Example 1: A 3-kg rock swings in a circle
of radius 5 m. If its constant speed is 8
m/s, what is the centripetal acceleration?
2
v
v
m
m = 3 kg
ac 
R
R
R = 5 m; v = 8 m/s
2
(8 m/s)
2
ac 
 12.8 m/s
5m
mv
Fc  mac 
R
2
F = (3 kg)(12.8 m/s2)
Fc =
Car Negotiating a Flat Turn
v
Fc
R
What is the direction of the
force ON the car?
Ans. Toward Center
This central force is exerted
BY the road ON the car.
Car Negotiating a Flat Turn
v
Fc
R
Is there also an outward force
acting ON the car?
Ans. No, but the car does exert a
outward reaction force ON the road.
Car Negotiating a Flat Turn
The centripetal force Fc is
that of static friction fs:
m
Fc
R
n
fs
Fc = fs
R
v
mg
The central force FC and the friction force fs
are not two different forces that are equal.
There is just one force on the car. The nature
of this central force is static friction.
Finding the maximum speed for
negotiating a turn without slipping.
n
fs
Fc = fs
m
v
R
Fc
R
mg
The car is on the verge of slipping when FC is
equal to the maximum force of static friction fs.
Fc = fs
Fc =
mv2
R
fs = msmg
Maximum speed without slipping (Cont.)
n
Fc = fs
fs
R
mv2
R
mg
v=
m
v
Fc
R
= msmg
msgR
Velocity v is maximum
speed for no slipping.
Example 4: A car negotiates a turn of
radius 70 m when the coefficient of static
friction is 0.7. What is the maximum
speed to avoid slipping?
m
v
Fc
Fc =
R
ms = 0.7
mv2
R
fs = msmg
From which: v =
msgR
g = 9.8 m/s2; R = 70 m
v  ms gR  (0.7)(9.8)(70m)
v=
Optimum Banking Angle
By banking a curve at the
optimum angle, the normal
Fc
m
R
v
fs
w
force n can provide the
necessary centripetal force
without the need for a
friction force.
n
q
slow speed
n
w
fs
q
fast speed
fs = 0
w
n
q
optimum
Free-body Diagram
Acceleration a is toward the
center. Set x axis along the
direction of ac , i. e.,
horizontal (left to right).
n
x
mg
q
n
q
n cos q
q
n
n sin q
mg
q
mg
+ ac
Optimum Banking Angle (Cont.)
n
mg
n cos q
n
n sin q
q
Apply
Newton’s 2nd
Law to x and y
axes.
q
mg
mv2
SFx = mac
n sin q 
SFy = 0
n cos q = mg
R
Optimum Banking Angle (Cont.)
n
mg
q
n sin q 
n cos q
q
n
n sin q
n
sin q
tan q 
n cos q
mg
mv2
R
n cos q = mg
2
mv
2
v
R
tan q 

mg gR
1
Optimum Banking Angle (Cont.)
n
mg
q
Optimum Banking
Angle q
n cos q
n
q
n sin q
mg
2
v
tan q 
gR
Example 5: A car negotiates a turn of
radius 80 m. What is the optimum
banking angle for this curve if the speed
is to be equal to 12 m/s?
n
tan q =
mg
n cos q
q
q
n
n sin q
mg
v2
gR
=
(12 m/s)2
(9.8 m/s2)(80 m)
tan q = 0.184
q=
How might you 2find the
mv
centripetal
FC force on the
car, knowing R
its mass?
The Conical Pendulum
A conical pendulum consists of a mass m
revolving in a horizontal circle of radius R
at the end of a cord of length L.
T cos q
L q
T
R
T
q
h
mg
T sin q
Note: The inward component of tension
T sin q gives the needed central force.
Angle q and velocity v:
T cos q
L q
T
q
h
mg
R
Solve two
equations
to find
angle q
T
T sin q 
T sin q
mv2
R
T cos q = mg
tan q =
v2
gR
Example 6: A 2-kg mass swings in a
horizontal circle at the end of a cord of
length 10 m. What is the constant
speed of the mass if the rope makes an
angle of 300 with the vertical?
q  300
L q
T
R
h
1. Draw & label sketch.
2. Recall formula for pendulum.
2
v
tan q 
gR
Find: v = ?
3. To use this formula, we need to find R = ?
R = L sin 300 = (10 m)(0.5)
R=
Example 6(Cont.): Find v for q = 300
4. Use given info to find the
velocity at 300.
R=5m
Solve for
v=?
g = 9.8 m/s2
L q
T
R=5m
h
R
v2
tan q 
gR
v  gR tan q
2
q  300
v  gR tan q
v  (9.8 m/s 2 )(5 m) tan 300
v=
Swinging Seats at the Fair
This problem is identical
to the other examples
except for finding R.
b
L q
T
h
d
R=d+b
R
tan q =
R = L sin q + b
v2
gR
and
v=
gR tan q
Example 9. If b = 5 m and L = 10 m, what
will be the speed if the angle q = 260?
v2
tan q =
R=d+b
gR
L q b
d = (10 m) sin 260 = 4.38 m T
d
R = 4.38 m + 5 m = 9.38 m
R
v  gR tan q
2
v  gR tan q
v  (9.8 m/s 2 )(9.38 m) tan 260
v=
Motion in a Vertical Circle
v
Consider the forces on a
ball attached to a string as
it moves in a vertical loop.
+
v
Note also that the positive
direction is always along
acceleration, i.e., toward
the center of the circle.
v
+
v
Bottom
Tmg
+
T
T
T
mg
+
+mg
mg
T
v
Top
ofSide
Path
Left
mg
Top Right
Right
Top
Tension is
Weight
hasas
no
Maximum
minimum
Weight
causes
Weight
has
no
effect
on
T
tension
T, W
Bottom
weight
helps
small
decrease
effect on T
opposes
Fc
Ftension
in
c force T
Note changes as you click
the mouse to show new
positions.
+
v
10 N
T
As an exercise, assume
that a central force of
Fc = 40 N is required to
maintain circular motion
of a ball and W = 10 N.
+
R
T
+
10 N
v
The tension T must
adjust so that central
resultant is 40 N.
At top: 10 N + T = 40 N
T=
Bottom: T – 10 N = 40 N
T = __?___
Motion in a Vertical Circle
v
mg
T
mv2
Resultant force
Fc =
toward center
R
R
v
mg + T =
AT TOP:
+
mg
T
Consider TOP of circle:
T=
mv2
R
mv2
R
- mg
Vertical Circle; Mass at bottom
v
T
Resultant force
toward center
R
v
mg
T - mg =
mg
+
R
Consider bottom of circle:
AT Bottom:
T
Fc =
mv2
T=
mv2
R
mv2
R
+ mg
Summary
Centripetal
acceleration:
v=
msgR
Conical
pendulum:
2
v
ac  ;
R
mv
Fc  mac 
R
v2
tan q = gR
v=
gR tan q
2
Summary: Motion in Circle
v
AT TOP:
+
mg
R
v
2
mv
T=
- mg
R
T
AT BOTTOM:
T
mg
+
2
mv
T=
+ mg
R
Summary: Ferris Wheel
v
AT TOP:
n
R
v
+
mg
AT BOTTOM:
n
mg
+
n=
mg
- n=
n = mg -
mv2
R
+ mg
mv2
R
mv2
R
CONCLUSION: Chapter 10
Uniform Circular Motion