Circular Motion
AIM: How is this even possible?????
But First…Facts about Circular Motion
• The distance around one lap of a circle is the
circumference (𝟐𝝅𝒓)
• What is ½ a lap?
• What is ¼ of a lap?
• The displacement for one lap around a circle is
zero.
• What is ½ a lap?
• What is ¼ of a lap?
• Time once around a circle is the Period (T)
• 𝑻=
𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 (𝒔𝒆𝒄𝒐𝒏𝒅𝒔)
𝑻𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒍𝒂𝒑𝒔
More Facts about Circular Motion
• Speed while moving in a circle
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒂𝒓𝒐𝒖𝒏𝒅 𝒐𝒏𝒄𝒆
𝟐𝝅𝒓
𝒗𝒄 =
=
𝒕𝒊𝒎𝒆 𝒕𝒐 𝒈𝒐 𝒂𝒓𝒐𝒖𝒏𝒅 𝒐𝒏𝒄𝒆
𝑻
• Acceleration while moving in a circle
𝒎𝒗𝟐
𝒂𝒄 =
𝒓
• Net force while moving in a circle
𝒎𝒗𝟐
𝑭𝒄 = 𝒎𝒂𝒄 =
𝒓
Even More Facts about Circular Motion
Velocity
(tangent to the circle at all times)
V
Fc
ac
Centripetal Acceleration
and
Centripetal Force
(always towards the center of the circle)
What Provides Centripetal Force?
• Centripetal force is a NET force which means it
is calculated
– either by Newton’s 2nd Law (Fc=mac)
– Or by drawing a free body diagram and
determining the net force from the orientation of
the forces on the diagram.
• Because it is a net force, you should use the
same steps you were using (hopefully) last
topic.
Friction as a Centripetal Force
(running, driving)
• On a level surface, friction IS the force that keeps the
object (car, motorcycle, person) moving in a circle.
– This is a STATIC friction force because the tires/shoes are not
skidding across the surface.
– Does the mass of the car affect its maximum turning speed?
Horizontal
JUSTIFY!
Vertical
𝐹𝑦 = 𝑚𝑎𝑦
FN
Ff
Fg
𝐹𝑁 − 𝐹𝑔 =0
𝐹𝑁 =𝑚𝑔
𝐹𝑥 = 𝑚𝑎𝑥
𝐹𝑓 = 𝑚𝑎𝑐
𝑚𝑣 2
𝜇𝑠 𝑚𝑔=
𝑟
𝑣𝑚𝑎𝑥 2
𝑟
= 𝜇𝑔
Examples
1. A 200Kg motorcycle with rubber tires wants to
complete a turn of radius 12m on a level dry asphalt
road as fast as possible.
a. What is the max static friction force between the road and
the tires?
b. What is the max speed the motorcycle can take the turn?
c. Compared to the speed of the motorcycle, what would be
the maximum speed a 400kg car could take the turn?
2. A 65Kg runner takes a turn of radius 15m at 7.5m/s.
a. What is the friction force required to make this turn?
b. What is the coefficient of friction between the shoe and the
track?
c. Is this a static or kinetic coefficient? WHY?
Friction on an Inclined Surface
Vertical
𝐹𝑦 = 𝑚𝑎𝑦
FN
Ff

𝐹𝑁 − 𝐹⊥ =0
𝐹𝑁 =𝐹𝑔 cos 𝜃
𝐹𝑁 =𝑚𝑔 cos 𝜃
Horizontal
𝐹𝑥 = 𝑚𝑎𝑥

𝐹𝑓 + 𝐹∥ = 𝑚𝑎𝑐
𝜇𝑠 𝐹𝑁 +
Fg
𝑚𝑣 2
𝐹𝑔 sin 𝜃=
𝑟
𝑚𝑣 2
𝜇𝑠 𝑚𝑔 cos 𝜃 + 𝑚𝑔 sin 𝜃=
𝑟
𝑣2
𝜇𝑠 𝑔 cos 𝜃 + 𝑔 sin 𝜃=
𝑟
Tension as the Centripetal Force
• An object moves in a horizontal circle on a
frictionless surface while attached to a string
of length l
𝐹 = 𝑚𝑎𝑥
T
Table top
𝑇 = 𝑚𝑎𝑐
𝑚𝑣 2
𝑇=
𝑟
𝑚𝑣 2
𝑇=
𝑙
Tension as the Centripetal Force
• An object moves in a vertical circle on a at a
constant speed while attached to a string of
length l
top
𝐹𝑦 = 𝑚𝑎𝑦
Fg
T
𝑇 + 𝐹𝑔 = 𝑚𝑎𝑐
𝑚𝑣 2
𝑇 + 𝑚𝑔 =
𝑟
𝑚𝑣 2
𝑇=
− 𝑚𝑔
𝑙
Tension as the Centripetal Force
• An object moves in a vertical circle on a at a
constant speed while attached to a string of
length l
bottom
𝐹𝑦 = 𝑚𝑎𝑦
T
Fg
𝑇 − 𝐹𝑔 = 𝑚𝑎𝑐
𝑚𝑣 2
𝑇 − 𝑚𝑔 =
𝑟
𝑚𝑣 2
𝑇=
+ 𝑚𝑔
𝑙
Tension as the Centripetal Force
• When an object is moving in a vertical circle, where in the
motion is the string most likely going to break? JUSTIFY!
𝑇𝑡𝑜𝑝
𝑚𝑣 2
=
− 𝑚𝑔
𝑙
𝑇𝑏𝑜𝑡𝑡𝑜𝑚
𝑚𝑣 2
=
+ 𝑚𝑔
𝑙
As seen in the equations above, the tension in the rope at the
top of the circle is the centripetal force minus the weight of the
object because at that point, gravity is providing some of the
centripetal force. At the bottom, the tension in the rope is the
centripetal force plus the weight because at that point, the
tension not only has to hold up the weight of the object, it must
also provide the centripetal force.
Tension as the Centripetal Force
• An object moves in a conical pendulum while
attached to a string of length l making an
angle  with the vertical
T

Ty
Tx
Fg
𝐹𝑦 = 𝑚𝑎𝑦
𝐹𝑥 = 𝑚𝑎𝑥
𝑇𝑦 − 𝐹𝑔 = 0
𝑇𝑐𝑜𝑠𝜃 = 𝑚𝑔
𝑇𝑥 = 𝑚𝑎𝑐
𝑚𝑣 2
𝑇𝑠𝑖𝑛𝜃 =
𝑟
𝑚𝑣 2
𝑇𝑠𝑖𝑛𝜃 =
𝑙𝑠𝑖𝑛𝜃
Normal Force as a Centripetal Force
(Gravitron Ride)
• When you are riding the gravitron, the centripetal force
is provided by the wall of the ride!
• Does the mass of the rider matter?
– PROVE IT MATHEMATICALLY!
Horizontal
Vertical
Ff
𝐹𝑥 = 𝑚𝑎𝑥
𝐹𝑦 = 𝑚𝑎𝑦
FN
Fg
𝐹𝑓 − 𝐹𝑔 = 0
𝜇𝑠 𝐹𝑁 =𝑚𝑔
𝐹𝑁 = 𝑚𝑎𝑐
𝑚𝑣 2
𝐹𝑁 =
𝑟
𝜇𝑠 𝑚𝑣 2
= 𝑚𝑔
𝑟
Examples
1. A 60Kg student rides a gravitron with a radius of 7m.
The ride completes 10 revolutions in 1 minute
a. What is the average speed of the rider?
b. What is the normal force provided by the wall?
c. What is the coefficient of friction between the rider and the
wall
2. An average person can tolerate an acceleration of 4g
before passing out. If you wanted to build a gravitron
ride with a radius of 12m
a. What is the maximum average speed you can set the ride
to?
b. What would have to be the coefficient of friction between
the wall and the person to make the ride successful?
Circular motion in a Vertical Loop
• At the TOP of the loop
𝐹𝑦 = 𝑚𝑎𝑦
𝐹𝑔 + 𝐹𝑁 = 𝑚𝑎𝑐
𝑚𝑣 2
𝑚𝑔 + 𝐹𝑁 =
𝑟
2
𝑚𝑣
𝐹𝑁 =
− 𝑚𝑔
𝑟
If you ‘just make it’ FN=0N
FN Fg
Circular motion in a Vertical Loop
• At the BOTTOM of the loop
𝐹𝑦 = 𝑚𝑎𝑦
𝐹𝑁 − 𝐹𝑔 = 𝑚𝑎𝑐
𝑚𝑣 2
𝐹𝑁 − 𝑚𝑔 =
𝑟
2
𝑚𝑣
𝐹𝑁 =
+ 𝑚𝑔
𝑟
FN
Fg
Circular motion in a Vertical Loop
• Where in the loop do you
feel the heaviest? JUSTIFY!
Top
𝑚𝑣 2
𝐹𝑁 =
− 𝑚𝑔
𝑟
Bottom
𝑚𝑣 2
𝐹𝑁 =
+ 𝑚𝑔
𝑟
You feel heaviest at the bottom
of the loop because your
‘apparent weight’ is the normal
force. At the bottom the normal
force has to counteract the
weight and provide the
centripetal force. At the top,
some of the centripetal force is
provided by gravity.
Examples
1. A 60Kg student rides a rollercoaster with a vertical
loop with a radius of 15m.
a. What is the minimum speed needed to complete the
vertical loop?
b. If the actual speed of the ride at the top is 20m/s, how
heavy (FN) does the rider feel at the top of the loop?
c. If the ride’s speed is 31m/s at the bottom, how heavy
do they feel?
2. A 3kg set of keys attached to a 0.75m long string in
being swung in a vertical loop.
a. Assuming the maximum tension the string can provide
is 140N, how fast can the keys be swung without breaking
the string?
Circular motion on a Hill
• At the TOP of the hill
𝐹𝑦 = 𝑚𝑎𝑦
𝐹𝑔 − 𝐹𝑁 = 𝑚𝑎𝑐
𝑚𝑣 2
𝑚𝑔−𝐹𝑁 =
𝑟
2
𝑚𝑣
𝐹𝑁 = 𝑚𝑔 −
𝑟
FN
Fg
If you ‘lose contact with the
seat’ FN=0N
Examples
1. A 50Kg student rides a rollercoaster with a circular
hill that has a radius of 20m
a. What is the maximum speed the coaster can take the
hill before the student loses contact with the seat?
2. If you want a 300kg motorcycle to be able to stay
in contact while driving over a circular hill at
40m/s. What is the radius of the hill?
Gravity as a Centripetal Force
Newton’s Law of Gravity
𝐺𝑚1 𝑚2
𝐹𝑔 =
𝑟2
• Fg Force of gravitational attraction
between any two objects.
• G Universal Gravitational Constant=6.67x10-11
Nm2/kg2
• m the mass of the objects (kg)
• r the distance between the CENTERS of the
objects.
Acceleration Due to Gravity
𝐹𝑔𝑜𝑛 𝑒𝑎𝑟𝑡ℎ = 𝐹𝑔 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙
𝑚𝑝𝑒𝑟𝑠𝑜𝑛 𝑔 =
𝑔=
𝐺𝑚𝑝𝑒𝑟𝑠𝑜𝑛 𝑀𝐸𝑎𝑟𝑡ℎ
𝑅𝐸𝑎𝑟𝑡ℎ 2
𝐺𝑀𝐸𝑎𝑟𝑡ℎ
𝑅𝐸𝑎𝑟𝑡ℎ 2
General g
𝑔=
𝐺𝑀𝑝𝑙𝑎𝑛𝑒𝑡
𝑅𝑝𝑙𝑎𝑛𝑒𝑡 2
Orbital Speed
𝐹 = 𝑚𝑎𝑐
𝑚𝑝𝑒𝑟𝑠𝑜𝑛 𝑣 2
𝐹𝑔 =
𝑟
𝐺𝑚𝑝𝑒𝑟𝑠𝑜𝑛 𝑀𝐸𝑎𝑟𝑡ℎ
𝑟2
=
𝑚𝑝𝑒𝑟𝑠𝑜𝑛 𝑣 2
𝑟
𝐺𝑀𝐸𝑎𝑟𝑡ℎ 𝑣 2
=
𝑟
1
𝑣𝑜𝑟𝑏𝑖𝑡 =
𝐺𝑀𝑒𝑎𝑟𝑡ℎ
𝑟
General Orbital Speed-> Kepler’s 3rd
Laws
𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑
𝑟𝑜𝑟𝑏𝑖𝑡
𝑣𝑜𝑟𝑏𝑖𝑡 =
2𝜋𝑟𝑜𝑟𝑏𝑖𝑡
=
𝑇
2𝜋𝑟𝑜𝑟𝑏𝑖𝑡
𝑇
4𝜋 2 𝑟𝑜𝑟𝑏𝑖𝑡 2 𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑
=
2
𝑇
𝑟𝑜𝑟𝑏𝑖𝑡
2
𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑
𝑟𝑜𝑟𝑏𝑖𝑡
𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑
=
𝑟𝑜𝑟𝑏𝑖𝑡
𝑟𝑜𝑟𝑏𝑖𝑡 3 𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑
=
2
𝑇
4𝜋 2
Kepler’s Laws of Planetary Motion
1. The law of ellipses: all planets orbit the sun in
elliptical paths with the sun at one focus.
2. The law of equal areas: As a planet orbits the sun,
it sweeps out equal areas in equal amounts of
time. This means the closer the planet is to the
sun, the faster it is moving.
3. The law of harmonies: the orbits of all the
planets’ orbits are related
3
𝑟𝑜𝑟𝑏𝑖𝑡
𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑
=
𝑇2
4𝜋 2
𝒓𝑨
𝟑
𝑻𝑨
𝟐
=
𝒓𝑩
𝟑
𝑻𝑩
𝟐
Questions
• How could have scientists figure out the mass of the
Earth?
𝑮𝑴𝒑𝒍𝒂𝒏𝒆𝒕
𝒈=
𝟐
𝑹𝒑𝒍𝒂𝒏𝒆𝒕
• How could have scientists figure out the mass of the
sun?
𝑮𝑴𝒐𝒓𝒃𝒊𝒕𝒆𝒅
𝒗𝒐𝒓𝒃𝒊𝒕 =
𝒓𝒐𝒓𝒃𝒊𝒕
• How could have scientists figure out the distances
to all the other planets in our solar system?
𝒓𝑨
𝟑
𝑻
𝟐
=
𝒓𝑩
𝟑
𝑻
𝟐