Circular Motion

Circular Motion

http://www.scicomics.com/uploads/centripetal_acceleration.jpg

Rotation vs Revolution

• Rotation occurs when an object rotates around an internal axis– ex: the earth rotates around its

axis (internal) every day

• Revolution occurs when an object rotates around an external axis– ex: the earth revolves

around the sun (external axis) every year

Rotational Speed• also called angular speed• equals the number of

rotations per time unit– All parts of the earth rotate about

its axis in the same amount of time…thus all parts of the earth have the same rotation …or the same number of rotations per time (revolutions per minute or RPM)

When a twirling ice skater extends her arms outward, her rotational speed decreases.

Tangential Velocity

• While the speed of the object is constant, its velocity is changing.

Velocity vector and position vector for an object executing uniform circular motion

Tangential Speed

• Tangential speed is directly proportional to rotational speed and the distance from the axis

Tangential ~ Radial x Rotational Speed distance Speed

-at the center of the earth…there is no tangential speed, (zero distance) but there is rotational speed

twice as fastfast

What type of speed is changing – rotational or tangential?

Mathematical Relationship between tangential and rotational

speed v = r ω

v = tangential speed

r = radial distance

ω = rotational speed

Accelerometer

• The deflection of the flame will be in the direction of the acceleration. This is because the hot gases of the flame are less massive and thus have less inertia than the cooler gases which surround

Centripetal Force

• the force required to change the direction of a moving object v = tangential speed

http://webpages.uah.edu/~wilderd/resources.htmlhttp://www2.ignatius.edu/faculty/decarlo/Centripetal%20force.htm

Fc = mv2 / r

Centripetal Force

• center–seeking or toward the center

• any force that causes an object to follow a circular path– think about the force you feel when riding a

fast circular ride at an amusement park…without that force you would continue to move in a straight line…being “launched” form the ride

Centripetal Force

As a car makes a turn, the force of friction acting upon the turned wheels of the car provide the centripetal force required for circular motion.

As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion.

As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.

Centrifugal Force…False Force

• center-fleeing or away from the center

• An apparent outward force that acts on a rotating object…it is due to inertia…an object’s desire to continue moving in a straight path.

• The sensation of being thrown outward is attributable to the idea of inertia, rather than the idea of force.

• Centrifugal force holds people on walls in Round Up ride

The Fictitious Force

• When a ball swings on a string, many people would say that "CENTRIFUGAL FORCE" keeps the string taut. – There is actually no force

pulling the ball out. – There is nothing out there

pulling the ball! – All the sideways and back and

forth forces you feel while riding in a car are FICTITIOUS, they are the results of your inertia resisting acceleration

http://www.scigolf.com/scigolf/myths/myth2.htm

The Spin Cycle in a Washing Machine

• The water goes through the holes at a tangent to the tub, while the clothes are moved inward in a circle.

http://www.scigolf.com/scigolf/myths/myth2.htm

The Fictitious “Force”

http://www.geo.umn.edu/courses/1006/Fall01_night/Coriolis.html

• The centripetal force of an aircraft is the resultant force of the two other forces acting on the aircraft: the weight of the aircraft and the lift of the aircraft.

Tether Ball

• Twirl a ball, at a constant speed, at the end of a rope.change of direction = acceleration

http://www2.ignatius.edu/faculty/decarlo/Centripetal%20force.htm

http://www.hoopsplus.com/nss-folder/outdoorequipment/tether.jpg

Direction of Accelerated Motion• Use vector addition to determine the

resultant velocity:– It is acting inward; the same as the force!

(Newton’s 2nd)


ac = v2 / r

• ac = centripetal acceleration

• v = tangential velocity

• r = radius

• The greater an object’s velocity, the greater the centripetal acceleration.

• The smaller the curve of the path being traveled, the greater the centripetal acceleration.

• http://library.thinkquest.org/27948/carousel.html • http://www.college-cram.com/study/physics/pres

entations/472


http://library.thinkquest.org/27948/carousel.html

http://www.college-cram.com/study/physics/presentations/472

http://www.college-cram.com/study/physics/presentations/472

Trial # 1• A 4.0 kg ball is attached to 0.70 meter

string and spun at 2.0 meters/sec. Calculate a) centripetal acceleration

• ac = v2/r• ac = (2.0 m/s)2/(0.7 m)• ac = 5.7 m/s2

b) centripetal force• Fc = mac • Fc = (4.0 kg)(5.71 m/s2)• Fc = 23 N

Double the speed from the previous problem and perform the same calculations

• A 4.0 kg ball is attached to 0.70 meter string and spun at 4.0 meters/sec. Calculate

a) centripetal acceleration (23 m/s2)

b) centripetal force (92 N)

c) what happened to the force when the speed was doubled?

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/rht.html

http://www.coolschool.ca/lor/PH12/unit4/U04L01.htm

FBD Consider an example of a person riding a

roller coaster through a circular section of the track, a "loop-the-loop."

At the TOP:net force to the center = N + mg

N + mg = m(v2/r)N = m(v2/r) - mg

At the BOTTOM:net force to the center = N - mg

N - mg = m(v2/r)N = m(v2/r) + mg

Roller Coaster Physics

– Many high-speed roller coasters use banked turns rather than the flat ones that are safe for slower speeds. Banking the turns in a roller coaster gives you the feeling of being pushed into your seat rather than being thrown to the side of the car.

– acceleration stress: http://library.thinkquest.org/TQ0313004/Accelertion_stress.html be sure to mouse over roller coaster photo

• Physics of Roller Coasters and look at the forces on a victim -- er, I mean "guest" -- as the roller coaster goes through the top of a hill, the bottom of a valley, or at the top of a loop.

FBD

Roller Coaster FBD

Opportunity:

While driving to work you pass over a "crest" in the road that has a radius of 30 meters. How fast would you need to be traveling to experience apparent "weightlessness" while passing over the crest?

Answer: As seen in the following diagram, the

NET FORCE to the center equals mg - N.

If we let the normal approach 0 to represent apparent weightlessness, then

Roller Coaster G Forces

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/rcd.html

Roller Coaster Problem• Anna Litical is riding on The Demon at Great America. Anna

experiences a downwards acceleration of 15.6 m/s2 at the top of the loop and an upwards acceleration of 26.3 m/s2 at the bottom of the loop. Use Newton's second law to determine the normal force acting upon Anna's 864 kg roller coaster car.

Given Info:m = 864 kgatop = 15.6 m/s2 , downabottom = 26.3 m/s2 , upFind:Fn at top and bottom

http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l2b.html

Solution to Roller Coaster Problem• Bottom of Loop

• Fnet=ma

• Fnet = (864 kg)(26.3 m/s2, up)

• Fnet = 22 723 N, up

• Top of Loop

• Fnet = ma

• Fnet = (864 kg)(15.6 m/s2, down)

• Fnet = 13478 N, down

Fn must be greater than the Fg by 22723 N in order to supply a net upwards force of 22723 N. Thus,Fn = Fg + Fnet

Fn = 31190 N

Fn and Fg together must combine together (i.e., add up) to supply the required inwards net force of 13478 N. Thus,Fn = Fnet - Fg

Fn = 5011 N

Various Directions of Acceleration

• http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l2b.html

Roller Coaster Problem # 2• Anna Litical is riding on The American Eagle at Great

America. Anna is moving at 18.9 m/s over the top of a hill which has a radius of curvature of 12.7 m. Use Newton's second law to determine the magnitude of the applied force of the track pulling down upon Anna's 621 kg roller coaster car.

Given Info:m = 621 kgv = 18.9 m/sr = 12.7 mFind:FA at top of hill


Solution: Accerleration

• a = v2 / r

• a = (18.9 m/s)2 / (12.7 m)

• a = 28.1 m/s2


Solution: Force

• Fnet = ma

• Fnet = (621 kg) • (28.1 m/s2, down)

• Fnet = 17467 N, down

• FA = Fnet - Fg

• FA = 11381 N

FA and Fg must combine together (i.e., add up) to

supply the required downwards net force of

17467 N.

Practice

• An air puck of mass 0.025 kg is tied to a string and allowed to revolve in a circle of radius 1.0 m on a frictionless horizontal surface. The other end of the string passes through a hole in the center of the surface, and a mass of 1.0 kg is tied to it. The suspended mass remains in equilibrium while the puck revolves on the surface.– A. what is the magnitude of the force that maintains circular

motion acting on the puck?– B. What is the linear speed of the puck?

http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch06/Ch6.html

Answer• mp = 0.025 kg• r = 1.0 m• m = 1.0 kg• g = 9.81 m/s2

• Fc = Fg = mg• = (1.0 kg)(9.91 m/s2)• = 9.8 N• Fc = mpv2r• v = 2.0 x 1.01 m/s

Practice • An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away. The coefficient of static friction between the person and the wall is s, and the radius of the cylinder is r.

• Draw the force diagram!


Force Diagram


Fn = m v2 / rFf = mgFf = Fn

http://visual.physics.tamu.edu/vp218/LabManual/Figure10_11.jpg

• Gravity: force of attraction between two objects.

• Newton noted that the moon was falling toward the Earth.

• The moon falls along a curved path. It acts as a projectile.

• The moon has a component of velocity parallel to the Earth’s surface known as tangential velocity.

• The force of gravitational attraction decreases as you move further from the center of the Earth.

• The moon orbits the Earth, just as the Earth & other planets orbit the Sun.

• The Sun is the center of the solar system.

Universal Gravitation• Law of Universal Gravitation:

Every object attracts every other object with a force that for any two objects is directly proportional to the mass of each object.

• The rest is just for Fun and Interest:

– Ocean Tides are due to the differences in the gravitational pull of the moon on opposite sides of the Earth. Variations in sun, moon and earth alignment causes low and high tides.

• http://oceanlink.island.net/oinfo/tides/tides.html• http://home.hiwaay.net/~krcool/Astro/moon/moontides/

Tides

SPRING TIDES: the sun and moon are aligned, there are exceptionally strong gravitational forces, causing very high and very low tides.

NEAP TIDES: the sun and moon are not aligned, the gravitational forces cancel each other out, and the tides are not as dramatically high and low.

Center of Gravity (CG)• Located at the object’s average

position of weight (or balance point).• For a symmetrical object such as a

ball. The CG is located in the center.• For an irregular shaped object (a

bat), the CG is located towards the more massive end.

Toppling (falling over)• If the CG is located above the area of

support, the object will not fall. If it is outside the support area, the object will fall over.

• Neutral, Stable and Unstable Equilibrium.

Torque

• The turning force used to turn a door knob or tighten a bolt produces torque.

• Torque is defined as force times length of lever arm. For greatest torque, force should be applied perpendicular to the lever arm.

• Notice that the unit of measurement for torque contains a distance (meter) and a force (Newton).

• To calculate the torque needed to turn something, multiply the force┴ and the distance from the center of the object you are trying to turn.

Torque = Force┴ x distance

Challenge:

• Our moon moves in a circle so its direction changes continually. Therefore the moon accelerates due to directional changes; this is the centripetal acceleration. – Calculate the speed and the centripetal

acceleration of the moon. • ac = v2 / r

– What do you know? What do you need to know? How can you find the unknown, needed information?

http://www.iop.org/activity/education/Teaching_Resources/Teaching%20Advanced%20Physics/Fields/Images%20400/img_tb_4641.gif

The Solution• Use available sources to find the radius of moon and

period of moon• Determine the moon’s speed:

d = 2πr therefore: v = d / T

(T = period or time for one revolution)v = 2πr / T

v = [2(3.14) ( 3.84 x 108 m ) ]/ (27.32 da)][da/24h][h/3600s]v = 1 022 m / s

• Calculate the acceleration of the moon:ac = v2 / r

ac = (1 022 m / s)2 / 3.84 x 108 mac = 2.72 x 10-3 m / s2

Documents

Circular Motion