Transcript Centripetal Force

Centripetal Force
Law of Action in Circles



Motion in a circle has a centripetal acceleration.
For every acceleration there is a net force.
There must be a centripetal force.
• Points to the center of the circle
• Magnitude is ma = mv2/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

Tension: FT = mg / cos q = 2.0 N

Centripetal force:
FTr = mg sin q / cos q = mg tan q
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
Vertical Curve

A loop-the-loop is a popular
rollercoaster feature.

There are only two forces
acting on the moving car.
• Gravity
• Normal force
FN

There is a centripetal
acceleration due to the loop.
• Not uniform circular motion
Fg
Staying on Track

If the normal force becomes
zero, the coaster will leave
the track in a parabolic
trajectory.
• Projectile motion

At any point there must be
enough velocity to maintain
pressure of the car on the
track.
Fg
Force at the Top

The forces of gravity and the
normal force are both
directed down.

Together these must match
the centripetal force.

The minimum occurs with
almost no normal force.

The maximum is at the
bottom: a = v2 / r.
Fg
FN
Fc  FN  Fg
mv2 / r  FN  mg
v
FN r
 gr
m
vmin  gr
Horizontal Curve


A vehicle on a horizontal curve has a centripetal
acceleration associated with the changing direction.
The curve doesn’t have to be a complete circle.
• There is still a radius (r) associated with the curve
• The force is still Fc = mv2/r directed inward
r
Fc
Curves and Friction

On a turn the force of static friction provides the
centripetal acceleration.

In the force diagram there is no other force acting in
the centripetal direction.
Fc  mv 2 / r
Fc  F f
r
Fc
F f   s mg
Skidding

The limit of steering in a
curve occurs when the
centripetal acceleration
equals the maximum static
friction.
2
Fc  mvmax
/ r  F f   s mg
v
2
max
/ r  s g
vmax   s gr
2
r  vmax
/ s g


A curve on a dry road (s =
1.0) is safe at a speed of 90
km/h.
What is the safe speed on
the same curve with ice (s =
0.2)?
•
•
•
•
90 km/h = 25 m/s
rdry = v2/ s g = 64 m
v2icy = s g r = 120 m2/s2
vicy = 11 m/s = 40 km/h
Banking
Fc  FN sin q  F f
2
mvmax
/ r  mg tan q   s mg
vmax  (  s  tan q ) gr

Curves intended for
higher speeds are
banked.

Without friction a curve
banked at an angle q can
supply a centripetal force
Fc = mg tan q.

The car can turn without
any friction.
next