Copyright Sautter 2003
Centripetal Force
“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.
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.
Distance traveled
Over angle 
THE INSTANTANEOUS VELOCITY
VECTOR IN CIRCULAR MOTION IS
TANGENTIAL TO THE CIRCULAR PATH
r 
r
V1
r 
r
The smaller  gets, the
better ac is approximated
by ac= V2 / r
S
Vector difference
V2

V2
V1
V
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
Displacement Vector
(Radius)
Velocity Vector
(Tangent to the path)
Acceleration Vector
(Towards the center)
The velocity vector
is always
tangential to the
circular path
The acceleration
vector is always
towards the center
of the circular path
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
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 ?
v = 9 m/s
r = 30 m
• (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
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 ?
v = 20mph
 = 0.80
r = ? ft
• 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
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.
reaction
to normal
force
centripetal
force
Vector diagram
Fc
normal
force
W
weight


Fn
tan  = Fc / W
Fcentripetal = W x tan 
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  = 73.5 2 / (32 x 1000) = 0.169
• tan-1 (0.169) = 9.60
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.
At the top of the path
Weight & Fc are
Opposite in direction
Centripetal force (Fc)
weight
Centripetal force (Fc)
weight
tension
tension
tension
tension
Centripetal force (Fc)
Centripetal force (Fc)
weight
At the bottom of the path
Weight & Fc are
weight
in the same direction
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.
Centripetal force (Fc)
weight
tension
tension
weight
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 + 9.8 = 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 / 0.50 + 1 (9.8)
T = 59.8 nt (upward) or + 59.8 nt
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 ?
The maximum tension
Occurs at the bottom
Of the circle & equals
The weight + centripetal
force
v = 10 ft/s
tension
weight = 2 lbs
Centripetal force (Fc)
• T = Fc + w = mv2 /r + w, w = mg, m =w/g
• m = 2/32 slugs
• T = ((2/32) x 102) / 3) + 2 = 4.08 lbs.
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.
1 /9
g
16.7 lbs
scale
12000 miles
¼g
37.5 lbs
scale
8000 miles
Normal
gravity
150 lbs
scale
Radius of Earth = 4000 miles
Center
of Earth
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)
g
Earth
Fc
moon
Fcentripetal = Fgravity
mv2/r = mg
V2/r = g
v = (r x g )1/2
(a) 242,000 miles = 242,000 x 5280 = 1.28 x 10 9 ft
v = (r x g )1/2 = ((1.28 x 109) x (0.00874))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, 8.04 x 109/ 3,340 = 2.41 x 106 seconds
2.41 x 106 / 3600 = 669 hours or 27.9 days
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 0.65. What is the turning radius of the car ?
(A) 9.4 m (B) 22 m (C) 44 m (D) 565 m
A 0.500 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