pilot is flying a small plane at 56.6 m/s in a
circular path with a radius of 188.5 m. The
centripetal force needed to maintain the
plane’s circular motion is 1.89 x 104 N. What is
the plane’s mass?
The tangent is
determined by the
object’s location along its
circular path.
Centripetal force acts at right angles to an object’s circular motion,
and therefore, changes the object’s direction.
If the force disappears, the object no longer moves in a circle, but
instead moves along its tangent line.
Rotational Motion and the Law of Gravity
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Centrifugal force – object is “forced” to the outside while spinning.
Not a true force.
Ex: a car exits to the right at a high rate of speed and encounters a sharp turn to
the left. The passenger slides right and hits the right door. “Centrifugal force”?



Newton’s 1st law – law of inertia
Not enough friction to hold the passenger stationary
Slides right, which was toward the original path (straight)
Newton’s Law of Universal Gravitation
Earth and many planets in our solar system travel in nearly circular orbits around the sun.
A centripetal force keeps the planets in orbit.
Newton realized that the same force that caused an apple to fall to the ground is the
same centripetal force that keeps planets in orbit – gravitational force.
Orbiting objects are in free-fall
Newton’s law of universal gravitation:
Fg = Gm1m2
r2
Gravitational = Constant x mass1 x mass2
Force
(distance between masses)2
[inverse square law]
G = 6.673 x 10-11N▪m2
kg2
Suppose a cannon were fired from
on top of a mountain
 Path of cannonball is a
parabola
 Horizontal distance increases
as the initial speed increases
 If shot at the right speed, the
cannonball would continue
falling while the Earth curved
out from under the it
o Cannonball would orbit
the Earth
o Gravitational force
between the cannonball
and the Earth is a
centripetal force that
keeps the cannonball in
orbit
Newton’s 3rd law also applies
FEm = –FmE
Also centripetal forces ― Earth
and the moon orbit around the
center of mass of the Earth-moon
system.
Rotational Motion and the Law of Gravity
Newton’s 2nd law
F = ma, so a = F
m
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The apple falls to the earth, the earth accelerates up to meet the
apple! But, acceleration is very small because earth’s mass is so
large; acceleration is virtually negligible.
Practice problem C (textbook)
Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude
of the gravitational force between them is 8.92 x 10-11N.
Shortcomings to Law of Universal
Gravitation. Not accurate for:
 Very small objects
 Objects moving at or close to the
speed of light
Gravity is a field force
Weight changes with location
Fw = Fg = mg (mass x gravitational field
strength)



Your weight depends on where you
are on Earth
As the distance from the center of
the Earth increases, the value of “g”
decreases, so your weight decreases
Gravitational mass = inertial mass
Newton’s 2nd law F = ma
m = inertial mass that resists
acceleration
Newton’s Law of Universal Gravitation:
Fg = Gm1m2
r2
m = gravitational mass – relates how
objects attract one another
How do we know this is true?
Acceleration of objects in free-fall is
always the same confirms that the two
types of masses are equal.


Masses create a gravitational field in
space
A gravitational force is an interaction
between a mass and the gravitational
field created by other masses
Ex. Raising an object above the Earth
increases its potential energy. The GPE is
stored in the gravitational field.
Gravitational field strength: g = Fg
m
Gravitational field is a vector that points in
the direction of the gravitational force.
g = gravitational field strength and
free-fall acceleration
 Free-fall acceleration (ag) always
equals gravitational field strength for
that location [equivalent but not the
same!]
 Spring scale shows gravitational field
strength, but not acceleration – the
object is at rest.
 Gravitational field strength decreases
rapidly as the distance from Earth
increases.
o Intuitively obvious since gravity
is an inverse-square law.
Rotational Motion and the Law of Gravity
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Motion in Space
Scientific thought prior to the mid-16th
century:
 Earth is at the center of the
universe
 All orbits (around Earth) are perfect
circles
Problem:
Didn’t account for all the observations
of planetary motion.
Ptolemy (2nd century CE)
 Earth is center of universe, but
planets travel in small circles
(epicycles) while simultaneously
traveling in larger circular orbits
Copernicus (1543)
 Earth and other planets orbit the
sun in perfect circles
o Not first to theorize a suncentered universe –
Aristarchus first theorized it
1700 years before Copernicus.
o Most scientists did not accept
his theory.
Kepler’s 3 laws describe planetary motion
1. Each planet travels in an elliptical
orbit around the sun, and the sun is at
one of the focal points.
2. An imaginary line drawn from the sun
to any planet sweeps out equal areas
in equal time intervals.
3. The square of a planet’s orbital
period (T2) is proportional to the cube
of the average distance (r3) between
the planet and the sun, or T2
r3
Kepler worked with astronomer Tycho
Brahe and used his data for the orbit of
Mars.
 Data only fit if the orbit is an
ellipse rather than a circle, with
the sun at one focal point of the
ellipse.
 Kepler’s 2nd law shows an
imaginary line from the sun to any
planet sweeps out equal areas in
equal times

Kepler’s law also applies to satellites
orbiting the Earth, including our
moon.
 r is the distance between the
orbiting satellite and Earth
 Proportionality constant
depends on the mass of the
central object

For the times to be equal, the
planet must travel faster when
they are closer to the sun.
3rd law relates orbital periods and
distances of one planet to those of
another planet.
o Period (T) = time it takes a
planet to finish 1 full
revolution
o Radius (r) = mean distance
between the planet and the
sun
o T12 = r13
T22 r23
Rotational Motion and the Law of Gravity
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Kepler and Newton’s Law of Gravitation
Kepler’s 3rd law describes orbital period
Newton used Kepler’s laws to support his
law of gravitation
 Force of gravity is inversely
proportional to the distance
squared, the resulting orbit must
be an ellipse or a circle
 Law of gravitation can be used to
derive Kepler’s 3rd law


Sample problem 7D
The color-enhanced image of Venus was
compiled from data taken by Magellan, the
first planetary spacecraft to be launched
from a space shuttle. During the
spacecraft’s 5th orbit around Venus,
Magellan traveled at a mean altitude of 361
km. If the orbit had been circular, what
would Magellan’s period and speed have
been?

T2
r3
Proportionality constant between these
2 variables is 4π2/Gm
o m = mass of the object being
orbited
o T2 = 4π2 x r3
Gm
o Take the square root of T and you
have the period of any object
that is in a circular orbit.
o The speed of an object in a
circular orbit depends on the
same factors as the period
o Most planets (except Mercury)
have a nearly circular orbit
T = 2π√r3/Gm
orbital period = 2π [sq. root radius3
]
constant x mass)
vt = √Gm/r
orbital speed = [sq.root constant x mass ]
radius



m in both equations is the mass of the
central object that is being orbited.
Mass of the object that is in orbit does
not affect its speed or period
Mean radius (r) is the distance between
the centers of the 2 bodies
o For an artificial satellite orbiting
Earth, r is equal to Earth’s mean
radius plus the satellite’s distance
from the Earth’s surface.
Weight and weightlessness



Bathroom scales measure the amount of
force exerted on it when you step on it
You exert the downward force, the
scale exerts an upward normal force
Riding in an elevator and measuring
your weight
o At rest = your weight
o Descending = normal force is less
o At free-fall, FN = 0 apparent
weightlessness
Rotational Motion and the Law of Gravity
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