Transcript PPT

Circular Motion
Rotational Quantities
A point on an object, located a
distance r from a fixed axis of
rotation, rotates in such a way
that it travels a distance d along
the circumference of a circle.
The ratio of d to r is defined to be
the angle q measured in radians.
θ = d / r and will be referred to as
the angular displacement
dA
A
r
q
O

Rotational Quantities
A point on an object, located a
distance r from a fixed axis of
rotation, rotates in such a way
that it travels a distance d along
the circumference of a circle.
The ratio of d to r is defined to be
the angle q measured in radians.
θ = d / r and will be referred to as
the angular displacement
dA
A
In a similar fashion to linear motion:
r
r q
w 
t is the angular velocity (w = v/r)
r

w
a
r
t
is the angular acceleration (a = a/r)
dB
B
q
O
Chart of the Quantities
A simple set of relationships exists between the linear and angular quantities.
Linear Quantities:
d (displacement)
v  d t
a  v t
Angular Quantities:
Conversions:
q (angular displ.)
w  q
t
a  w t
Linear Kinematic Expressions
d qr
v  wr
a  ar
Angular Kinematic Expressions
d f  di  v i t  1 2 at 2
q f  q i  wi t  1 2 a t 2
v 2f  v i2  2ad
w 2f  wi2  2aq
Monster Trucks
A car’s speedometer is set for the tire
that is designed for use with that
particular car.
Suppose that you put tires on that are
twice the radius of the normal
tires. Does your speedometer:
1.
2.
3.
4.
Read a slower speed than is actually the case?
Read a higher speed than is actually the case?
Read the correct speed?
Doesn’t matter because you can outrun anything else on the road.
The speedometer reads ω and converts to a speed using v = ωr.
If, in reality, r is twice as big as expected, your speed (v) will be twice as big as
the speedometer calculates.
Bicycle Wheel
A bicycle tire, initially rotating at 12 rad/s, is brought to rest by
applying a brake. The tire takes 4 seconds to come to a halt.
What was the acceleration of the wheel and how many
rotations did it make before stopping? Assume the acceleration
is constant over the 4 seconds.
a  w t 
a  slope 

0 12
0 12
 3
4 0
rad
4  3
sec
q  w i t  12 a t 
2
 124  12 34  24 rad
2

ang. disp.  area  12 412  24 rad


24 rad
1 rot.
 3.82 rotations
2 rad
Uniform Circular Motion
Types of motion we’ve discussed:
Uniform Motion (constant speed)
Uniform Acceleration (constant acceleration)
Projectile Motion (2D combination of first two types)
New type of motion: Uniform Circular Motion
Object moves in circular path at a constant speed
Object of mass m swings in circle of radius R and completes one
revolution in time T.
What is its velocity?
Magnitude of velocity = speed = constant
r
v  d t  2R T
T is the period (time it takes for a complete revolution)
 Direction = ?

Direction of the Velocity
r
r d
v
t
r
Direction of v
r
is same as direction of d .
The mass moves from point
 A to point B with
 the change in displacement given by the red
vector.
This change in displacement divided by the time it
takes to travel between points A and B is the
average velocity of the mass and is equal to the
velocity at point P, the point midway between A
and B (blue vector).
Velocity Changes Directions
As the mass moves around the circle the magnitude of the velocity (it's speed)
remains constant while the direction of the velocity constantly changes, as
shown in the diagram below.
One more term to discuss: frequency.
Number of revolutions per unit time
1
f 
T
Our equation for the speed of an object can be extended
to include frequency:

v  2R T  2Rf
Angular velocity is also sometimes called angular
frequency:

w  2 T  2f
Centripetal Acceleration
From Newton’s Principia
An object shot at a low horizontal speed will
land very close to the bottom of the mountain
top from which it was shot.
If the object is shot at an increasingly higher
horizontal speed it will land farther and farther
from the base of the mountain.
Centripetal Acceleration
From Newton’s Principia
An object shot at a low horizontal speed will
land very close to the bottom of the mountain
top from which it was shot.
If the object is shot at an increasingly higher
horizontal speed it will land farther and farther
from the base of the mountain.
Newton envisioned a horizontal velocity at
which the curvature of the projectile's
trajectory matched the curvature of the
earth.
Centripetal Acceleration
x = vx t
y = ½ at
A
R
If it were not for gravity, the object would travel in a
straight line (at constant speed) and reach point X in
time t. The distance that it would travel would be given
by the horizontal motion equation:
x
2
x  vx t
R
Due to gravity, the object arrives at point A having
fallen a distance y given by the vertical equation:
From the Pythagorean Theorem
R 2  x 2  R  y 

y  1 2 at 2
2
R 2  x 2  R 2  2Ry  y 2
If t is allowed to go to zero (instantaneous
acceleration),
 then we have
x 2  2Ry  y 2
v x t 
2
 2R 2 at   2 at
1
2
1
v t  Rat  1 4 a t
2 2
x
2
v x2  Ra  1 4 a 2 t 2
2 4

2 2
v x2  Ra
v x2
a
R
What is direction?
Centrifuge
The centrifuge is spinning so that the
bottom of the test tube has a tangential
speed of 90 m/s. If the test tube is 8.5cm
long, what acceleration is experienced at
the bottom of the tube?
2
v
a
R
902
a
 95294
0.085
m
s2
~ 9724 g ' s
How many rotations does it
make per minute?
2 R
v
 2 Rf
T
v
90
f 

 169
2 R 2 0.085
 10111 rev min
rev
s
Dynamics of Circular Motion
Newton's 2nd Law provides us the insight we need to explain
why circular motion occurs.
•An object accelerates toward the center of a circle due to
the action of a net force in that direction.
•This force is referred to as the Centripetal Force.
•THIS IS NOT A NEW FORCE!
•Can simultaneously have tangential forces.
Vertical Circles
A cup supported by a platform and suspended
by strings is being swung in a vertical circle.
Take a look at the forces at each of the
indicated positions.
At the 3 and 9 o'clock position there are
two forces acting on the cup.
The weight is acting straight down
and does not contribute to the circular
motion of the cup.
The normal force is the only force
pointing toward the center of the
circle and must be responsible for the
centripetal acceleration.
r
r
FN  FC
At the 6 and 12 o'clock position the
weight and the normal force are in
the same dimension and must
together add up to be responsible
for the centripetal acceleration.
r
r
r
FN  W  FC
Limiting Case
Consider object at top. What is the normal force?
r
r
r
FN  W  FC
r
r
r
FN  FC  W
mv 2
FN 
 mg
R
v 2

FN  m  g
 R

What does it mean physically if FN = 0?

2
v min
g0
R
v min  gR
Minimum speed required to not fall off
and turn into a projectile!
Independent of mass.