Uniform Circular Motion
Centripetal Force
Torque
El Paso Independent School District
Uniform Circular Motion
Uniform circular motion is motion along a
circular path in which there is no change in
speed, only a change in direction.
Fc
v
Constant velocity
V tangent to
path.
Constant force Fc
toward center.
Question: Is there an outward force on the ball?
Uniform Circular Motion (Cont.)
The question of an outward force can be
resolved by asking what happens when the
string breaks!
Ball moves tangent to
v
path, NOT outward as
might be expected.
When central force is removed,
ball continues in straight line.
Centripetal force is needed to change direction.
Examples of Centripetal Force
You are sitting on the seat next to
the outside door. What is the
direction of the resultant force on
you as you turn? Is it away from
center or toward center of the turn?
• Car going around a
curve.
Fc
Force ON you is toward the center.
Deriving Central Acceleration
Consider initial velocity at A and final velocity at B:
vf
vf B
R
vo
A
PAP
-vo Dv
R
s v
o
Deriving Acceleration (Cont.)
Dv
Definition: ac =
Similar
Triangles
ac =
PAP
Dv
t
=
Centripetal
acceleration:
Dv
vf
vf
Dv -vo
t
R
=
d v
o
mass m
v2
R
2
v
ac  ;
R
mv
Fc  mac 
R
2
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?
v
m
R
2
v
ac 
R
m = 3 kg
R = 5 m; v = 8 m/s
ac = 12.8 m/s2
mv
Fc  mac 
R
2
F = (3 kg)(12.8 m/s2)
Fc = 38.4 N
Example 2: A skater moves with 15 m/s in a circle of
radius 30 m. The ice exerts a central force of 450 N.
What is the mass of the skater?
Draw and label a sketch
v = 15 m/s
Fc R
450 N
Solve for mass(m)
• F = mv2/R
• FR = mv2
• FR/v2 = m, substitute
m=(450N)(30m)/(15m/s)2
30 m
m=?
Speed skater
m = 60.0 kg
Car Negotiating a Flat Turn
The centripetal force Fc is
that of static friction fs:
m
Fc
n
R
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.
PAP
PAP
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 = 21.9 m/s
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
Visual Aid: Assume that the centripetal force required to
maintain circular motion is 20 N. Further assume that the
weight is 5 N.
v
Resultant central force FC
at every point in path!
R
v
FC = 20 N at top
AND at bottom.
FC = 20 N
Weight vector W is
downward at every point.
W = 5 N, down
Visual Aid: The resultant force (20 N) is the vector sum
of T and W at ANY point in path.
W
T
+
T
W
+
v
Top: T + W = FC
T + 5 N = 20 N
T = 20 N - 5 N = 15 N
R
v
FC = 20 N at top
AND at bottom.
Bottom:
T - W = FC
T - 5 N = 20 N
T = 20 N + 5 N = 25 N
For Motion in Circle
v
AT TOP:
R
+ T=
mg
mv2
R
- mg
T
v
AT BOTTOM:
T
mg
+
T=
mv2
R
+ mg
Example 6: A 2-kg rock swings in a vertical circle of
radius 8 m. The speed of the rock as it passes its
highest point is 10 m/s. What is tension T in rope?
At Top:
v
mg
T
T=
R
v
mg + T =
mv2
mv2
- mg
R
R
2
(2 kg)(10 m/s)
2
T
 2 kg(9.8 m/s )
8m
T = 25 N - 19.6 N
T = 5.40 N
Example 6 (cont): A 2-kg rock swings in a vertical
circle of radius 8 m. The speed of the rock as it passes
its lowest point is 10 m/s. What is tension T in rope?
At Bottom:
v
R
T
mg
T=
v
T - mg =
mv2
mv2
+ mg
R
R
2
(2 kg)(10 m/s)
2
T
 2 kg(9.8 m/s )
8m
T = 25 N + 19.6 N
T = 44.6 N
Example 7: What is the critical speed vc at the top, if
the 2-kg mass is to continue in a circle of radius 8 m?
0
v
At Top:
mg
T
vc occurs when T = 0
R
v
v=
gR =
mg + T =
mg =
mv2
R
(9.8 m/s2)(8 m)
vc =
mv2
R
gR
vc = 8.85 m/s
Definition of Torque
Torque is defined as the tendency to
produce a change in rotational motion.
Examples:
Units for Torque
Torque is proportional to the magnitude of
F and to the distance r from the axis. Thus,
a tentative formula might be:
t = Fr
Units: Nm or lbft
t = (40 N)(0.60 m)
= 24.0 Nm, cw
t = 24.0 Nm, cw
6 cm
40 N
Torque is determined by Three Factors:
• The magnitude of the applied force.
• The direction of the applied force.
• The location of the applied force.
Each
The
40-N
of the
force
20-Nthe
The
forces
nearer
forces
produces
different
the
end ofhas
theatwice
wrench
torque
torque
asdue
does
to the
the
have
greater
torques.
direction
20-N force.
of force.
Magnitude
Locationofofof
force
force
Direction
Force
20 N q
2020
N
20NN
20
40NN
20 N
20 N
q
Direction of Torque
Torque is a vector quantity that has
direction as well as magnitude.
Turning the handle of a
screwdriver clockwise and
then counterclockwise will
advance the screw first
inward and then outward.
Sign Convention for Torque
By convention, counterclockwise torques are
positive and clockwise torques are negative.
Positive torque:
Counter-clockwise,
out of page
ccw
cw
Negative torque:
clockwise, into page
Calculating Torque
•
•
•
•
•
Read problem and draw a rough figure.
Extend line of action of the force.
Draw and label moment arm.
Calculate the moment arm if necessary.
Apply definition of torque:
t = Fr
Torque = force x moment arm
Example 1: An 80-N force acts at the end of
a 12-cm wrench as shown. Find the torque.
• Extend line of action, draw, calculate r.
r = 12 cm sin 600
= 10.4 cm
t = (80 N)(0.104 m) =
8.31 N m
Alternate: An 80-N force acts at the end of
a 12-cm wrench as shown. Find the torque.
positive
12 cm
Resolve 80-N force into components as shown.
Note from figure: rx = 0 and ry = 12 cm
t = (69.3 N)(0.12 m)
t = 8.31 N m as before
Universal Law of Gravitation
So because of Newton’s 3rd law every body in the
universe exerts a force of attraction on every other
body.
• This is Newton’s Universal Law of Gravitation
The force between two
objects, due to their masses,
is called the gravitational force
(Fg)- in this case it’s not Earth
specific and is not 9.81 m/s2).
GmM
Fg 
2
r
Universal Gravitational Constant6.67
2
m
x10-11 N  kg 2
Masses of the 2 objects
GmM
Fg 
2
r
Distance between
the objects
Example 1:What is the gravitational force
between the Earth and the Moon?
mEarth = M = 6.0 x 1024 kg
mMoon = m = 7.4 x 1022 kg
r = 3.8 x 108 m
2
m
G = 6.67 x 10-11 N  2
kg
Example 2: What is the gravitational force between the
Earth and Venus?
mEarth = M = 6.0 x 1024 kg
mVenus = m = 5.0 x 1024 kg
r = 3.8 x 1010 m
2
m
G = 6.67 x 10-11 N  2
kg
Answer:
F = 1.386x1018 N
There is also a way to determine the
gravitational field around one object:
GmM
Fg 
2
r
GmM

mg
 
r2
GmM
w
2
r
GmM
mg 
r2
GM
g 2
r
This is now
the gravitational
Field Strength(GFS)
Example 3: What is the Gravitational Field Strength in
Earth?
• Radius of the Earth – 6.37 x 106 m
Example 4: What is the Gravitational field strength on
the moon?
• Radius of the Moon – 1.7 x 106 m
What if the distance between the Earth and the
Moon was doubled?
What if the distance between the Earth and the
Moon was tripled?
What if the distance between the Earth and
the Moon was quadrupled?
Both the ULG and the GFS follow the
Inverse square law:
• ULG-If the distance between two objects is
doubled the gravitational attraction is (1/4) of
the original.
OR
• GFS-If we travel beyond the Earth by a
distance that is double it’s radius than we will
only feel a quarter of Earth’s gravitational pull
(9.81 m/s2/4 = 2.45 m/s2).
re = 6.4 x 106 m
1x
1 re =
2x
3x
¼ re=
1/9 re =
9.81 m/s2 2.24 m/s2
4x
1/16 re =
1.09 m/s2 .61 m/s2
Example 5: The gravitational attraction between the
Earth and Mars is 8.7 x 1016 N. The distance between
the two planets is 5.5 x 1010m. Earth has a mass of
6.0 x 1024 kg. What’s the mass of Mars?