Transcript Circular Motion PPT

Rotational Motion
Uniform Circular Motion
Using similar triangles abc def
Where r is proportional to v1 or
v
r
7
vt
---- = -----------Since a =
7v
r
r
2
7v
------
t
c
Acceleration
2
centripetal acceleration is
always toward the center of
the circle
e
b
r
2
v
ac = ----r
r
ac = Centripetal
vThe direction of the
a
7
vxv
---- = -----------
vt
d
2
Chord ab = d = v t
vv
1
2
7
Chord ab
---- = -------------
vv
v
f
Difficult to directly measure velocity of an object moving in a
circle.
Useful to measure the “time for one complete revolution”.
Defined as the Period (T).
The distance traveled in a circle is the circumference 2 r of
the circle.

2r
v
T
d
v
t
d  2r
t T
Centripetal Acceleration
Can now calculate the acceleration of an object moving in
a circle knowing the Period and radius of the circle.
2r
( v42 )2 r 2r
T
aa  2 v 
T
rTr
2
Centripetal Force
If an object is moving in a circular path, there must be a
change in its velocity and hence an acceleration.
If there is an acceleration, a force must be present to cause
this acceleration. It also must be in the same direction.
mv4 r
F  m
ma 2
r T
2
2
Circular Motion
2r
 2rf Equations
v
T
2
4 r
2
2
a  2  4 rf
Known as frequency
T
1
1
2
f

T
4 r
T 2
f2
F  m 2  m4 rf
T
Period = The time for one revolution
Sometimes useful to show the number of
revolutions per time
Frequency is the inverse of the Period
Units are
1
 s 1  Hertz  Hz
sec
Problem:
A 75.0 kg person is attached to a pole by a 5.00 meter
rope. He makes one revolution in 4.50 seconds.
Find:
The speed of the person
2r
v
His AccelerationT
r m)
24(5.00
av  2
The Force required to 4
him on
s his
Tkeep.50
2
path 2
(m5s2.00
m)
v  64.984
r
aF  m
2
2
(4T
.50s)
a  9.75 m s2
4 (5.00m)
F  75.0kg
2
(4.50s)
F  731N
2
Problem 2:
A car approaches a level, circular curve with a radius of 45.0 m. If the concrete pavement is dry, with a coefficient of
friction of 1.20 between the rubber and concrete, what is the maximum speed at which the car can negotiate the
curve at a constant speed?
The car will be in uniform circular motion on the curve, so there must be a
centripetal force. This force is supplied by friction, so the maximum frictional
force provides the centripetal (net) force when the car is at its maximum
tangential speed.
2
v
2  m
f max  Fc v F
c
r

g

v

r

g
f max  N r 2
v. )(9.80 m s )
v  N(45
.
0
m
)(
120
 mg
mg
m
f max  vmg
 23.0rm s
2