Explains centripetal and centrifugal force and acceleration. **More good stuff available at: www.wsautter.com

1. Copyright Sautter 2015

2. Centripetal Force
• Newton’s First Law tells us that only a force can cause a
body to move out of a straight line path. In circular
motion the direction of the body is continually changing
at every instant. Therefore a force must be acting. That
force is called centripetal (central) force since it acts
toward the circle of the circular path.
“Objects in motion tend to remain in motion, at the same rate,
And in the same direction, unless acted on by an outside force”
ALL CIRCULAR MOTION REQUIRES A
CENTRIPETAL FORCE, OTHERWISE THE
BODY CONTINUES IN A STRAIGHT LINE PATH.
2

3. Centripetal Force
• All circular motion requires a centripetal force.
Newton’s Second Law of Motion tells us that force
equal mass times acceleration. Therefore, centripetal
force must produce an acceleration (centripetal
acceleration). Since the force acts towards the center of
the circular path, the acceleration must also be
towards the center !
ALL CIRCULAR MOTION IS ACCELERATED
MOTION. THE ACCELERATION IS ALWAYS
TOWARDS THE CENTER OF THE CIRCULAR PATH.
3

4. THE INSTANTANEOUS VELOCITY
VECTOR IN CIRCULAR MOTION IS
TANGENTIAL TO THE CIRCULAR PATH
V1
V2
r
r
r
r
Distance traveled
Over angle 
S
Similar triangles give S/ r = V / V
Distance traveled (S) = V t
Therefore V t / r = V / V
Rearranging the equation gives
V / t = V x V / r
a = V / t = V 2 / r
V1
V
V2

Vector difference
The smaller  gets, the
better ac is approximated
by ac= V2 / r
4

5. Displacement Vector
(Radius)
Velocity Vector
(Tangent to the path)
Acceleration Vector
(Towards the center)
5

6. The velocity vector
is always
tangential to the
circular path
6

7. The acceleration
vector is always
towards the center
of the circular path
7

8. AVERAGE =  /  t = (2 + 1) / 2
 = o t + ½ t2
i = o + t
i = ½ (i
2 - o
2) / 
s =  R
Vlinear =  R
alinear =  R
f = 1/ T, T = 1 / f
1 revolution = 360 degrees = 2  radians
 = 2  f ,  = 2  / T
acentripetal = V2 / r
Fcentripetal = m V2 / r 8

9. Centripetal Force & Acceleration Problems
A 1000 kg car rounds a turn of 30 meter radius at 9 m/s. (a)
What is its acceleration ? (b) What is the centripetal force ?
• (a) acentripetal = V2 / r = 9 2 / 30 = 2.7 m /s2
• (b) Fcentripetal = m V2 / r = m x ac = 1000 x 2.7 = 2700 nt
r = 30 mv = 9 m/s
9

10. Centripetal Force & Acceleration Problems
A car is traveling at 20 mph on a level road with a coefficient of
friction of 0.80. What is the maximum curve radius ?
• In the English system velocity must be in ft/sec.
• 20 mph x 5280ft / 3600 sec = 29.4 ft/sec.
• The centripetal force which allows the car to round the curve is
supplied by friction.
• Ff =  Fn when the car is on level ground the normal force is the
car’s weight w = mg
• Centripetal force is given by mv2/r
• Fc = Ff ,  mg = mv2/r , canceling mass from both sides leaves
 g = v2/r and rearranging the equation, r = v2/ ( g)
• R = (29.4)2/ ( 0.80 x 32) = 34 ft
r = ? ftv = 20mph
 = 0.80
10

11. Centripetal Force & Acceleration Problems
• Roadways are often banked to help supply
centripetal force thereby allowing cars to
execute curves more readily.
• Without banking, the friction between the tires
and the road, are the only source of centripetal
force.
• As the banking angle increases the amount of
centripetal force the roadway supplies
increases.
• The following slide analyzes the relationship
between the angle of banking and the
centripetal force.
11

12. 12
Click Here