Transcript (uniform) speed on a circular path.

Uniform Circular
Motion is the motion
of an object traveling
at a constant (uniform)
speed on a circular
path.
v = 2πr/T
The velocity of uniform
circular motion is the
circumference divided
by the period of one
revolution.
Ex. 1 - The wheel of a car has a
radius of 0.29 m and is being
rotated at 830 revolutions per
minute on a tire-balancing
machine. Find the speed in m/s
at which the outer edge of the
wheel is moving.
In uniform circular motion the
magnitude of the velocity vector
is constant. The direction;
however, is constantly changing.
This change is an acceleration
which is called
“centripetal acceleration”.
These are isosceles triangles;
therefore,
∆v/ v = v∆t / r
∆v/ v = v∆t / r
This can be solved for
∆v/∆t, to show that
ac = v2/r.
The direction of
the vector quantity
ac is toward the
center of the circle.
Ex. 3 - The bobsled track
at the Olympics contained turns
with radii of 33 m and 24 m.
Find the centripetal
acceleration for each turn for a
speed of 34 m/s.
Express the answers as
multiples of g = 9.8 m/s2.
An object in uniform circular
motion is constantly being
accelerated. This is a noninertial frame of reference; and
an object in uniform circular
motion can never be in
equilibrium.
The force that pulls an
object into a circular path
is called a centripetal
force. F = ma, so Fc = mac;
therefore,
Fc =
2
mv /r
Ex. 5 - A model airplane has
a mass of 0.90 kg and moves
at a constant speed on a
circle that is parallel
to the ground. Find the
tension T in the guideline
(length = 17 m) for speeds of
19 and 38 m/s.
When a car moves around
an unbanked curve, the
static friction of the tires on
the road provides the
centripetal force that keeps
the car from sliding, at least,
we hope so......
Ex. 7 - Compare the maximum
speeds at which a car can safely
negotiate an unbanked turn
(radius = 50.0 m) in dry weather
(coefficient of static friction =
0.900) and icy weather
(coefficient of static friction =
0.100).
Banked curves on roadways
provide a centripetal force
to help the car into a curved
path. The inward pointing
component of the normal
force provides the
centripetal force.
FC = FNsin q =
r
Also, FNcos q = mg
2
mv /
(vertical component must equal
weight)
FNsin q =
--------------------FNcos q = mg
2
mv /r
tan q =
2
v /rg
Ex. 8 - The turns at the Daytona
500 have a maximum radius (at
the top) of r = 316 m and are
banked steeply, with q = 31°.
Suppose these turns were
frictionless. At what speed
would the race cars have
to travel around them?