1. Centripetal Force
Copyright Sautter 2003

2. Centripetal Force ALL CIRCULAR MOTION REQUIRES A
“Objects in motion tend to remain in motion, at the same rate,
And in the same direction, unless acted on by an outside 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.
ALL CIRCULAR MOTION REQUIRES A
CENTRIPETAL FORCE, OTHERWISE THE
BODY CONTINUES IN A STRAIGHT LINE PATH.

3. The Second Law F = MA 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.

4. Deriving the Centripetal Acceleration Equation
Distance traveled
Over angle  S
THE INSTANTANEOUS VELOCITY
VECTOR IN CIRCULAR MOTION IS
TANGENTIAL TO THE CIRCULAR PATH V1 V2 r  V1 V V2 
Vector difference
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
The smaller  gets, the
better ac is approximated
by ac= V2 / r

5. Vectors in Circular Motion
Displacement Vector
(Radius)
Acceleration Vector
(Towards the center)
Vectors in
Circular Motion
Velocity Vector
(Tangent to the path)

6. Circular Motion with Velocity Vector
The velocity vector
is always
tangential to the
circular path

7. Circular Motion with Velocity & Acceleration Vectors
The acceleration
vector is always
towards the center
of the circular path

8. Summary of Rotational Motion Equations
AVERAGE =  /  t = (2 + 1) / 2
 = o t + ½ t2
i = o + t
i = ½ (i2 - o2) / 
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

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 ?
r = 30 m
v = 9 m/s
(a) acentripetal = V2 / r = 9 2 / 30 = 2.7 m /s2
(b) Fcentripetal = m V2 / r = m x ac = 1000 x 2.7 = 2700 N

10. Centripetal Force & Acceleration Problems A car is traveling at 20 mph on a level road with a coefficient of friction of What is the maximum curve radius ?
r = ? ft
v = 20mph
 = 0.80
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

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.

12. Banked Curves & Centripetal Force
reaction
to normal
force
centripetal
force Fc W Fn 
Vector diagram
normal
force
weight
tan  = Fc / W
Fcentripetal = W x tan 

13. Centripetal Force & Acceleration Problems What is the angle of banking which would allow vehicles to round a curve of 1000 foot radius at 50 mph without friction ?
r = 1000 ft
v =50 mph 
50 mph x 5280 ft / 3600 sec = 73.5 ft/sec
The centripetal force supplied by the banking.
Fbanking = w x tan  = mg tan , Fc = m v2/r
Fbanking = Fc , mg tan  = m v2/r , canceling mass gives g tan  = v2/r and tan  = v2 / (g x r)
tan  = / (32 x 1000) = 0.169
tan-1 (0.169) = 9.60

14. Centripetal Force & Acceleration Problems
Centripetal force problems often involve objects whirled in a circle. An example is a rock tied to a string. The circular path may be in the horizontal plane or the vertical plane.
The rock moves in a circular path because of the pull of the string to which it is tied. The string supplies the centripetal force necessary for the rock to be moved from its inertial straight line path into a circular path. The pull of the string is referred to as tension in the string.
When moving in a horizontal path the forces acting (the tension, the weight and centripetal force) remain constant in magnitude
When moving in a vertical path the forces acting vary from point to point during the travel.

15. Motion in a Vertical Circle
Centripetal force (Fc)
At the top of the path
Weight & Fc are
Opposite in direction
weight
tension
Centripetal force (Fc)
Centripetal force (Fc)
tension
tension
tension
weight
weight
At the bottom of the path
Weight & Fc are
in the same direction
weight
Centripetal force (Fc)

16. Centripetal force (Fc)
Circular Motion in a Vertical Plane A string 0.50 meters long is used to whirl a 1 kg rock in a vertical circle at 5.0 m/s. (a) Find the tension at the top of the circle (b) Find the tension at the bottom.
weight
tension
Centripetal force (Fc)
At the top
T + w = Fc
At the bottom
T = Fc + w
Downward vectors are –
Upward vectors are +
Fc = m v2 / r , w = mg
(a) T + w = Fc , T + mg = mv2/r, T + (1 x 9.8) = 1 (5.0)2 / 0.50
T = 25 / 0.50, T = 40.2 nt (downward) or – 40.2 nt
(b) T = Fc + w , T = mv2/r + mg = T = 1 (5.0)2 / (9.8)
T = 59.8 nt (upward) or nt

17. Centripetal force (Fc)
Circular Motion in a Vertical Plane A 2 lb ball on the end of a string is spun in a vertical circle at 10 ft/s. The radius of the circle is 3.0 ft. What is the maximum tension in the string ?
weight = 2 lbs
tension
Centripetal force (Fc)
v = 10 ft/s
The maximum tension
Occurs at the bottom
Of the circle & equals
The weight + centripetal
force
T = Fc + w = mv2 /r + w, w = mg, m =w/g
m = 2/32 slugs
T = ((2/32) x 102) / 3) + 2 = lbs.

18. Circular Motion & Satellites
Satellites orbiting the earth are held in their circular path by the force of gravity. Gravity supplies the centripetal force required for satellite motion.
Gravity at the Earth’s surface is, on average, 9.8 meters per second2 or 32 feet per second2. As we move away from the Earth’s surface, the force of gravity decreases.
Since the force of gravity is the centripetal force acting on a satellite, the velocity of the satellite depends on the distance of the satellite from the Earth.

19. Distance , Gravity 16.7 lbs scale 1/9 g 12000 miles 37.5 lbs ¼ g scale
Normal
gravity
scale
Center
of Earth
Radius of Earth = 4000 miles

20. Fcentripetal = Fgravity
Circular Motion & Satellites (a) What is the velocity of moon orbiting the Earth ? The moon is 242,000 miles from earth. (b) What is the period of the moon’s rotation about the Earth ?(g at the moon’s location = 8.74 x 10-3 ft/s2)
Earth Fc g
moon
Fcentripetal = Fgravity
mv2/r = mg
V2/r = g
v = (r x g )1/2
242,000 miles = 242,000 x 5280 = 1.28 x 10 9 ft
v = (r x g )1/2 = ((1.28 x 109) x ( ))1/2 = 3.34 x 103 ft/sec
(3.34 x 103) x 3600 x (1/5280) = 2,280 mph
(b) Circumference of moon’s orbit = 2  1.28 x 109 = 8.04 x 109 ft
Distance / velocity = time, x 109/ 3,340 = 2.41 x 106 seconds
2.41 x 106 / 3600 = 669 hours or 27.9 days

21. Now it's time for you to try some problems on your own !
The problems are similar to the ones which have been solved
so look back and review the appropriate problem
if you get stuck !

22. A 1200 kg car rounds a turn of 30 meter radius at 6.0 m/s
What is the centripetal acting on the car ?
(A) 48 nt (B) 147 nt (C) 240 nt (D) 1440 nt
Click
here for
answers
A car just rounds a turn at 60 km/hr. The coefficient of friction between the
and the road is What is the turning radius of the car ?
(A) 9.4 m (B) 22 m (C) 44 m (D) 565 m
A kg rock is whirled in a vertical circle of radius 60 cm. The velocity
of the rock at the bottom of the swing is 4.0 m/s. What is the string tension ?
(A) 4.9 nt (B) 8.4 nt (C) 13.3 nt (D) 18.2 nt
A satellite orbits the Earth at 8000 km from the Earth’s center. Gravity
At this location is 6.2 m/s2 . What is the velocity of the satelliter ?
(A) 0.90 km /s (B) 7.0 km /s (C) 8.9 km /s (D) cannot be determined
A road is banked at 14 degrees and has a radius of 250 meters. What is
maximum speed of the cars rounding the turn without considering friction ?
(A) 2.47 m/s (B) 9.8 m /s (C) 32 m /s (D) 14 m/s