Download What causes UCM?

What causes UCM?
What causes UCM?

UCM involves circular motion but at a constant
speed.
C
What causes UCM?

UCM involves circular motion but at a constant
speed. What is the direction of the instantaneous
acceleration at points A, B and D ?
A
B
C
D
What causes UCM?

UCM involves circular motion but at a constant
speed. The direction of the instantaneous acceleration at
points A, B and D is “toward the centre”.
A
a
B
a
C
a
D
What causes UCM?

UCM involves circular motion but at a constant
speed. The direction of the instantaneous acceleration at
points A, B and D is “toward the centre”.
What is the direction of the
net or unbalanced force at
points A, B and D and why?
A
a
B
a
C
a
D
What causes UCM?

UCM involves circular motion but at a constant
speed. The direction of the instantaneous acceleration at
points A, B and D is “toward the centre”.
The net force is also
“towards the centre” because
Newton's second law says
that a and Fnet are always in
the same direction. a=Fnet /m
A
Fnet
a
Fnet
B
a
C
Fnet
a
D
What causes UCM?

UCM involves circular motion but at a constant
speed. The direction of the instantaneous acceleration at
points A, B and D is “toward the centre”.
The net force is also
“towards the centre” because
Newton's second law says
that a and Fnet are always in
the same direction. a=Fnet /m
The net force that keeps a body in
UCM is always directed toward the
centre and is therefore called a
_____________ force.
A
Fnet
a
Fnet
B
a
C
Fnet
a
D
What causes UCM?

UCM involves circular motion but at a constant
speed. The direction of the instantaneous acceleration at
points A, B and D is “toward the centre”.
The net force is also
“towards the centre” because
Newton's second law says
that a and Fnet are always in
the same direction. a=Fnet /m
The net force that keeps a body in
UCM is always directed toward the
centre and is therefore called a
centripetal force.
A
Fnet
a
Fnet
B
a
C
Fnet
a
D
Centripetal Force and Velocity
Centripetal Force and Velocity

What is the direction of the instantaneous velocity at points A, B
and D ?
A
Fnet
a
Fnet
B
a
C
up
Fnet
a
D
right
Centripetal Force and Velocity

The directions of the instantaneous velocities at points A, B and
D are given by the tangent-lines to the circle at those points.
v
A
Fnet
a
Fnet
B
a
C
up
Fnet
a
D
right
v
v
Centripetal Force and Velocity

The directions of the instantaneous velocities at points A, B and
D are given by the tangent-lines to the circle at those points.
v
A
The “net” or “centripetal”
force for UCM is always
90° to the velocity. What
effect does this “net” or
“centripetal” force have
on the object's velocity?
Fnet
a
Fnet
B
a
C
up
Fnet
a
D
right
v
v
Centripetal Force and Velocity

The directions of the instantaneous velocities at points A, B and
D are given by the tangent-lines to the circle at those points.
v
A
The “net” or “centripetal”
force for UCM is always
90° to the velocity. This
centripetal force only
changes the direction of
motion, but not its speed.
Fnet
a
Fnet
B
a
C
Fnet
up
a
D
right
v
v
Is centripetal force a “Real” Force?
Is centripetal force a “Real” Force?

Centripetal Force is not one of the four “real”
forces of nature.
Is centripetal force a “Real” Force?

Centripetal Force is not one of the four “real”
forces of nature. It is just a special ______
force that keeps an object moving in uniform
circular motion. (UCM)
Is centripetal force a “Real” Force?

Centripetal Force is not one of the four “real”
forces of nature. It is just a special net or
unbalanced force that keeps an object moving
in uniform circular motion. (UCM)
Is centripetal force a “Real” Force?


Centripetal Force is not one of the four “real”
forces of nature. It is just a special net or
unbalanced force that keeps an object moving
in uniform circular motion. (UCM)
For UCM, the centripetal force must be
supplied by one or more of the four real
forces of nature.
What real force supplies the
centripetal force for these cases?

Case #1: An electron is revolving in UCM
around a proton in a hydrogen atom. (simple
planetary model)
?
= Fc
+
Fnet or FC =
?
What real force supplies the
centripetal force for these cases?

Case #1: An electron is revolving in UCM
around a proton in a hydrogen atom. (simple
planetary model)
Fel
= Fc
The electrical force of
the proton supplies
the net or centripetal
force on the electron.
+
Fnet or FC = Fel
What real force supplies the
centripetal force for these cases?
Case #1: An electron is revolving in UCM
around a proton in a hydrogen atom. (simple
planetary model)
Fel
= Fc
The electrical force of
+
the proton supplies
Fnet or FC = Fel
the net or centripetal
force on the electron.

This is a vector statement and a good starting
point in centripetal force problems
What real force supplies the
centripetal force for these cases?

Case #2: The moon revolves around the earth
in approximate UCM every 27.3 days.
?
= Fc
Moon
Earth
Fnet or FC =
?
What real force supplies the
centripetal force for these cases?

Case #2: The moon revolves around the earth
in approximate UCM every 27.3 days.
Fg
= Fc
The gravitational force
of the earth supplies
the net or centripetal
force on the moon.
Moon
Earth
Fnet or FC = Fg
What real force supplies the
centripetal force for these cases?

Case #2: The moon revolves around the earth
in approximate UCM every 27.3 days.
Fg
= Fc
The gravitational force
of the earth supplies
the net or centripetal
force on the moon.
Moon
Earth
Fnet or FC = Fg
This is a vector statement and a good starting
point in centripetal force problems
What real force supplies the
centripetal force for these cases?

Case #3: A car rounds a unbanked flat circular
curve at a constant speed. (See side view)
car
road
Fnet or FC =
?
Flat road
surface
?
= Fc
Centre of
curve
What real force supplies the
centripetal force for these cases?

Case #3: A car rounds a unbanked flat circular
curve at a constant speed. (See side view)
car
road
Fnet or FC = fs
Flat road
surface
Centre of
curve
fs = Fc
The static friction force of the
road supplies the net or
centripetal force on the car.
Centripetal Force Formulas
Centripetal Force Formulas

Review☺: The centripetal force is just a special
_______ force to keep something in UCM.
Centripetal Force Formulas

Review☺: The centripetal force is just a special
net force to keep something in UCM.
Centripetal Force Formulas


Review☺: The centripetal force is just a special
net force to keep something in UCM.
In terms of mass and acceleration, what is Fnet
?
Centripetal Force Formulas


Review☺: The centripetal force is just a special
net force to keep something in UCM.
│Fnet │= m │ ac │
Centripetal Force Formulas



Review☺: The centripetal force is just a special
net force to keep something in UCM.
│Fnet │= m │ ac │
Fc = m ( ? )
in terms of v and r ??
Centripetal Force Formulas



Review☺: The centripetal force is just a special
net force to keep something in UCM.
│Fnet │= m │ ac │
Fc = m v2/r
Equation #1
Centripetal Force Formulas


Review☺: The centripetal force is just a special
net force to keep something in UCM.
│Fnet │= m │ ac │

Fc = m v2/r

Fc = m ( ? )
Equation #1
in terms of T and r ??
Centripetal Force Formulas


Review☺: The centripetal force is just a special
net force to keep something in UCM.
│Fnet │= m │ ac │

Fc = m v2/r

Fc = m 4 π2 r / T2
Equation #1
Equation #2
Centripetal Force Formulas


Review☺: The centripetal force is just a special
net force to keep something in UCM.
│Fnet │= m │ ac │

Fc = m v2/r

Fc = m 4 π2 r / T2

Fc = m ( ? )
Equation #1
Equation #2
in terms of f and r ??
Centripetal Force Formulas


Review☺: The centripetal force is just a special
net force to keep something in UCM.
│Fnet │= m │ ac │

Fc = m v2/r

Fc = m 4 π2 r / T2

Fc = m 4 π2 r f2
Equation #1
Equation #2
Equation #3
Centripetal Force Formulas


Review☺: The centripetal force is just a special
net force to keep something in UCM.
│Fnet │= m │ ac │

Fc = m v2/r

Fc = m 4 π2 r / T2

Fc = m 4 π2 r f2
Equation #1
Memorize please!
Equation #2
Equation #3
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: ?
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: ? = 1525 kg
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
? = 72.00 km/h
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = ?
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
? = 300.0 m
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: ?
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?

Formula: ?
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?

Formula:
Fc = ?
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?

Formula:
Fc = mv2/r
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?

Formula:

Sub: ?
Fc = mv2/r
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?

Formula:

Sub: Fc = (1525)(20.0)2/(300.0) = ?
Fc = mv2/r
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?

Formula:

Sub: Fc = (1525)(20.0)2/(300.0) = 2033 N
Fc = mv2/r
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?

Formula:

Sub: Fc = (1525)(20.0)2/(300.0) = 2033 N
b) ?
Fc = mv2/r
Ex #1: A 1525 kg car is moving around a circular track of
radius 300.0 m at a constant speed of 72.00 km/h. Find:
a) the magnitude of the centripetal force required b) the
real force that supplies the centripetal force on the car

Given: m = 1525 kg
v = 72.00 km/h = 20.0 m/s
r = 300.0 m

Unknown: Fc = ?

Formula:

Sub: Fc = (1525)(20.0)2/(300.0) = 2033 N
Fc = mv2/r
b) The static friction of the road on the car
supplies the centripetal force and is
2033 N.
Try!: The earth has a mass of 6.0 x 1024 kg. It goes around the
sun in approximate UCM with an orbital radius of 1.5 X 1011 m.
The orbital period is 365.25 days. Find:
a) the size of the centripetal force b) the real force that
supplies Fc
c) the mass of the sun ( G = 6.67 X 10-11 mks units )
Try!: The earth has a mass of 6.0 x 1024 kg. It goes around the
sun in approximate UCM with an orbital radius of 1.5 X 1011 m.
The orbital period is 365.25 days. Find: a) the size of the
centripetal force
b) the real force that supplies Fc
c) the mass of the sun ( G = 6.67 X 10-11 in mks units )

Given: m = 6.0 x 1024 kg r = 1.5 X 1011 m
T = 365.25 d X 24 h/d X 3600 s/ h =3.2 X 107 seconds

Unknown: Fc = ? Formula: Fc = m 4 π2 r / T2
Fc = (6.0 x 1024 kg) 4 π2 (1.5 X 1011 m) / (3.2 X 107 s)2 = 3.5 X 1022 N
The sun exerts a gravitational force of 3.5 X 1022 N on the earth and
supplies the centripetal force that keeps the earth in UCM.

Fg = GMm/r2 so M = Fgr2/Gm
= ( 3.5 X 1022 ) (1.5 X 1011 )2 / (6.0 x 1024 X 6.67 X 10-11) mks

Fg = 2.0 X 1030 kg
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s

What real forces supply the centripetal force to keep
the mass in UCM?
m = 5.00 kg
String of
r = 1.60 m
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s

The tension force of the string and the earth's force of
gravity provide the centripetal force to keep the mass
m = 5.00 kg
in UCM.
String of
r = 1.60 m
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s

The tension force of the string and the earth's force of
gravity provide the centripetal force to keep the mass
m = 5.00 kg
in UCM.
String of
r = 1.60 m
a) Find the magnitude of the
centripetal force needed to
keep the mass in UCM.
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s

The tension force of the string and the earth's force of
gravity provide the centripetal force to keep the mass
m = 5.00 kg
in UCM.
String of
r = 1.60 m
a) Find the magnitude of the
centripetal force needed to
keep the mass in UCM.
v= 6.00 m/s

Fc = mv2/r = 5(6)2 /1.6
= 113 N
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: ?
m = 5.00 kg
String of
r = 1.60 m
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = ?
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50.0 N
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
Step 2: ?
T
Fg = 50 N
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50 N
Step 2: Write a vector statement of the real
v= 6.00 m/s
forces supplying the centripetal force.
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50 N
Step 2: Write a vector statement of the real
v= 6.00 m/s
forces supplying the centripetal force.
T + F g = Fc
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50 N
Step 2: Write a vector statement of the real
v= 6.00 m/s
forces supplying the centripetal force.
Y+ up
T + F g = Fc
?
Step 3 Scalar statement
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50 N
Step 2: Write a vector statement of the real
v= 6.00 m/s
forces supplying the centripetal force.
Y+ up
T + F g = Fc
T - 50 = - 113
Step 3 Scalar statement
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50 N
Step 2: Write a vector statement of the real
v= 6.00 m/s
forces supplying the centripetal force.
Y+ up
T + F g = Fc
T - 50 = - 113
T=?
Step 3 Scalar statement
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50 N
Step 2: Write a vector statement of the real
v= 6.00 m/s
forces supplying the centripetal force.
Y+ up
T + F g = Fc
T - 50 = - 113
T = 50 – 113 = ?
Step 3 Scalar statement
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50 N
Step 2: Write a vector statement of the real
v= 6.00 m/s
forces supplying the centripetal force.
Y+ up
T + F g = Fc
T - 50 = - 113
Step 3 Scalar statement
T = 50 – 113 = - 63.0 N
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
b) Find the tension force of the string at the top of the circle.
Step 1: Draw an FBD at the top of the circle.
m = 5.00 kg
String of
r = 1.60 m
a
T
Fg = 50 N
Step 2: Write a vector statement of the real
v= 6.00 m/s
forces supplying the centripetal force.
Y+ up
T + F g = Fc
T - 50 = - 113
Step 3 Scalar statement
X + right
T = 50 – 113 = - 63.0 N At the top of the circle, the tension force of the
string on the mass is 63.0 N [down]
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = ?
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = 50 N
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = 50 N
v= 6.00 m/s
Vector Statement
?
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
T
a
String of
r = 1.60 m
Fg = 50 N
v= 6.00 m/s
Vector Statement
T + Fg = Fc
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = 50 N
v= 6.00 m/s
Vector Statement
Scalar statement
T + Fg = Fc
?
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = 50 N
v= 6.00 m/s
Vector Statement
T + Fg = Fc
Scalar statement T – 50 = 113
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = 50 N
v= 6.00 m/s
Vector Statement
T + Fg = Fc
Scalar statement T – 50 = 113
T= ?
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = 50 N
v= 6.00 m/s
Vector Statement
T + Fg = Fc
Scalar statement T – 50 = 113
T = 50 + 113
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = 50 N
Vector Statement
T + Fg = Fc
Scalar statement T – 50 = 113
T = 50 + 113
T = 163 N
v= 6.00 m/s
Y+ up
X + right
UCM Dynamics in a Vertical Plane: Using a string, a
mass of 5.00 kg is twirled in a circle of radius 1.60 m on a
vertical plane with a constant speed of 6.00 m/s
c) Find the tension force of the string at the bottom of the circle.
m = 5.00 kg
String of
r = 1.60 m
T
a
Fg = 50 N
Vector Statement
T + Fg = Fc
Scalar statement T – 50 = 113
T = 50 + 113
T = 163 N
At the bottom of the circle, the
tension force of the string on the
mass is 163 N [up]
v= 6.00 m/s
Y+ up
X + right
Try this yourself: Using a string, a mass of 8.00 kg is
twirled in a circle of radius 0.900 m on a vertical plane
with a constant speed of 4.00 m/s
a) Find the magnitude of the centripetal force required for UCM
b) Find the tension force of the string...
i) at the bottom of the circle
m = 8.00 kg
String of
r = 0.900 m
ii) at the top of the circle
v= 4.00 m/s
Y+ up
X + right
Try this yourself: Using a string, a mass of 8.00 kg is
twirled in a circle of radius 0.900 m on a vertical plane
with a constant speed of 4.00 m/s
a) Find the magnitude of the centripetal force required for UCM
b) Find the tension force of the string...
i) at the bottom of the circle
m = 8.00 kg
String of
r = 0.900 m
ii) at the top of the circle
a) Fc = mv2/r = 8(4)2/0.9 = 142 N
v= 4.00 m/s
Y+ up
X + right
Try this yourself: Using a string, a mass of 8.00 kg is
twirled in a circle of radius 0.900 m on a vertical plane
with a constant speed of 4.00 m/s
a) Find the magnitude of the centripetal force required for UCM
b) Find the tension force of the string...
m = 8.00 kg
i) at the bottom of the circle
String of
r = 0.900 m
ii) at the top of the circle
a) Fc = mv2/r = 8(4)2/0.9 = 142 N
b) part i) T + Fg= Fc
T – 80 = 142
T = 222 N [up]
T
a
v= 4.00 m/s
Fg= 80 N
Y+ up
c) part ii) T + Fg= Fc
T – 80 = -142
T = - 62 = 62 N [d]
T
Fg= 80 N
a
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
m = ? kg
String of
r = 0.900 m
v= ?
Y+ up
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top
of the circle. It is at that
point that tension has it
minimum value. When
moving at minimum speed
to go around in UCM, the
tension force will be zero
at the top of the circle.
m = ? kg
String of
r = 0.900 m
v= ?
Y+ up
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
m = ? kg
String of
r = 0.900 m
FBD = ?
v= ?
Y+ up
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = ?
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
Vector Statement
m = ? kg
String of
r = 0.900 m
?
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
m = ? kg
String of
r = 0.900 m
Fg = FC
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
Fg = FC
Scalar statement = ?
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
Fg = FC
-mg = -mv2/r
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
Fg = FC
-mg = -mv2/r
v=?
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
Fg = FC
-mg = -mv2/r
v = ( r g )1/2
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
Fg = FC
-mg = -mv2/r
v = ( r g )1/2
v = ( 0.900 X 10 )1/2
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
Fg = FC
-mg = -mv2/r
v = ( r g )1/2
v = ( 0.900 X 10 )1/2
v = 3.00 m/s
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = mg
X + right
Minimum speed example: Using a string, an unknown
mass is twirled in a circle of radius 0.900 m on a vertical
plane with just enough speed to go around in UCM. Find
the minimum speed required.
Strategy: Focus at the top of the
circle. It is at that point that
tension has it minimum value.
When moving at minimum
speed to go around in UCM, the
tension force will be zero at the
top of the circle.
Fg = FC
-mg = -mv2/r
v = ( r g )1/2
v = ( 0.900 X 10 )1/2
v = 3.00 m/s
m = ? kg
String of
r = 0.900 m
a
v= ?
Y+ up
Fg = mg
X + right
The minimum speed for the mass to undergo UCM is 3.00 m/s. Note the formula only
depends on the radius and not the mass.
Critical Thinking Problems

Critical Thinking: will take-up in class
1. An airplane is moving in “loop the loop” in a vertical circle with radius
500.0 m and a constant speed of 540.0 km/h. At the bottom of the
loop, the 80.0 kg pilot has an apparent weight not equal to her actual
weight of 800.0 N. What is her apparent weight? (Ans 4400 N or 5.5
times her actual weight)
2. Using a string, a mass “ m “ kg is twirled in a circle of radius “ r “ m
on a vertical plane with a constant speed of “ v ” m/s. As it twirls, the
string makes an angle Θ with a vertical line from the center to the top
of the circle. Derive or create a formula to find the tension force in
terms of m, g , v and Θ. We will take up in class.
3. The sun has a mass of 2.0 X 1030 kg. The earth orbits the sun in
approximate UCM with an orbital radius of 1.5 X 1011 m. Find the
orbital speed of the earth. G = 6.67 x 10-11 Ans 30 km/s
More practice next slide.
Centripetal Force Practice




Read over your notes on this lesson or review
the power point presentation on Moodle.
Do the centripetal force handouts.
New textbook: p123 Q1,Q3,Q4 ans same
page p124 Q1,Q2, Q4,Q6 ans p715
Old textbook: p 133 Q3,Q4, Q6, Q7, Q8 ans
same page p 138 Q2-Q7 ans p782