Transcript Circular Motion - Paso Robles High School

Circular Motion
Objectives:
Use vectors to pictorially analyze
problems for circular and rotary
motion.
Solve circular motion word
problems.
Interpret the relationship between
the force on an object and the
velocity of the object for circular
motion through the analysis of
experimental data.
v
Circular Motion
v
v
Suppose you drive a go cart in a perfect
circle at a constant speed. Even though
your speed isn’t changing, you are
accelerating. This is because acceleration
is the rate of change of velocity (not
speed), and your velocity is changing
because your direction is changing!
Remember, a velocity vector is always
tangent to the path of motion.
Centripetal Acceleration
Formula for centripetal acceleration
in uniform circular motion
vf = v
r
v0 = v

r
vf - v0
v0

vf
vt
r

r
Acceleration, a = (vf - v0) / t. We are
subtracting vectors here, not speeds,
otherwise a would be zero. (v0 and
vf have the same magnitudes.) The
smaller t is, the smaller  will be,
and the more the blue sector will
approximate a triangle. The blue
“triangle” has sides r, r, and v t
(from d = v t ). The vector triangle
has sides v, v, and | vf - v0 |. The
two triangles are similar (side-angleside similarity).
continued on next slide
CentripetalBy similar
Acceleration
triangles,
vf = v
r
v
v0 = v

r
r
=
(cont.)
| vf - v0 |
vt
So, multiplying both sides above by v, we
have
| vf - v0 |
ac =
t
v2
r
=
vt
v

v
r

r
Unit check:
(m/s)2
m
m
=
s2
m2 / s 2
=
m
Tangential vs. Centripetal Acceleration
Suppose now you drive your go cart faster and
faster in a circle. Now your velocity vector
changes in both magnitude and direction. If
you go from start to finish in 4 s, your average
tangential acceleration is:
15 m/s
18 m/s
finish
start
10 m/s
at = (18 m/s - 10 m/s) / 4 s = 2 m/s2
So you’re speeding up at a rate of 2 m/s per second. This is the rate at which your velocity
changes tangentially. But what about the rate at which your velocity changes radially, due to
its changing direction? This is your centripetal (or radial) acceleration.
So how do we calculate the centripetal acceleration ? ? ?
Stay tuned!
Centripetal acceleration vector always
points toward center of circle.
v
v
ac
ac
at
moving counterclockwise;
speeding up
at
moving counterclockwise;
slowing down
ac depends on both v and r.
However, regardless of speed or tangential acceleration, ac always points toward the
center. That is, ac is always radial (along the radius).
“Centripetal” means “center-seeking.” The magnitude of
Resultant Acceleration
The overall acceleration is the vector sum
of the centripetal acceleration and the
tangential acceleration. That is,
a = ac + a t
This is true regardless of the direction of
motion. It holds true even when an object
is not moving in a perfect circle. Note: The
equation above does not include v.
Vectors of different quantities cannot be
added!
moving counterclockwise while
speeding up or moving clockwise
while slowing down
Non-circular paths
Here we have an object moving along the brown path at a constant speed (at = 0). ac
changes, though, since the radius of curvature changes. At P1 the path is approximated by the
large green circle, at P2 by the smaller orange one. The smaller r is, the bigger ac is.
ac =
v2
r
R2
v
R1
ac
ac
ac
P1
v
P2
Friction and ac
You’re cruising at a constant 20 m/s on a winding
highway. The radius of curvature where you are is
60 m. Your centripetal acceleration is:
ac
20 m/s
= (20 m/s)2 / (60 m) = 6.67 m/s2
The force that causes this acceleration is friction,
which is why it’s hard to turn on ice. Friction, in
this case, is the “centripetal force.” The sharper
the turn or the greater your speed, the greater the
frictional force must be.
overhead view
continued on next slide
Friction
and
a
(cont.)
c
Since you’re not speeding up, f is the net force, so F
net
= f = s N = s mg = ma. We use s because you’re
not sliding (or even moving) radially. Thus,
s mg = ma.
f
v
Mass cancels out, showing that your centripetal
acceleration doesn’t depend on how heavy your vehicle
is. Solving for a we have
a = s g.
m
overhead view
In the diagram N is pointing out of the slide. Also, a = ac
in this case since
at = 0.
Centripetal Force, Fc
From F = ma, we get Fc = mac = mv2 / r.
2
mv
Fc =
r
If a body is turning, look at all forces acting on it, and find the
net force. The component of the net force that acts toward
the center of curvature (perpendicular to the body’s motion)
is the centripetal force. The component that acts parallel to
its motion (forward or backwards) is the tangential
component of the net force.
Forces that can provide a centripetal
force
Friction, as in the turning car example
•
• Tension, as in a rock whirling around while attached to a string,
or the tension in the chains on a swing at the park.*
• Normal Force, as in a “round-up ride” at an amusement park
(that spins & the floor drops out), or the component of normal
force on a car on a banked track that acts toward the center.*
• Gravity: The force of gravity between the Earth and sun keeps
the Earth moving in a nearly circular orbit.
• Any force directed toward your center of curvature, such as an
applied force.
* Picture on upcoming slides
Loop-the-loop
in
a
Plane
A plane flies in a vertical circle, so it’s
v
upside down at the high point. Its
speed is constant, but because of its
nonlinear motion, the pilot must
experience centripetal acceleration.
This ac is provided by a combination
of her weight and her normal force.
mg is constant; N is not. N is the
force her pilot’s chair exerts on her
body.
Ntop
mg
Nbot
v
mg
continued on next slide
Top: Normal force and weight team up to
provide centripetal force: Ntop + mg =
mv2 / r. If the pilot were sitting on a scale,
it would say she’s very light.
Loop-the-loop
v
Ntop
(cont.)
mg
r
Bottom: Weight works against normal
force, so N must be bigger down here to
provide the same centripetal force:
Nbot - mg = mv2 / r. (Fc has a constant
magnitude since m, v, and r are
constants.) Here a scale would say that
the pilot is very heavy.
Nbot
v
continued on next slide
mg
Loop-the-loop
(cont.)
The normal force (force on pilot due to seat)
changes throughout the loop. This case is
similar to the simple pendulum (the only
difference being that speed is constant here).
Part of the weight opposes N, and the net
radial force is the centripetal force:
r
N - mg cos = mac = mv2 / r
N
We’ve been discussing the pilot,
but what force causes the plane to
turn?
mg cos

mg sin
mg
Answer:
The air provides the centripetal force on plane

Simple Pendulum
string of
length L

Two forces act on a swinging pendulum, tension and weight.
Tension acts radially. We break the weight vector into a radial
component (green) and tangential one (violet). Blue is bigger
than green, otherwise there would be no net centripetal force,
and the mass couldn’t turn.
Fc = T - mg cos = mac = mv2 / L.
Ft = mg sin = mat
T
Ft is the force that speeds the mass up or slows it down. The
weight is a constant, but since  changes, so do the weight’s
components.
mg cos is greatest when the mass is at its low point. This is
also where it’s moving its fastest. For these two reasons T is the
greatest when the mass is at the low point.
m

mg
Facts about the Simple Pendulum
• A pendulum’s period is the same for big arcs as it is for little
arcs, so long as the angle through which it swings isn’t real
large. (The average speed, therefore, is greater when the arc is
bigger, because it must cover a bigger distance in the same
time.)
• The period is independent of the mass.
• The period depends only on the pendulum’s length.
• The period = T = 2
(proven in advanced physics)
L
g
• This formula gives us a way to measure the acceleration due to
gravity (which varies slightly with location) by measuring the
period and length of a pendulum.
• Don’t confuse the symbol T, which is used for both period and
tension.
Conical Pendulum
Like a tether ball, the mass hangs from a
rope and sweeps out a circular path, and
the rope a cone.  is a constant. The
vertical component of T balances mg.
The horiz. comp. of T is the centripetal
force.


T
T
ac
m
ac
m
v
mg
mg
continued on next slide
v
Conical Pendulum

T cos
(cont.)
All vectors shown are forces.
v points into or out of the slide, depending on
the direction in which the mass moves. ac is
parallel to
T sin, which serves as the centripetal force.
T

T cos = mg
T sin = mv2 / r
T sin
m
mg
See it in action
Dividing equations:
tan = v2 / rg
Gravity
Newton’s Law of Gravitation
Kepler’s Laws of Planetary
Motion
Gravitational Fields
Newton’s Law of Gravitation
r
m2
m1
There is a force of gravity between any pair of objects
anywhere. The force is proportional to each mass and
inversely proportional to the square of the distance between
the two objects. Its equation is:
FG
G m1 m2
=
r2
The constant of proportionality is G, the universal gravitation
constant. G = 6.67 · 10-11 N·m2 / kg2. Note how the units of G all
cancel out except for the Newtons, which is the unit needed on the
left side of the equation.
Gravity Example
How hard do two planets pull on each other if their masses are
1.231026 kg and 5.211022 kg and they 230 million kilometers
apart?
G m1 m2
FG =
r2
-11 N·m2 / kg2) (1.23 · 1026 kg) (5.21 · 1022 kg)
(6.67
·
10
=
(230 · 103 · 106 m) 2
= 8.08 · 1015 N
This is the force each planet exerts on the other. Note the denominator
is has a factor of 103 to convert to meters and a factor of 106 to
account for the million. It doesn’t matter which way or how fast the
planets are moving.
3rd Law: Action-Reaction
In the last example the force on each planet is the same. This is due to
to Newton’s third law of motion: the force on Planet 1 due to Planet 2
is just as strong but in the opposite direction as the force on Planet 2
due to Planet 1. The effects of these forces are not the same, however,
since the planets have different masses.
For the big planet: a = (8.08 · 1015 N) / (1.23 · 1026 kg)
= 6.57 · 10-11 m/s2.
For the little planet: a = (8.08 · 1015 N) / (5.21 · 1022 kg)
= 1.55 · 10-7 m/s2.
8.08 · 1015 N
5.21
· 1022 kg
8.08 · 1015 N
1.23 · 1026 kg
Inverse Square Law
The law of gravitation is called an inverse square law because the
magnitude of the force is inversely proportional to the square of the
separation. If the masses are moved twice as far apart, the force of
gravity between is cut by a factor of four. Triple the separation and
the force is nine times weaker.
FG
G m1 m2
=
r2
What if each mass and the separation were all quadrupled?
answer:
no change in the force
Calculating the Gravitational Constant
In 1798 Sir Henry Cavendish suspended a rod with two small masses
(red) from a thin wire. Two larger mass (green) attract the small
masses and cause the wire to twist slightly, since each force of
attraction produces a torque in the same direction. By varying the
masses and measuring the separations and the amount of twist,
Cavendish was the first to calculate G.
Since G is only
6.67 · 10-11 N·m2 / kg2,
the measurements had
to be very precise.
Calculating the mass of the Earth
Knowing G, we can now actually calculate the mass of the Earth. All
we do is write the weight of any object in two different ways and
equate them. Its weight is the force of gravity between it and the
Earth, which is FG in the equation below. ME is the mass of the
Earth, RE is the radius of the Earth, and m is the mass of the object.
The object’s weight can also be written as mg.
FG
G m1 m2
=
r2
=
G ME m
RE 2
= mg
The m’s cancel in the last equation. g can be measured experimentally;
Cavendish determined G’s value; and RE can be calculated at
6.37 · 106 m (see next slide). ME is the only unknown. Solving for
ME we have:
g RE 2
= 5.98 · 1024 kg
ME =
G
Rotational Motion
s

r
Arc length: s = r 
If  is in radians, then the arc length, s, is 
times r. This follows directly from the
definition of a radian. One radian is the angle
made when the radius of a circle is wrapped
along the circle.
r
When the arc length is as long as the radius,
the angle subtended is one radian. (A radian
is really dimensionless, since it’s found by
dividing a length by a length.)
1 radian
r
B
Angular Speed, 
110
3m
A
Linear speed is how fast you move, measured
as distance per unit time. Angular speed is
how fast you turn, measured as an angle per
unit time. The symbol for angular speed is the
small Greek letter omega, , which looks like a
curvy “w”. Units for angular speed include:
degrees per second; radians per second; and
rpm (revolutions per minute).
Suppose an object moves steadily from A to B along the circle in 5 s. Then  = 110 / 5 s =
22 / s. The distance it covers is
(110 / 360)(2)(3) = 5.7596 m. So its linear speed is
v = (5.7596 m) / (5 s) = 1.1519 m / s.
v = r

v /2
v
s = r
r

s
r
=
=
t
t
t
v = r
The Three Stooges go to the park. Moe and Larry
are on a merry-go-round ride of radius r that
Larry is pushing counterclockwise, running at a
speed v. In a time t, Moe goes from M to M´
and Curly goes from C to C´. Moe is twice as far
from the center as Curly. The distance Moe
travels is r , where  is in radians. Curly’s
distance is ½ r . Both stooges sweep out the
same angle in the same time, so each has the
same angular speed. However, since Moe travels
twice as far, his linear speed in twice as great.
Ferris Wheel
Problem
Schmedrick is working as a miniature ferris wheel
operator (radius 2.1 m). He gets a little overzealous
and cranks it up to 75 rpm. His little brother
Poindexter flies out at point P, when he is 35 from
the low point. At the low point the wheel is 1 m off
the ground. A 1.5 m high wall is 27 m from the low
point of the wheel. Does Poindexter clear the wall?
P
Strategy outlined on next slide
Ferris Wheel Problem-Solving Strategy
1. Based on his angular speed and the radius,
calculate Poindexter’s linear speed.
16.4934 m/s
2. Break his launch velocity down into vertical
and horizontal components.
vx = 13.5106 m/s
vy0 = 9.4602 m/s
3. Use trig to find the height of his launch.
4. Use trig to find the horizontal distance from
the launch site to the wall.
1.3798 m
5. Calculate the time it takes him to go that far
horizontally.
25.7955 m
6. Calculate height at that time.
7. Draw your conclusion.
1.9093 s
1.5798 m
He just makes it by about 8 cm !