Transcript Banked Turns PPT

Chapter 5
Circular Motion; Gravitation
Units of Chapter 5
•Kinematics of Uniform Circular Motion
•Dynamics of Uniform Circular Motion
•Highway Curves, Banked and Unbanked
•Nonuniform Circular Motion
•Centrifugation
•Newton’s Law of Universal Gravitation
•Gravity Near the Earth’s Surface; Geophysical Applications
•Satellites and “Weightlessness”
•Kepler’s Laws and Newton’s Synthesis
•Types of Forces in Nature
5-1 Kinematics of Uniform Circular Motion
Uniform circular motion: motion in a circle of
constant radius at constant speed
Instantaneous velocity is always tangent to
circle.
5-1 Kinematics of Uniform Circular Motion
Looking at the change in velocity in the limit that the
time interval becomes infinitesimally small, we see that
(5-1)
This acceleration is called the centripetal, or radial,
acceleration, and it points towards the center of the
circle.
Example 5-1
A 150 kg 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?
v2
aR =
r
2r 2(3.14)(0.600 m)
v=
=
= 7.54 m/s
T
(0.500 s)
v 2 (7.54 m/s) 2
aR =
=
= 94.7 m/s 2
r
(0.600 m)

Example 5-2
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.
r = 3.84x10 8 m, T = (27.3 d)(24.0 h/d)(3600 s/h) = 2.36x10 6 s
2r
v 2 4 2r 2 4  2r 4 2 (3.84x10 8 m)
v=
a =
= 2 = 2 =
= 0.00272 m/s 2
6
2
T
r
Tr
T
(2.36x10 s)

5-2 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:
(5-1)
We can see that the force
must be inward by
thinking about a ball on a
string:
5-2 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 tangent to the circle.
Example 5-3
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 (T=0.500 s).
v2
(2r/T) 2
(F)R = ma R = m = m
r
r
4  2 (0.600 m/0.500 s) 2
= (0.150 kg)
(0.600 m)
=14 N

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.
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:
Example 5-6
A 1000. kg car rounds a curve o a flat road of radius
50. m at a speed of 50. km/h (14 m/s). Will the car
follow the curve, or it will it skid? Assume (a) the
pavement is dry and the coefficient of static friction is
s=0.60; (b) the pavement is icy and s=0.25.
FN = mg = (1000. kg)(9.8 m/s 2 ) = 9800 N
v2
(14 m/s) 2
(F) R = ma R = m = (1000 kg)
= 3900 N
r
(50 m)
(a) (Ffr ) max = sFN = (0.60)(9800 N) = 5900 N
Since this is greater than 3900 N, the car will
follow the curve.
(b) (Ffr ) max = sFN = (0.25)(9800 N) = 2500 N
Since this is less than 3900 N, the car will not
follow the curve, so it will skid.

Example 5-7
(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 the angle for an expressway
off-ramp curve of 50 m at a design speed of 50
km/h?
mv 2
FR = ma R  FNsin  =
r
Fy = ma y = 0  FNcos - mg = 0  FN =
mg
cos
mv 2
mg
mv 2
mv 2
FNsin  =

sin  =
 mgtan  =
r
cos
r
r
v2
tan  =
rg
For r = 50 m and v = 14 m/s, we have
(14 m/s) 2
tan  =
= 0.40   = 22
(50 m)(9.8 m/s 2 )
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.
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 5-9
The rotor of an ultracentrifuge rotates at 50,000
rpm (revolutions per minute). The top of a 4.00
cm long test tube is 6.00 cm from the rotation axis
and is perpendicular to it. The bottom of the tube
is 10.00 cm from the axis of rotation. Calculate
the centripetal accelerations in “g’s”, at the top and
the bottom of the tube.
At top, 2r = (2 )(0.0600 m) = 0.377 m per revolution
1
50,000 rpm = 833 rev/s so T =
= 1.20x10 -3 s/rev
833 rev/s
2r
0.377 m/rev
2
v=
=
=
3.14x10
m/s
-3
T 1.20x10 s/rev
v 2 (3.14x10 2 m/s) 2
aR =
=
= 1.64x10 6 m/s 2 = 1.67x10 5 g' s
r
0.0600 m
2r (2 )(0.1000 m)
At the bottom, v =
=
= 523.6 m/s
-3
T
1.20x10 s/rev
v 2 (523.6 m/s) 2
aR =
=
= 2.74x10 6 m/s 2 = 2.80x10 5 g' s
r
0.1000 m