Circular Motion

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Transcript Circular Motion

Circular Motion
AIM: How is this even possible?????
But Firstโ€ฆFacts about Circular Motion
โ€ข The distance around one lap of a circle is the
circumference (๐Ÿ๐…๐’“)
โ€ข What is ½ a lap?
โ€ข What is ¼ of a lap?
โ€ข The displacement for one lap around a circle is
zero.
โ€ข What is ½ a lap?
โ€ข What is ¼ of a lap?
โ€ข Time once around a circle is the Period (T)
โ€ข ๐‘ป=
๐’•๐’๐’•๐’‚๐’ ๐’•๐’Š๐’Ž๐’† (๐’”๐’†๐’„๐’๐’๐’…๐’”)
๐‘ป๐’๐’•๐’‚๐’ ๐’๐’–๐’Ž๐’ƒ๐’†๐’“ ๐’๐’‡ ๐’๐’‚๐’‘๐’”
More Facts about Circular Motion
โ€ข Speed while moving in a circle
๐’…๐’Š๐’”๐’•๐’‚๐’๐’„๐’† ๐’‚๐’“๐’๐’–๐’๐’… ๐’๐’๐’„๐’†
๐Ÿ๐…๐’“
๐’—๐’„ =
=
๐’•๐’Š๐’Ž๐’† ๐’•๐’ ๐’ˆ๐’ ๐’‚๐’“๐’๐’–๐’๐’… ๐’๐’๐’„๐’†
๐‘ป
โ€ข Acceleration while moving in a circle
๐’Ž๐’—๐Ÿ
๐’‚๐’„ =
๐’“
โ€ข Net force while moving in a circle
๐’Ž๐’—๐Ÿ
๐‘ญ๐’„ = ๐’Ž๐’‚๐’„ =
๐’“
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?
โ€ข Centripetal force is a NET force which means it
is calculated
โ€“ either by Newtonโ€™s 2nd Law (Fc=mac)
โ€“ Or by drawing a free body diagram and
determining the net force from the orientation of
the forces on the diagram.
โ€ข 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)
โ€ข On a level surface, friction IS the force that keeps the
object (car, motorcycle, person) moving in a circle.
โ€“ This is a STATIC friction force because the tires/shoes are not
skidding across the surface.
โ€“ Does the mass of the car affect its maximum turning speed?
Horizontal
JUSTIFY!
Vertical
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
FN
Ff
Fg
๐น๐‘ โˆ’ ๐น๐‘” =0
๐น๐‘ =๐‘š๐‘”
๐น๐‘ฅ = ๐‘š๐‘Ž๐‘ฅ
๐น๐‘“ = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐œ‡๐‘  ๐‘š๐‘”=
๐‘Ÿ
๐‘ฃ๐‘š๐‘Ž๐‘ฅ 2
๐‘Ÿ
= ๐œ‡๐‘”
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
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
FN
Ff
๏ฑ
๐น๐‘ โˆ’ ๐นโŠฅ =0
๐น๐‘ =๐น๐‘” cos ๐œƒ
๐น๐‘ =๐‘š๐‘” cos ๐œƒ
Horizontal
๐น๐‘ฅ = ๐‘š๐‘Ž๐‘ฅ
๏ฑ
๐น๐‘“ + ๐นโˆฅ = ๐‘š๐‘Ž๐‘
๐œ‡๐‘  ๐น๐‘ +
Fg
๐‘š๐‘ฃ 2
๐น๐‘” sin ๐œƒ=
๐‘Ÿ
๐‘š๐‘ฃ 2
๐œ‡๐‘  ๐‘š๐‘” cos ๐œƒ + ๐‘š๐‘” sin ๐œƒ=
๐‘Ÿ
๐‘ฃ2
๐œ‡๐‘  ๐‘” cos ๐œƒ + ๐‘” sin ๐œƒ=
๐‘Ÿ
Tension as the Centripetal Force
โ€ข An object moves in a horizontal circle on a
frictionless surface while attached to a string
of length l
๐น = ๐‘š๐‘Ž๐‘ฅ
T
Table top
๐‘‡ = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐‘‡=
๐‘Ÿ
๐‘š๐‘ฃ 2
๐‘‡=
๐‘™
Tension as the Centripetal Force
โ€ข An object moves in a vertical circle on a at a
constant speed while attached to a string of
length l
top
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
Fg
T
๐‘‡ + ๐น๐‘” = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐‘‡ + ๐‘š๐‘” =
๐‘Ÿ
๐‘š๐‘ฃ 2
๐‘‡=
โˆ’ ๐‘š๐‘”
๐‘™
Tension as the Centripetal Force
โ€ข An object moves in a vertical circle on a at a
constant speed while attached to a string of
length l
bottom
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
T
Fg
๐‘‡ โˆ’ ๐น๐‘” = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐‘‡ โˆ’ ๐‘š๐‘” =
๐‘Ÿ
๐‘š๐‘ฃ 2
๐‘‡=
+ ๐‘š๐‘”
๐‘™
Tension as the Centripetal Force
โ€ข When an object is moving in a vertical circle, where in the
motion is the string most likely going to break? JUSTIFY!
๐‘‡๐‘ก๐‘œ๐‘
๐‘š๐‘ฃ 2
=
โˆ’ ๐‘š๐‘”
๐‘™
๐‘‡๐‘๐‘œ๐‘ก๐‘ก๐‘œ๐‘š
๐‘š๐‘ฃ 2
=
+ ๐‘š๐‘”
๐‘™
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
โ€ข An object moves in a conical pendulum while
attached to a string of length l making an
angle ๏ฑ with the vertical
T
๏ฑ
Ty
Tx
Fg
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
๐น๐‘ฅ = ๐‘š๐‘Ž๐‘ฅ
๐‘‡๐‘ฆ โˆ’ ๐น๐‘” = 0
๐‘‡๐‘๐‘œ๐‘ ๐œƒ = ๐‘š๐‘”
๐‘‡๐‘ฅ = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐‘‡๐‘ ๐‘–๐‘›๐œƒ =
๐‘Ÿ
๐‘š๐‘ฃ 2
๐‘‡๐‘ ๐‘–๐‘›๐œƒ =
๐‘™๐‘ ๐‘–๐‘›๐œƒ
Normal Force as a Centripetal Force
(Gravitron Ride)
โ€ข When you are riding the gravitron, the centripetal force
is provided by the wall of the ride!
โ€ข Does the mass of the rider matter?
โ€“ PROVE IT MATHEMATICALLY!
Horizontal
Vertical
Ff
๐น๐‘ฅ = ๐‘š๐‘Ž๐‘ฅ
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
FN
Fg
๐น๐‘“ โˆ’ ๐น๐‘” = 0
๐œ‡๐‘  ๐น๐‘ =๐‘š๐‘”
๐น๐‘ = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐น๐‘ =
๐‘Ÿ
๐œ‡๐‘  ๐‘š๐‘ฃ 2
= ๐‘š๐‘”
๐‘Ÿ
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
โ€ข At the TOP of the loop
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
๐น๐‘” + ๐น๐‘ = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐‘š๐‘” + ๐น๐‘ =
๐‘Ÿ
2
๐‘š๐‘ฃ
๐น๐‘ =
โˆ’ ๐‘š๐‘”
๐‘Ÿ
If you โ€˜just make itโ€™ FN=0N
FN Fg
Circular motion in a Vertical Loop
โ€ข At the BOTTOM of the loop
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
๐น๐‘ โˆ’ ๐น๐‘” = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐น๐‘ โˆ’ ๐‘š๐‘” =
๐‘Ÿ
2
๐‘š๐‘ฃ
๐น๐‘ =
+ ๐‘š๐‘”
๐‘Ÿ
FN
Fg
Circular motion in a Vertical Loop
โ€ข Where in the loop do you
feel the heaviest? JUSTIFY!
Top
๐‘š๐‘ฃ 2
๐น๐‘ =
โˆ’ ๐‘š๐‘”
๐‘Ÿ
Bottom
๐‘š๐‘ฃ 2
๐น๐‘ =
+ ๐‘š๐‘”
๐‘Ÿ
You feel heaviest at the bottom
of the loop because your
โ€˜apparent weightโ€™ 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 rideโ€™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
โ€ข At the TOP of the hill
๐น๐‘ฆ = ๐‘š๐‘Ž๐‘ฆ
๐น๐‘” โˆ’ ๐น๐‘ = ๐‘š๐‘Ž๐‘
๐‘š๐‘ฃ 2
๐‘š๐‘”โˆ’๐น๐‘ =
๐‘Ÿ
2
๐‘š๐‘ฃ
๐น๐‘ = ๐‘š๐‘” โˆ’
๐‘Ÿ
FN
Fg
If you โ€˜lose contact with the
seatโ€™ 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
Newtonโ€™s Law of Gravity
๐บ๐‘š1 ๐‘š2
๐น๐‘” =
๐‘Ÿ2
โ€ข Fg Force of gravitational attraction
between any two objects.
โ€ข G Universal Gravitational Constant=6.67x10-11
Nm2/kg2
โ€ข m the mass of the objects (kg)
โ€ข r the distance between the CENTERS of the
objects.
Acceleration Due to Gravity
๐น๐‘”๐‘œ๐‘› ๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž = ๐น๐‘” ๐‘ข๐‘›๐‘–๐‘ฃ๐‘’๐‘Ÿ๐‘ ๐‘Ž๐‘™
๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘œ๐‘› ๐‘” =
๐‘”=
๐บ๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘œ๐‘› ๐‘€๐ธ๐‘Ž๐‘Ÿ๐‘กโ„Ž
๐‘…๐ธ๐‘Ž๐‘Ÿ๐‘กโ„Ž 2
๐บ๐‘€๐ธ๐‘Ž๐‘Ÿ๐‘กโ„Ž
๐‘…๐ธ๐‘Ž๐‘Ÿ๐‘กโ„Ž 2
General g
๐‘”=
๐บ๐‘€๐‘๐‘™๐‘Ž๐‘›๐‘’๐‘ก
๐‘…๐‘๐‘™๐‘Ž๐‘›๐‘’๐‘ก 2
Orbital Speed
๐น = ๐‘š๐‘Ž๐‘
๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘œ๐‘› ๐‘ฃ 2
๐น๐‘” =
๐‘Ÿ
๐บ๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘œ๐‘› ๐‘€๐ธ๐‘Ž๐‘Ÿ๐‘กโ„Ž
๐‘Ÿ2
=
๐‘š๐‘๐‘’๐‘Ÿ๐‘ ๐‘œ๐‘› ๐‘ฃ 2
๐‘Ÿ
๐บ๐‘€๐ธ๐‘Ž๐‘Ÿ๐‘กโ„Ž ๐‘ฃ 2
=
๐‘Ÿ
1
๐‘ฃ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก =
๐บ๐‘€๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž
๐‘Ÿ
General Orbital Speed-> Keplerโ€™s 3rd
Laws
๐บ๐‘€๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก๐‘’๐‘‘
๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก
๐‘ฃ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก =
2๐œ‹๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก
=
๐‘‡
2๐œ‹๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก
๐‘‡
4๐œ‹ 2 ๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก 2 ๐บ๐‘€๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก๐‘’๐‘‘
=
2
๐‘‡
๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก
2
๐บ๐‘€๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก๐‘’๐‘‘
๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก
๐บ๐‘€๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก๐‘’๐‘‘
=
๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก
๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก 3 ๐บ๐‘€๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก๐‘’๐‘‘
=
2
๐‘‡
4๐œ‹ 2
Keplerโ€™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
planetsโ€™ orbits are related
3
๐‘Ÿ๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก
๐บ๐‘€๐‘œ๐‘Ÿ๐‘๐‘–๐‘ก๐‘’๐‘‘
=
๐‘‡2
4๐œ‹ 2
๐’“๐‘จ
๐Ÿ‘
๐‘ป๐‘จ
๐Ÿ
=
๐’“๐‘ฉ
๐Ÿ‘
๐‘ป๐‘ฉ
๐Ÿ
Questions
โ€ข How could have scientists figure out the mass of the
Earth?
๐‘ฎ๐‘ด๐’‘๐’๐’‚๐’๐’†๐’•
๐’ˆ=
๐Ÿ
๐‘น๐’‘๐’๐’‚๐’๐’†๐’•
โ€ข How could have scientists figure out the mass of the
sun?
๐‘ฎ๐‘ด๐’๐’“๐’ƒ๐’Š๐’•๐’†๐’…
๐’—๐’๐’“๐’ƒ๐’Š๐’• =
๐’“๐’๐’“๐’ƒ๐’Š๐’•
โ€ข How could have scientists figure out the distances
to all the other planets in our solar system?
๐’“๐‘จ
๐Ÿ‘
๐‘ป
๐Ÿ
=
๐’“๐‘ฉ
๐Ÿ‘
๐‘ป
๐Ÿ