circular motion

LuisLuis AnchordoquiAnchordoqui

Kinematics of Uniform Circular Motion Kinematics of Uniform Circular Motion An object that moves in a circle at constant speedAn object that moves in a circle at constant speed

is said to experience uniform circular motionis said to experience uniform circular motionThe magnitude of the velocity remains constant in this case The magnitude of the velocity remains constant in this case

but the direction of the velocity continuously changes but the direction of the velocity continuously changes as the object moves around the circleas the object moves around the circle

An object revolving in a circle is continuously acceleratingAn object revolving in a circle is continuously acceleratingeven when the speed remains constanteven when the speed remains constant


Kinematics of Uniform Circular Motion (contKinematics of Uniform Circular Motion (cont’’d) d) The acceleration is defined asThe acceleration is defined as for a short interval of timefor a short interval of time


We will eventually consider the situation when We will eventually consider the situation when ∆∆t approaches to zero t approaches to zero to obtain the instantaneous accelerationto obtain the instantaneous acceleration

For the purposes of making a clear drawing consider a nonFor the purposes of making a clear drawing consider a non--zero time intervalzero time interval

If we let If we let ∆∆tt be very small be very small ∆∆l l and and ∆Θ∆Θ are also very small are also very small and will be almost parallel toand will be almost parallel to

will be essentially perpendicular to & will be essentially perpendicular to & pointing to the center of the circlepointing to the center of the circle

Kinematics of Uniform Circular Motion (contKinematics of Uniform Circular Motion (cont’’d) d)


must too point to the center of the circlemust too point to the center of the circle

Kinematics of Uniform Circular Motion (contKinematics of Uniform Circular Motion (cont’’d) d) The magnitude of the velocity is not changing we can writeThe magnitude of the velocity is not changing we can write


This is an exact equality when This is an exact equality when ∆∆t approaches zero t approaches zero Let Let ∆∆t approach zero and solve for t approach zero and solve for ∆∆vv

To get the centripetal acceleration we divide by To get the centripetal acceleration we divide by ∆∆tt

is just the linear speedis just the linear speed

Kinematics of Uniform Circular Motion (contKinematics of Uniform Circular Motion (cont’’d) d)


The acceleration vector points towards the center of the circleThe acceleration vector points towards the center of the circleThe velocity vector always points in the direction of motionThe velocity vector always points in the direction of motion

Circular motion is often described in terms Circular motion is often described in terms of the frequency (f)of the frequency (f)

the number of revolutions per secondthe number of revolutions per second

The period of an object (T) revolving in a circle is the time reThe period of an object (T) revolving in a circle is the time required for one quired for one complete revolutioncomplete revolution

For an object revolving in a circle at constant speed we haveFor an object revolving in a circle at constant speed we have

A satellite moves at constant speed in a circular orbit about thA satellite moves at constant speed in a circular orbit about the center e center of Earth near the surface of Earth. of Earth near the surface of Earth.

If the magnitude of its acceleration is g = 9.81 m/sIf the magnitude of its acceleration is g = 9.81 m/s²², find , find (a)(a) its speed and its speed and (b) the time for one complete revolution(b) the time for one complete revolution

T = 5060 s = 84.3 minT = 5060 s = 84.3 min

A Satellite's MotionA Satellite's Motion

v = 7.91 km/sv = 7.91 km/s



Moon's Centripetal AccelerationMoon's Centripetal Acceleration

Earth as seen from Apollo 11 orbiting the Moon on July 16, 1969 Earth as seen from Apollo 11 orbiting the Moon on July 16, 1969 (NASA)(NASA)

The Moon's nearly circular orbit about the Earth has a radius ofThe Moon's nearly circular orbit about the Earth has a radius ofabout 384,000km and a period T of 27.3 days.about 384,000km and a period T of 27.3 days.

Determine the acceleration of the Moon towards the Earth.Determine the acceleration of the Moon towards the Earth.

a = 2.78 x 10 g a = 2.78 x 10 g --44


Dynamics of Uniform Circular Motion Dynamics of Uniform Circular Motion According to Newton's second law

an object that is accelerating must have a net force acting on it An object moving in a circle such as a ball on the end of a string

must therefore have a force applied to it to keep it moving on that circle The magnitude of the force can be calculated using Newton's second law

for the radial component


Dynamics of Uniform Circular Motion (contDynamics of Uniform Circular Motion (cont’’d) d) There is a common misconception that an object moving in a circlThere is a common misconception that an object moving in a circle has e has

an outward force acting on it: centrifugal ("center feeling") foan outward force acting on it: centrifugal ("center feeling") forcerce

There is no outward force on the There is no outward force on the revolving objectrevolving object


Dynamics of Uniform Circular Motion (contDynamics of Uniform Circular Motion (cont’’d) d) Consider for example a person swinging a ball on the end of a stConsider for example a person swinging a ball on the end of a string around her head ring around her head

If you ever done this yourself you know that you feel a force puIf you ever done this yourself you know that you feel a force pulling lling outward on your hand outward on your hand

The misconception arises when this pull is interpreted as an outThe misconception arises when this pull is interpreted as an outward "centrifugal" ward "centrifugal" force pulling on the ball that is force pulling on the ball that is transmitedtransmited along the string to your hand along the string to your hand

To keep the ball moving on a circle you pull inwardly on the strTo keep the ball moving on a circle you pull inwardly on the string ing and the string exerts this force on the ball and the string exerts this force on the ball

The ball exerts an equal and opposite force on the string (NewtoThe ball exerts an equal and opposite force on the string (Newton's third law) n's third law) and this is the outward force your hand feels and this is the outward force your hand feels


Dynamics of Uniform Circular Motion (contDynamics of Uniform Circular Motion (cont’’d) d) To see even more convincing evidence that a "centrifugal force" To see even more convincing evidence that a "centrifugal force" does not does not

act on the ball consider what happens when you let go of thact on the ball consider what happens when you let go of the string e string If a centrifugal force were acting the ball would fly outward If a centrifugal force were acting the ball would fly outward

The ball flies off tangentially in the direction of the velocityThe ball flies off tangentially in the direction of the velocity it had at it had at the moment it was released because the inward force no longer acthe moment it was released because the inward force no longer acts ts

Try it and see! Try it and see!


Sparks fly in straight lines tangentially from the edge of a rotSparks fly in straight lines tangentially from the edge of a rotating grinding wheelating grinding wheel

Dynamics of Uniform Circular Motion (contDynamics of Uniform Circular Motion (cont’’d) d)


We assume the weight is small and set We assume the weight is small and set ΦΦ approximately zeroapproximately zero

Revolving ball (horizontal circle) Revolving ball (horizontal circle) Estimate the force a person must exert on a string attached to aEstimate the force a person must exert on a string attached to a 0.15 kg 0.15 kg

ball to make the ball revolve in a horizontal circle of radius 0ball to make the ball revolve in a horizontal circle of radius 0.6 m..6 m.

The ball makes 2 revolutions per secondThe ball makes 2 revolutions per second

The ball's weight complicates matters and makes it impossible toThe ball's weight complicates matters and makes it impossible torevolve the ball with a cord perfectly horizontal.revolve the ball with a cord perfectly horizontal.

TT


F = 14 NF = 14 N

(b) Calculate the tension in the cord at the bottom of the arc, (b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice speed of partassuming the ball is moving at twice speed of part

RevolvingRevolving ballball (vertical (vertical circlecircle))A 0.15 kg ball on the end of a 1 m long cord (of negligible massA 0.15 kg ball on the end of a 1 m long cord (of negligible mass) is ) is

swung in a vertical circle.swung in a vertical circle.(a) Determine the minimum speed the ball must have at the top of(a) Determine the minimum speed the ball must have at the top of

its arc so that the ball continues moving in a circleits arc so that the ball continues moving in a circle

v = 3.28 v = 3.28 m/sm/s

F = 7.34 N F = 7.34 N

toptop

T2T2


A tetherball of mass m is suspended from a length rope and A tetherball of mass m is suspended from a length rope and travels at constant speed v in a horizontal circle of radius rtravels at constant speed v in a horizontal circle of radius r

TetherballTetherball

FindFind (a) the tension of the rope(a) the tension of the rope

(b) the speed of the ball(b) the speed of the ball

T = mg/T = mg/coscos θθ

v=(r g tan v=(r g tan θθ))1/21/2


Tarzan plans to cross a gorge by swinging in an arc from a hangiTarzan plans to cross a gorge by swinging in an arc from a hanging vine.ng vine.

If his arms are capable of exerting a force of 1400 N on the vinIf his arms are capable of exerting a force of 1400 N on the vine, what e, what is the maximum speed he can tolerate at the lowest point of his is the maximum speed he can tolerate at the lowest point of his swing?swing?

His mass is 80 kg and His mass is 80 kg and the vine is 5.5m longthe vine is 5.5m long

v = 6.5 v = 6.5 m/sm/smaxmax


A flat puck (mass M) is rotated in a circle on a A flat puck (mass M) is rotated in a circle on a frictionalessfrictionalessairair--hockey tabletop, and is held in this orbit by a light cord hockey tabletop, and is held in this orbit by a light cord

connected to a dangling block (mass m) through a central hole.connected to a dangling block (mass m) through a central hole.

Show that the speed of the puck is given by v = (Show that the speed of the puck is given by v = (mgRmgR/M) /M)


½½

A projected space station consists of a circular tube that will A projected space station consists of a circular tube that will rotate rotate about its center (like a tubular bicycle tire)about its center (like a tubular bicycle tire)

1.8 x 10 rev/d1.8 x 10 rev/d33

The circle formed by the tube has a diameter of about 1.1 km.The circle formed by the tube has a diameter of about 1.1 km.

What must be the rotation speed (revolutions per day) if an effeWhat must be the rotation speed (revolutions per day) if an effect ct equal to gravity at the surface of the earth is to be felt?equal to gravity at the surface of the earth is to be felt?


A jet pilot takes his aircraft in a vertical loopA jet pilot takes his aircraft in a vertical loop

(a) If the jet is moving at a speed of 1300 km/h at the lowest p(a) If the jet is moving at a speed of 1300 km/h at the lowest point of oint of the loop, determine the minimum radius of the circle so that thethe loop, determine the minimum radius of the circle so that thecentripetal acceleration at lowest point does not exceed 6.0 g'scentripetal acceleration at lowest point does not exceed 6.0 g's

(b) Calculate the 78 kg pilot's effective weight (the force with(b) Calculate the 78 kg pilot's effective weight (the force with which which the seat pushes up on him at the bottom of the circle)the seat pushes up on him at the bottom of the circle)

(c) at the top of the circle (assume the same speed)(c) at the top of the circle (assume the same speed)

r = 2.2 x 10 mr = 2.2 x 10 m33

F = 5.4 x 10 NF = 5.4 x 10 NNN

33

F = 3.8 x 10 NF = 3.8 x 10 NNN33


v = 8.5 v = 8.5 m/sm/sminmin

At what minimum speed must a roller coaster be At what minimum speed must a roller coaster be travellingtravelling when upside when upside down at the top of a circle so that passengers will not fall outdown at the top of a circle so that passengers will not fall out. .

Assume a radius of curvature of 7.4 mAssume a radius of curvature of 7.4 m


11 rpm11 rpm

How many revolutions per minute would a 15How many revolutions per minute would a 15--mm--diameter Ferrisdiameter Ferriswhellwhell need to make for the passengers to feel need to make for the passengers to feel ‘‘ ‘‘ weightless weightless ´´´´ at the topmost point.at the topmost point.


When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc.

If the road is flat, that force is supplied by friction.


If the frictional force is insufficient, the car will tend to If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.move more nearly in a straight line, as the skid marks show.

Highway Curves, Banked and Highway Curves, Banked and UnbankedUnbanked (cont(cont’’d)d)



As long as the tires do not slip, the friction is static. As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, If the tires do start to slip, the friction is kinetic,

which is bad in two ways:which is bad in two ways:

1. 1. The kinetic frictional force is smaller than the static.The kinetic frictional force is smaller than the static.

2. The static frictional force can point towards the center2. The static frictional force can point towards the centerof the circle, but the kinetic frictional force opposes theof the circle, but the kinetic frictional force opposes the

direction of motion, making it very difficult to regain direction of motion, making it very difficult to regain control of the car and continue around the curve.control of the car and continue around the curve.


Banking the curve can help keep cars from skidding. Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed where In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the horizontal the entire centripetal force is supplied by the horizontal

component of the normal force, and no friction is required.component of the normal force, and no friction is required.


This occurs when:This occurs when:


Rounding a Banked Curve Rounding a Banked Curve A curve of radius 30.0 m is banked at an angle A curve of radius 30.0 m is banked at an angle θθ. .

That is the normal to the road surface makes an angle That is the normal to the road surface makes an angle of 30 degrees with the vertical. of 30 degrees with the vertical.

Find Find θθ such that a car can round the curve at 40 km/h even if the such that a car can round the curve at 40 km/h even if the road is coated with ice, making the road essentially frictionlesroad is coated with ice, making the road essentially frictionless s

θθ= 22.8 degrees = 22.8 degrees LuisLuis AnchordoquiAnchordoqui

A Road Test A Road Test You have a summer job with NASCAR as part of an automobile tire You have a summer job with NASCAR as part of an automobile tire testing. testing.

You are testing a new model of racing tires to see whether or noYou are testing a new model of racing tires to see whether or not t the coefficient of static friction between the tires and the coefficient of static friction between the tires and

dry concrete pavement is 0.90 as claimed by the manufacturer. dry concrete pavement is 0.90 as claimed by the manufacturer. In a In a skidpadskidpad test, a racecar is able to travel at constant speed in a test, a racecar is able to travel at constant speed in a

circle of radius 45.7 m in 15.2 s without skidding. circle of radius 45.7 m in 15.2 s without skidding. Assume air drag and rolling friction are negligible and assume tAssume air drag and rolling friction are negligible and assume that the hat the

road surface is horizontal.road surface is horizontal.In a In a skidpadskidpad test a car travels in a circle on a flat, horizontal surface attest a car travels in a circle on a flat, horizontal surface at

maximum possible speed without skidding.maximum possible speed without skidding.

v = 18.9 v = 18.9 m/sm/s

μμ = 0.796 = 0.796

(a) What was its speed?(a) What was its speed? (b) What was its acceleration(b) What was its acceleration(c) What was the minimum value for the coefficient of static fri(c) What was the minimum value for the coefficient of static friction ction

between the tires and the road?between the tires and the road?

ss

a = 7.81 m/sa = 7.81 m/s²²

a = 0 a = 0 TT

CC


CentrifugationCentrifugationThese devices are used to sediment materials quickly or to separThese devices are used to sediment materials quickly or to separate materials ate materials

Test tubes are held in the centrifugal rotor which is accelerateTest tubes are held in the centrifugal rotor which is accelerated to very high d to very high rotational speedsrotational speeds

The small green dot represents a small particle The small green dot represents a small particle (macromolecule) in a fluid filled test tube.(macromolecule) in a fluid filled test tube.

When the tube is at position A and the rotor is When the tube is at position A and the rotor is turning the particle has a tendency to move in a turning the particle has a tendency to move in a

straight line in the direction of the dashed arrow.straight line in the direction of the dashed arrow.

But the fluid that resists the motion of these But the fluid that resists the motion of these particles exerts a centripetal force that keeps particles exerts a centripetal force that keeps

the particles moving nearly in a circle.the particles moving nearly in a circle.Usually the resistance of the tube does not quite equal mvUsually the resistance of the tube does not quite equal mv²²/r and the /r and the

particles eventually reach the bottom of the tube.particles eventually reach the bottom of the tube.The purpose of a centrifuge is to provide and The purpose of a centrifuge is to provide and ‘‘ ‘‘ effective gravityeffective gravity´´´´ much much

larger than normal gravity because of the high rotational speedslarger than normal gravity because of the high rotational speeds, thus , thus causing more rapid sedimentation causing more rapid sedimentation


In 1993 a descendent probe containing instruments went deepinto the Jovian atmosphere of Jupiter.

The fully assemble probed was tested at accelerations up to200 g’s in this large centrifuge at Sandia National Laboratories


Ultracentrifuge Ultracentrifuge The rotor of an ultracentrifuge rotates at 50,000 rpm. The rotor of an ultracentrifuge rotates at 50,000 rpm.

The top of the 4cm long test tube is 6cm from the rotation axis The top of the 4cm long test tube is 6cm from the rotation axis and is perpendicular to it. and is perpendicular to it.

The bottom of the tube is 10 cm from the axis of rotation. The bottom of the tube is 10 cm from the axis of rotation.

a = 1.67 x 10 ga = 1.67 x 10 g

a = 2.8 x 10 g a = 2.8 x 10 g

Calculate the centripetal acceleration in g's at Calculate the centripetal acceleration in g's at (a)(a) the top the top (b) the bottom of the tube(b) the bottom of the tube

RR

RR55

55


Nonuniform Circular Motion

If an object is moving in a circular path but at varying speeds,If an object is moving in a circular path but at varying speeds, it must it must have a tangential component to its acceleration as well as the rhave a tangential component to its acceleration as well as the radial one.adial one.


This concept can be used for an object moving along any curved pThis concept can be used for an object moving along any curved path, ath, as a small segment of the path will be approximately circular.as a small segment of the path will be approximately circular.

Nonuniform Circular Motion (cont’d)


A car at Indianapolis accelerates uniformly from the pit area, gA car at Indianapolis accelerates uniformly from the pit area, going oing from rest to 320 km/h in a semicircular arc with a radius of 220from rest to 320 km/h in a semicircular arc with a radius of 220 m.m.

Determine the tangential and radial acceleration of the car whenDetermine the tangential and radial acceleration of the car when it is it is halfway through the turn, assuming constant tangential accelerathalfway through the turn, assuming constant tangential acceleration.ion.If the curve where flat, what would the coefficient of static frIf the curve where flat, what would the coefficient of static friction iction

have to be between the tires and the road to provide this have to be between the tires and the road to provide this acceleration with no slipping or skidding?acceleration with no slipping or skidding?

a = 5.72 m/sa = 5.72 m/s²²

a = 18 m/sa = 18 m/s²²

μμ = 1.9= 1.9

tantan

RR

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circular motion