Transcript n - mrhsluniewskiscience

Chapter 5
Circular Motion; Gravitation
Objectives
• Identify the force that is the cause of the
centripetal acceleration and determine the
direction of the acceleration vector.
• Use Newton's laws of motion and the
concept of centripetal acceleration to solve
word problems.
5-3 Highway Curves, Banked and Unbanked
When a car goes around a curve, there must be
a net force towards the center of the circle of
which the curve is an arc. If the road is flat, that
force is supplied by friction.
5-3 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.
5-3 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 towards
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.
Car Negotiating a Flat Turn
v
Fc
R
What is the direction of the
force ON the car?
Ans. Toward Center
This central force is exerted
BY the road ON the car.
Car Negotiating a Flat Turn
v
Fc
R
Is there also an outward force
acting ON the car?
Ans. No, but the car does exert a
outward reaction force ON the road.
Car Negotiating a Flat Turn
The centripetal force Fc is
that of static friction fs:
m
Fc
R
n
fs
Fc = fs
R
v
mg
The central force FC and the friction force fs
are not two different forces that are equal.
There is just one force on the car. The nature
of this central force is static friction.
Finding the maximum speed for
negotiating a turn without slipping.
n
fs
Fc = fs
m
v
R
Fc
R
mg
The car is on the verge of slipping when FC is
equal to the maximum force of static friction fs.
Fc = fs
Fc =
mv2
R
fs = msmg
Maximum speed without slipping (Cont.)
n
Fc = fs
fs
R
mv2
R
mg
v=
m
v
Fc
R
= msmg
msgR
Velocity v is maximum
speed for no slipping.
Example: A car negotiates a turn of
radius 70 m when the coefficient of static
friction is 0.7. What is the maximum
speed to avoid slipping?
m
v
Fc
Fc =
R
ms = 0.7
mv2
R
fs = msmg
From which: v =
msgR
g = 9.8 m/s2; R = 70 m
v  ms gR  (0.7)(9.8)(70 m) v = 21.9 m/s
Level Curves – Try this one.
• A 1500 kg car moving on a flat,
horizontal road negotiates a
curve as shown. If the radius of
the curve is 35.0 m and the
coefficient of static friction
between the tires and dry
pavement is 0.523, find the
maximum speed the car can
have and still make the turn
successfully.
v  mrg
April 8, 2016
Level Curves
• The force of static friction directed toward the
center of the curve keeps the car moving in a
circular path.
2
vmax
f s ,max  m s N  m
r
 Fy  N  mg  0
N  mg
vmax 
m s Nr
m

m s mgr
m
 m s gr
 (0.523)(9.8m / s 2 )(35.0m)  13.4m / s
v  mrg
April 8, 2016
Period 1 stopped here
Optimum Banking Angle
By banking a curve at the
optimum angle, the normal
Fc
m
R
v
fs
w
force n can provide the
necessary centripetal force
without the need for a
friction force.
n
q
slow speed
n
w
fs
q
fast speed
fs = 0
w
n
q
optimum
Free-body Diagram
Acceleration a is toward the
center. Set x axis along the
direction of ac , i. e.,
horizontal (left to right).
n
x
mg
q
n
q
n cos q
q
n
n sin q
mg
q
mg
+ ac
Optimum Banking Angle (Cont.)
n
mg
n cos q
n
n sin q
q
Apply
Newton’s 2nd
Law to x and y
axes.
q
mg
mv2
SFx = mac
n sin q 
SFy = 0
n cos q = mg
R
Optimum Banking Angle (Cont.)
n
mg
q
n sin q 
n cos q
q
n
n sin q
n
sin q
tan q 
n cos q
mg
mv2
R
n cos q = mg
2
mv
2
v
R
tan q 

mg
gR
1
Optimum Banking Angle (Cont.)
n
mg
q
Optimum Banking
Angle q
n cos q
n
q
n sin q
mg
2
v
tan q 
gR
Example 5: A car negotiates a turn of
radius 80 m. What is the optimum
banking angle for this curve if the speed
is to be equal to 12 m/s?
n
tan q =
mg
n cos q
q
q
n
n sin q
mg
v2
gR
=
(12 m/s)2
(9.8 m/s2)(80 m)
tan q = 0.184
q = 10.40
How might you 2find the
mv
centripetal
FC force on the
car, knowing R
its mass?
5-3 Highway Curves, Banked and Unbanked
Banking the curve can help keep
cars from skidding. In fact, for
every banked curve, there is one
speed where the entire centripetal
force is supplied by the
horizontal component of
the normal force, and no
friction is required. This
occurs when:
Banked Curves
• A car moving at the designated
speed can negotiate the curve.
Such a ramp is usually banked,
which means that the roadway
is tilted toward the inside of
the curve. Suppose the
designated speed for the ramp
is to be 13.4 m/s and the
radius of the curve is 35.0 m.
At what angle should the curve
be banked?
April 8, 2016
Banked Curves
v  13.4 m/s
r  35.0 m
mv 2
 Fr  n sin q  mac  r
 Fy  n cos q  mg  0
n cos q  mg
v2
tan q 
rg
13.4 m/s

q  tan (
)

27
.
6
(35.0 m)(9.8 m/s 2 )
1
Show work for problem
April 8, 2016
A car of mass, m, is traveling at a constant speed, v,
along a curve that is now banked and has a radius, R.
What bank angle, q, makes reliance on friction
unnecessary?
N
q
 Fy  0 N cosq  mg
2
F

ma

N
sin
q

mv
/R
 r
mg
tan q 
R
2
v
2
v
gR
g tan q
homework
• Questions p.129 # 8, 9
• Problems 9, 10, and do example 5-7 on
page 114 (cover up answer)
5-4 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.
5-4 Nonuniform Circular Motion
This concept can be used for an object moving
along any curved path, as a small segment of the
path will be approximately circular.
Tangential & Total Acceleration
An object may be changing
its speed (speeding up or
slowing down) as it moves in a
circular path. In that case,
there is a tangential
acceleration as well as a
centripetal acceleration.
The total acceleration atotal
is the vector sum of the
centripetal acceleration acp,
which points toward the center
of rotation, and the tangential
acceleration at, which points in
the direction of speed
increase.
Tangential Acceleration
• Tangential Acceleration – The
instantaneous linear acceleration of an
object directed along the tangent to the
object’s circular path.
• atan = Δv/Δt
Example 5-8
Example 5-8 Solution
The Centrifuge
The apparent weight of an object
can be greatly increased using
circular motion.
A centrifuge is a laboratory
device used in chemistry, biology, and
medicine for increasing the
sedimentation rate and separation of
a sample by subjecting it to a very
high centripetal acceleration.
Accelerations on the order of 10,000
g can be achieved. This can give a 12
g sample can be given an apparent
weight of about 1130 N = 250 lb.
5-5 Centrifugation
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.
Example: Big Gees
A centrifuge rotates at a rate such
that the bottom of a test tube travels at a
speed of 89.3 m/s. The bottom of the test
tube is 8.50 cm from the axis of rotation.
What is the centripetal acceleration acp
at the bottom of the test tube in m/s and
in g (where 1 g = 9.81 m/s2)?
v 2 (89.3 m/s)2
2
acp  
 93,800 m/s  9,560 g
r (0.0850 m)
“Centrifugal Force”
• “Centrifugal force” is a fictitious force - it
is not an interaction between 2 objects,
and therefore not a real force.
• Nothing pulls an object away from the
center of the circle.
“Centrifugal Force”
• What is erroneously attributed to
“centrifugal force” is actually the action of
the object’s inertia - whatever velocity it
has (speed + direction) it wants to keep.