Chapter 5
Circular Motion and Gravitation
Kinematics of Uniform Circular motion
 An object that moves in a circle at constant speed, v, is said to be
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under uniform circular motion.
The magnitude of velocity remains constant.
The direction of velocity changes continuously.
Rate of change in direction is acceleration.
Objects revolving in a circle are continuously accelerating.
What causes acceleration? A net force. What direction? Toward
the center.
Centripetal Acceleration
 Acceleration toward the center of a circular path is called
“centripetal” or “radial” acceleration, aR.
 Since acceleration depends on velocity and distance,
2
v
aR 
r
Acceleration and Velocity for circular
motion
The acceleration vector points toward the
center of the circle. The velocity vector
points in the direction of the motion.
Acceleration and velocity are always
perpendicular at each point in the path of
uniform circular motion.
Velocity is tangential to the
Path of the circular motion.
Centripetal Acceleration
 Let’s take a look at some examples we see in everyday life.
 http://www.youtube.com/watch?v=-G7tjiMNVlc
Period and Frequency
 The frequency, f, of a revolving object is the number of
revolutions it makes each second.
 The period, T, of an object is the time for one complete
revolution.
1
T
f
 If an object revolves at 3 rev/s, then each revolution takes 1/3 s.
For an object revolving in a circle at constant speed, v, we say
o Since an object travels one circumference in one revolution.

Recall:
d
v
t
So:
2r
v
t
Example 1
A 150 g ball at the end of a string is revolving uniformly in a
horizontal circle of radius 0.600m. The ball makes 2.00
revolutions in a second. What is its centripetal acceleration?
Solution: The centripetal acceleration is aR = v2/r. First we
determine the speed of the ball, v. If the ball makes 2
complete revolutions each second, then the ball travels in a
complete circle in 0.500s, which is its period, T. Since the
distance equals 2πr, where r is the radius of the circle, the
speed is 2πr/T.

Solution Example 1
 Therefore the ball has a speed
2r 2(3.14)(0.600m)
v

 7.54m /s
T
(0.500s)
 The centripetal acceleration is
v 2 (7.54m /s) 2
aR  
 94.8m /s2
r
(0.600m)
Example 2
 The moon’s nearly circular orbit about the Earth has a radius of
about 384,000km and a period T of 27.3 days. Determine the
acceleration of the Moon toward the Earth.
Solution example 2
In orbit around the Earth, the Moon travels a distance of 2πr,
where r = 3.84 x 108m is the radius of its circular path. The
speed of the Moon in its orbit about the Earth is v = 2πr/T. The
period T in seconds is
T = (27.3d)(24.0h/d)(3600s/h) = 2.36x106s.
Therefore,
v 2 (2r) 2
[2(3.14)(3.84 x10 8 m)]2
aR   2 
r
T r
(2.36x10 6 s) 2 (3.84 x10 8 m)
 0.00272m /s2  2.72x103 m /s2
In terms of g = 9.80m/s2, aR = 2.78x10-4g.

Dynamics of Uniform Circular Motion
 Circular motion still follows Newton’s laws, especially Newton’s
Second. An object moving in a circle must have a force applied to it
to keep it moving in that circle.
 A net force gives a circularly moving object, centripetal
acceleration.
2
v
 FR  maR  m r
 Since aR is toward the center, the net force must be directed toward
the center of the circle.

Newton’s First law revisited
 If NO net force acted on the circling object, then it would
continue in a STRAIGHT LINE path, NOT in a circle!
 Since the direction of the straight line path continually
changes, the direction of the force must continually change
so that it is always directed toward the center of the circle.
This is the centripetal force, the net force.
What applies the force?
 The centripetal force on an object must be applied by a
different object.
 In a rock circling at the end of a string over a person’s head,
the person pulls on the string and the string exerts the
centripetal force on the rock.
How does the object stay out there?
 Is there a force keeping the revolving object out there?
 A common misconception is the sense of a “center fleeing force,
or centrifugal force”. This is incorrect! The inertia of the
circling object causes it to continue in a straight line.
 You keep pulling inward, changing the path, but what if the
string breaks?
 The rock NOT flying outward from the center disproves the
“centrifugal force” idea…
Example 3
 Estimate the force a person must exert on a string attached to a
0.150kg ball to make the ball revolve in a horizontal circle of
radius 0.600m. The ball makes 2.00 revolutions per second.
Example 3 Solution
 First draw a free body diagram for the ball showing the 2 forces
acting on the ball, Fg = mg and the tension force, FT from the string.
(the ball’s weight makes it impossible to twirl the string truly
horizontal, but if it was small enough we could ignore it and FT can
act horizontally and provide the force to give the centripetal accel.
v2
(7.54m /s) 2
 ΣFx = max or FTx  m
 (0.150kg)
 14N
r
(0.600m)
Where we round off because we ignore ball’s mass.

Example 4: Tetherball anyone?
 The game of tetherball is played with a ball tied to a pole with a
string. When the ball is struck, it whirls around the pole. In
what direction is the acceleration of the ball, and what causes the
acceleration?
Example 4 Solution
 The acceleration points horizontally toward the center of the ball’s
circular path. The force responsible for the acceleration may not be
obvious at first, since there seems to be no force pointing directly
horizontal. But it is the net force (sum of mg and FT) that must point
in the direction of the acceleration. The vertical component of the
string tension balances the ball’s weight, mg. The horizontal
component of the string tension, FTx, is the force that produces the
centripetal acceleration.
Example 4 Illustrated
FTy
FT
FTx
mg
5-3: A car rounding a curve
 One example of centripetal acceleration occurs when an
automobile rounds a curve. The car must have an inward force
exerted on it if it is to move in a curve. On a flat road, this force
is supplied by friction between the tires and the pavement.
 As long as the tires are not slipping, the friction is static as one
part is stationary for an instant.
 If friction is not great enough, the car will skid out of the curve
in a nearly straight path.
Skidding on a curve problem
 A 1000 kg car rounds a curve on a flat road of radius 50m at a
speed of 50km/h (14m/s). Will the car make the turn, or will it
skid if: (a) the pavement is dry and the coefficient of static
friction, μs = 0.60; (b) the pavement is icy and μs = 0.25?
Racecar solution
 The normal force on the car = its weight since the road is flat and no
vertical acceleration:
 FN = mg = (1000 kg) (9.8 m/s2) = 9800 N
 In the horizontal direction, the only force is friction and we must
compare it to the force needed to produce the centripetal acceleration.
mv 2
(14m /s) 2
 FR  maR  r  (1000kg) 50m  3900N
 (a) (FFR)max=μsFN = (0.60)(9800N) = 5900 N, so the car can make the
turn.
Racecar solution part b
 (b): (FFR)max=μsFN=(0.25)(9800N) = 2500 N
 The car will skid because the ground cannot exert sufficient force
(3900 N is needed) to keep it moving in a curve of radius 50 m.
 If the wheels lock (stop rotating) when the brakes are applied too hard,
the situation gets worse! Tires slide and the friction force is now
kinetic which is less than static.
The banking of curves helps reduce
chance of skidding because there is a
component of the normal force toward
the center of the circle.
The banking angle of a
road, θ, is chosen so the
horizontal component of
the normal force, Fnsinθ,
is just equal to the force
required to give
centripetal acceleration,
mv2/r.
Your turn to Practice
 Please do Chapter 5 Review pg 138 #s 1,2,4,5,6
 Please do Ch 5 Rev p 139 #s 1-4,6,7,9,14