Circular Motion AIM: How is this even possible?????
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Transcript of Circular Motion AIM: How is this even possible?????
But FirstโฆFacts about Circular Motionโข The distance around one lap of a circle is the
circumference ()โข What is ยฝ a lap?โข What is ยผ of a lap?
โข The displacement for one lap around a circle is zero.โข What is ยฝ a lap?โข What is ยผ of a lap?
โข Time once around a circle is the Period (T)
More Facts about Circular Motionโข Speed while moving in a circle
โข Acceleration while moving in a circle
โข Net force while moving in a circle
Even More Facts about Circular MotionVelocity
(tangent to the circle at all times)
Centripetal Acceleration and
Centripetal Force(always towards the center of the circle)
V
Fc ac
What Provides Centripetal Force?
โข Centripetal force is a NET force which means it is calculated โ either by Newtonโs 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.
โข 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)
โข 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? JUSTIFY!
0=
Vertical
=
Horizontal
Ff
FN
Fg
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
Ff
FN
Fg
๐ญโฅ=๐๐cos๐ฝ
๐ญโฅ=๐๐ sin ๐ฝ
0=
Vertical
===
Horizontal
Tension as the Centripetal Forceโข An object moves in a horizontal circle on a
frictionless surface while attached to a string of length l
Table top
T
Tension as the Centripetal Forceโข An object moves in a vertical circle on a at a
constant speed while attached to a string of length l top
TFg
Tension as the Centripetal Forceโข An object moves in a vertical circle on a at a
constant speed while attached to a string of length l bottom
T
Fg
Tension as the Centripetal Forceโข When an object is moving in a vertical circle, where in the
motion is the string most likely going to break? JUSTIFY!
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.
๐ ๐ก๐๐=๐๐ฃ2
๐โ๐๐ ๐ ๐๐๐ก๐ก๐๐=๐๐ฃ2
๐+๐๐
Tension as the Centripetal Forceโข An object moves in a conical pendulum while
attached to a string of length l making an angle with the vertical
T
Fg
Ty
Tx
Normal Force as a Centripetal Force(Gravitron Ride)
โข When you are riding the gravitron, the centripetal force is provided by the wall of the ride!
โข Does the mass of the rider matter? โ PROVE IT MATHEMATICALLY!
Ff
FN
Fg
=
Vertical Horizontal
๐๐ ๐๐ฃ2
๐=๐๐
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 Loopโข At the TOP of the loop
FN Fg
If you โjust make itโ FN=0N
Circular motion in a Vertical Loopโข At the BOTTOM of the loop
FN
Fg
Circular motion in a Vertical Loopโข Where in the loop do you
feel the heaviest? JUSTIFY!TopBottom
You feel heaviest at the bottom of the loop because your โapparent 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 rideโs 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 Hillโข At the TOP of the hill
FN
Fg
If you โlose 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
Newtonโs Law of Gravity
โข Fg Force of gravitational attraction between any two objects.
โข G Universal Gravitational Constant=6.67x10-11 Nm2/kg2
โข m the mass of the objects (kg)โข r the distance between the CENTERS of the
objects.
๐น ๐=๐บ๐1๐2
๐ 2
Acceleration Due to Gravity
General g
Orbital Speed
=
General Orbital Speed-> Keplerโs 3rd Laws
๐ฃ๐๐๐๐๐ก=โ๐บ๐๐๐๐๐๐ก๐๐
๐ ๐๐๐๐๐ก
2๐ ๐๐๐๐๐๐ก๐
=โ๐บ๐๐๐๐๐๐ก๐๐
๐๐๐๐๐๐ก
( 2๐๐๐๐๐๐๐ก๐ )2
=๐บ๐๐๐๐๐๐ก๐๐
๐๐๐๐๐๐ก4๐ 2๐๐๐๐๐๐ก
2
๐ 2 =๐บ๐๐๐๐๐๐ก๐๐
๐๐๐๐๐๐ก๐๐๐๐๐๐ก
3
๐ 2 =๐บ๐๐๐๐๐๐ก๐๐
4๐ 2
Keplerโs Laws of Planetary Motion1. 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
๐๐๐๐๐๐ก3
๐ 2 =๐บ๐๐๐๐๐๐ก๐๐
4๐ 2
๐ ๐จ๐
๐ป ๐จ๐=
๐ ๐ฉ๐
๐ป๐ฉ๐
Questionsโข How could have scientists figure out the mass of the
Earth?
โข How could have scientists figure out the mass of the sun?
โข How could have scientists figure out the distances to all the other planets in our solar system?
๐=๐ฎ๐ด๐๐๐๐๐๐
๐น๐๐๐๐๐๐๐
๐๐๐๐๐๐=โ๐ฎ๐ด๐๐๐๐๐๐๐
๐ ๐๐๐๐๐
๐ ๐จ๐
๐ป ๐จ๐=
๐ ๐ฉ๐
๐ป๐ฉ๐