Transcript Centripetal Force

Centripetal Force
Acceleration in a Circle

v2
dq

dv

v1

Acceleration is a vector
change in velocity compared
to time.

For small angle changes the
acceleration vector points
directly inward.

This is called centripetal
acceleration.

a
Centripetal Acceleration

Uniform circular motion takes place with a constant
speed but changing velocity direction.

The acceleration always is directed toward the center
of the circle and has a constant magnitude.
v2
ar 
  2r
r
Buzz Saw

A circular saw is designed
with teeth that will move at
40. m/s.

• r = v2/a.


The bonds that hold the
cutting tips can withstand a
maximum acceleration of
2.0 x 104 m/s2.
Start with a = v2/r.
Substitute values:
•
•

Find the diameter:
•

Find the maximum diameter
of the blade.
r = (40. m/s)2/(2.0 x 104 m/s2)
r = 0.080 m.
d = 0.16 m = 16 cm.
Law of Acceleration in Circles


Motion in a circle has a
centripetal acceleration.
There must be a centripetal
force.
• Vector points to the center

F

a
mv2
Fr 
 m 2 r
r

The centrifugal force that we
describe is just inertia.
• It points in the opposite
direction – to the outside
• It isn’t a real force
Conical Pendulum

A 200. g mass hung is from a 50. cm string as a
conical pendulum. The period of the pendulum in a
perfect circle is 1.4 s. What is the angle of the
pendulum? What is the tension on the string?
q
FT
Radial Net Force


The mass has a downward
gravitational force, -mg.
There is tension in the string.
• The vertical component must
cancel gravity
FTy = mg
FT = mg / cos q
FTr = mg sin q / cos q = mg tan q

This is the net radial force – the
centripetal force.
q
FT
FT cos q
FT sin q
mg
Acceleration to Velocity

The acceleration and
velocity on a circular path
are related.
a  F / m  g tan q
a  v2 / r
v 2  gr tan q
v  gr tan q
q
FT
r
mg tan q
mg
Period of Revolution

The pendulum period is
related to the speed and
radius.
r  L sin q
v  2r / T  2L sin q / T
v 2  gr tan q
4 L sin q / T  gL sin q sin q / cosq
2 2
2
2
cos q = 0.973
q = 13°
L
FT
r
2
cos q  T g / 4 L
2
q
mg tan q
Radial Tension

The tension on the string can be found using the
angle and mass.

FT = mg / cos q = 2.0 N

If the tension is too high the string will break!
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