Transcript Unit 4 - PowerPoint

CIRCULAR MOTION
Chapter 5 – Section 5.2 to 5.5
5-2 Uniform Circular Motion—Kinematics
Uniform circular motion: motion in a circle of
constant radius at constant speed
Instantaneous velocity is always tangent to the
circle.
SPEED/VELOCITY IN A CIRCLE
Consider an object moving in a circle
around a specific origin. The DISTANCE the
object covers in ONE REVOLUTION is
called the CIRCUMFERENCE. The TIME
that it takes to cover this distance is called
the PERIOD.
scircle 
d 2r

T
T
Speed is the MAGNITUDE of the
velocity. And while the speed may be
constant, the VELOCITY is NOT. Since
velocity is a vector with BOTH
magnitude AND direction, we see that
the direction o the velocity is ALWAYS
changing.
We call this velocity, TANGENTIAL velocity as its
direction is draw TANGENT to the circle.
DERIVATION OF CENTRIPETAL ACCELERATION
Initially, the object is at a position given by vector ri and has velocity
vi. The final position and velocity vectors are rf and vf
These are similar triangles
DERIVATION- CONTINUED
Initial velocity and position vectors vi and ri are
perpendicular, as are the final vectors vf and rf.
This means the angle θ between vi and vf is the same as the
angle between ri and rf.
Since we have two isosceles triangles, and the angle between
the equal sides is identical, the two triangles must be similar.
STEP BY STEP DERIVATION
DRAWING THE DIRECTIONS CORRECTLY
So for an object traveling in a
counter-clockwise path. The
velocity would be drawn
TANGENT to the circle and the
acceleration would be drawn
TOWARDS the CENTER.
To find the MAGNITUDES of
each we have:
2r
vc 
T
2
v
ac 
r
5-2 Uniform Circular Motion—Kinematics
Looking at the change in velocity in the limit that the
time interval becomes infinitesimally small, we see
that
.
5-2 Uniform Circular Motion—Kinematics
This acceleration is called the centripetal, or
radial, acceleration, and it points toward the center
of the circle.
5-2 Uniform Circular Motion—Kinematics
Example 5-8: Acceleration of a revolving ball.
A 150-g ball at the end of a string is revolving
uniformly in a horizontal circle of radius 0.600
m. The ball makes 2.00 revolutions in a second.
What is its centripetal acceleration?
5-2 Uniform Circular Motion—Kinematics
Example 5-9: Moon’s centripetal acceleration.
The Moon’s nearly circular orbit about the Earth
has a radius of about 384,000 km and a period T
of 27.3 days. Determine the acceleration of the
Moon toward the Earth.
5-2 Uniform Circular Motion—Kinematics
A centrifuge works by
spinning very fast. This
means there must be a
very large centripetal force.
The object at A would go in
a straight line but for this
force; as it is, it winds up at
B.
5-2 Uniform Circular Motion—Kinematics
Example 5-10: Ultracentrifuge.
The rotor of an ultracentrifuge rotates at
50,000 rpm (revolutions per minute). A
particle at the top of a test tube is 6.00
cm from the rotation axis. Calculate its
centripetal acceleration, in “g’s.”
CIRCULAR MOTION AND N.S.L
2
Recall that according to
Newton’s Second Law, the
acceleration is directly
proportional to the Force. If
this is true:
v
FNET  ma ac 
r
2
mv
FNET  Fc 
r
Fc  Centripetal Force
Since the acceleration and the force are directly
related, the force must ALSO point towards the
center. This is called CENTRIPETAL FORCE.
NOTE: The centripetal force is a NET FORCE. It
could be represented by one or more forces. So
NEVER draw it in an F.B.D.
5-3 Dynamics of Uniform Circular Motion
For an object to be in uniform circular motion, there
must be a net force acting on it.
We already know the
acceleration, so can
immediately write the force:
5-3 Dynamics of Uniform Circular Motion
We can see that the
force must be inward
by thinking about a ball
on a string. Strings
only pull; they never
push.
EXAMPLES
2r
vc 
T
The blade of a windshield wiper moves
through an angle of 90 degrees in 0.28
seconds. The tip of the blade moves on
the arc of a circle that has a radius of
0.76m. What is the magnitude of the
centripetal acceleration of the tip of the
blade?
2 (.76)
vc 
 4.26 m / s
(.28 * 4)
v 2 (4.26) 2
ac 

 23.92 m / s 2
r
0.76
EXAMPLES
Top view
What is the minimum coefficient of static friction
necessary to allow a penny to rotate along a 33
1/3 rpm record (diameter= 0.300 m), when
the penny is placed at the outer edge of the
record?
F f  Fc
FN
mg
Side view
mv
r
mv 2
mg 
r
v2

rg
FN 
Ff
2
rev 1 min
33.3
*
 0.555 rev
sec
min 60 sec
1sec
 1.80 sec
T
rev
0.555 rev
2r 2 (0.15)
vc 

 0.524 m / s
T
1.80
v2
(0.524) 2


 0.187
rg (0.15)(9.8)
EXAMPLES
The maximum tension that a 0.50 m string
can tolerate is 14 N. A 0.25-kg ball attached
to this string is being whirled in a vertical
circle. What is the maximum speed the ball
can have
 (a) the top of the circle,
 (b)at the bottom of the circle?
mv 2
FNET  Fc  mac 
r
mv 2
T  mg 
 r (T  mg )  mv 2
r
r (T  mg )
0.5(14  (0.25)(9.8))
v

m
0.25
v  5.74 m / s
T mg
EXAMPLES
mv 2
FNET  Fc  mac 
r
mv 2
T  mg 
 r (T  mg )  mv 2
r
r (T  mg )
0.5(14  (0.25)(9.8))
v

m
0.25
v  4.81 m / s
At the bottom?
T
mg
5-3 Dynamics of Uniform Circular Motion
Example 5-12: Revolving ball (vertical circle).
A 0.150-kg ball on the end of a
1.10-m-long cord (negligible
mass) is swung in a vertical
circle. (a) Determine the
minimum speed the ball must
have at the top of its arc so that
the ball continues moving in a
circle. (b) Calculate the tension
in the cord at the bottom of the
arc, assuming the ball is moving
at twice the speed of part (a).
5-3 Dynamics of Uniform Circular Motion
Example 5-13: Conical pendulum.
A small ball of mass m,
suspended by a cord of length l,
revolves in a circle of radius r =
l sin θ, where θ is the angle the
string makes with the vertical.
(a) In what direction is the
acceleration of the ball, and
what causes the acceleration?
(b) Calculate the speed and
period (time required for one
revolution) of the ball in terms of
l, θ, g, and m.
5-3 Dynamics of Uniform Circular Motion
Example 5-11: Force on revolving ball
(horizontal).
Estimate the force a person must exert on a
string attached to a 0.150-kg ball to make the
ball revolve in a horizontal circle of radius 0.600
m. The ball makes 2.00 revolutions per second.
Ignore the string’s mass.
5-5 Nonuniform Circular Motion
This concept can be used for an object
moving along any curved path, as any small
segment of the path will be approximately
circular.
5-5 Nonuniform Circular Motion
If an object is moving in a
circular path but at varying
speeds, it must have a
tangential component to its
acceleration as well as the
radial one.
UNBANKED & BANKED CURVES
Unbanked Curves
When a car goes around a curve, there must be a
net force toward the center of the circle of which the
curve is an arc. If the road is flat, that force is
supplied by friction.
5-4 Highway Curves: Banked and Unbanked
If the frictional force is
insufficient, the car will
tend to move more nearly
in a straight line, as the
skid marks show.
5-4 Highway Curves: Banked and Unbanked
As long as the tires do not slip, the friction is static. If
the tires do start to slip, the friction is kinetic, which
is bad in two ways:
1. The kinetic frictional force is smaller than the
static.
2. The static frictional force can point toward the
center of the circle, but the kinetic frictional force
opposes the direction of motion, making it very
difficult to regain control of the car and continue
around the curve.
Unbanked Curve Example
A 1000-kg car rounds a curve on a flat road of
radius 50 m at a speed of 15 m/s (54 km/h).
Will the car follow the curve, or will it skid?
Assume:
(a) the pavement is dry and the coefficient of
static friction is μs = 0.60;
(b) (b) the pavement is icy and μs = 0.25.
BANKED CURVES
Curves that are tilted at an angle instead of
being flat. Why?
Allows a car to go around the curve without needing
friction to make the turn
Normal force provides the necessary centripetal force to
complete the turn
5-4 Highway Curves: Banked and Unbanked
Banking the curve can help keep
cars from skidding. In fact, for every
banked curve, there is one speed at
which the entire centripetal force is
supplied by the
horizontal component of the
normal force, and no friction
is required. This occurs
when:
BANKED TURNS: FBD
oUnlike inclines, keep the
axis straight up and
down NOT parallel to
the surface.
oFor ideal banking, FNX
serves as the only
centripetal force
BANKED CURVES: N2L
Fc  mac
FNX
mv 2

r
mv 2
FN sin  
r
mg
mv 2
sin  
cos
r
mv 2
mg tan  
r
F
y
 may
FNY  mg  0
FN cos  mg
FN 
mg
cos 
v 
  tan  
 gr 
1
2
5-4 Highway Curves: Banked and Unbanked
Example 5-15: Banking angle.
(a) For a car traveling with speed v around a
curve of radius r, determine a formula for the
angle at which a road should be banked so
that no friction is required. (b) What is this
angle for an expressway off-ramp curve of
radius 50 m at a design speed of 50 km/h?
BANKED CURVES WITH FRICTION
A banked curve allows a car to complete the turn at a given speed
without the need of friction
If a car goes above the ideal speed of a banked curve, friction needs
to act in order to allow the car to safely complete the turn.
BANKED CURVES W/ FRICTION FBD
FfX
Friction allows for a range
of speeds.
Ffy
BANKED CURVES W/ FRICTION N2L
F
y
 may
FNY  F fy  mg  ma y
FfX
FN cos  F f sin   mg
FN cos  u s FN sin   mg
mg
FN 
cos    s sin 
Ffy
Notice that friction actually
increases the normal force
the car experiences…
BANKED CURVES W/ FRICTION N2L
F
c
FNX
 mac
mv 2
 F fx 
r
FfX
mv2
FN sin   us FN cos  
r
mv2
FN sin    s cos   
r
Ffy
v

sin    s cos  
gr
cos  s sin  
v

tan    s 
gr
1   s tan  

sin    s cos   v 2
g

cos    s sin 
r
BANKED CURVE:
IDEAL VS. MAX SPEED
Ideal Banking for 45 mph (~20m/s) and radius of 100 m.
2


20
1
  22.2
  tan 
 (9.8)(100) 
With friction between rubber tires and cement road (us=1.00),
the max speed becomes 26.6 m/s or about 60 mph.
v

tan 22.2  1.0
(9.8)(100)
 26.6m / s
1  1.0 tan 22.2
5-3 Dynamics of Uniform Circular Motion
There is no centrifugal force pointing outward; what
happens is that the natural tendency of the object to
move in a straight line must be overcome.
If the centripetal force vanishes, the object flies off at
a tangent to the circle.
FICTITIOUS FORCES
A perception of a force caused by the acceleration of a reference frame.
Looking at a washing
machine, it appears there is a
centrifugal (center-fleeing)
force that causes the clothes
to be pressed against the wall
of the washing machine.
FICTITIOUS FORCES CONT.
There is no force pushing
outward
The normal force from the
dryer is pushing them inward.
the clothes
(no source).
walls of the
That normal force causes the clothes to accelerate in a circle
as opposed to flying off in a straight line.
ARTIFICIAL GRAVITY
When a spacecraft is far away from Earth or any other
body, the force of gravity is near zero.
 No gravity- astronauts and objects float
Rotating the spacecraft creates this artificial gravity with
effects similar to true gravity
 people in rotating spacecraft believe they are indeed experiencing
true gravitational force
ARTIFICIAL GRAVITY
In order to feel like spacecraft has gravity, it must
experience a centripetal acceleration of 10 m/s2.
2r
v
T
2
v
ac   g  9.8 m s 2
r
2
gT
r
2
4