a2 h 41b Circularmotion

8/12/2019 a2 h 41b Circularmotion

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4.1b Further Mechanics

Circular MotionBreithaupt pages 22 to 33

December 25th, 2011

CLASS NOTES HANDOUT VERSION


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AQA A2 Specification

Lessons Topics

1 to 3 Circular motion

Motion in a circular path at constant speed implies

there is an acceleration and requires a centripetal force.

Angular speed = v / r = 2 f

Centripetal acceleration a = v2/ r = 2 r

Centripetal forceF = mv2/ r = m2 r

The derivation of a = v2

/ r will not be examined.


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Uniform Circular MotionConsider an object moving around acircular path of radius, r with aconstant linear speed , v

The circumference of this circle is 2r.

The time taken to complete one circle,the period, is T.

Therefore:

v = 2r / T

But frequency, f = 1 / Tand so also:v = 2r f

v

r

v

r

v

r

v

r

Note:The arrows represent the

velocityof the object. As the

directionis continually

changing, so is the velocity.


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Question

The tyre of a car, radius 40cm, rotates with afrequency of 20 Hz. Calculate (a) the period ofrotation and (b) the linear speed at the tyres edge.

(a)

(b)


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Angular displacement,

Angular displacement, is equal

to the angle swept out at the

centre of the circular path.

An object completing a complete

circle will therefore undergo an

angular displacement of 360.

circle = 180.

circle = 90.


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Angles in radians

The radian (rad) is defined as the angle

swept out at the centre of a circle whenthe arc length, sis equal to the radius, rof the circle.

If s= r

then = 1 radian

The circumference of a circle = 2r

Therefore 1 radian = 360/ 2= 57.3

And so:

360= 2radian (6.28 rad)180= radian (3.14 rad)

90= / 2 radian (1.57 rad)

r

r

s

Also: s = r


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Angular speed ()

angular speed = angular displacementtime

= / t

units:

angular displacement () inradians (rad)

time (t ) inseconds (s)

angular speed () inradians per second (rad s-1)


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Angular speed can also be measured inrevolutions per second (rev s-1) or revolutions perminute (r.p.m.)

Question:

Calculate the angular speed in rad s-1of an oldvinyl record player set at 78 r.p.m.


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Angular frequency ()

Angular frequency is the same as angular speed.

For an object taking time, Tto complete one circle ofangular displacement 2:

= 2/ T

but T = 1 / f

therefore: = 2f

that is:angular frequency = 2x frequency


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Relationship between angular

and linear speed

For an object taking timeperiod, Tto complete a circleradius r:

= 2/ Trearranging: T = 2/

but: v = 2r / T

= 2r / (2/ )

Therefore:

v = r

and:

= v / r


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Question

A hard disc drive, radius 50.0 mm, spins at7200 r.p.m. Calculate (a) its angular speedin rad s-1; (b) its outer edge linear speed.

(a)

(b)


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Complete

angular speed linear speed radius

6 ms-1 0.20 m

40 rad s-1 0.50 m

6 rad s-1 18 ms-1

48 cms-1 4.0 m

45 r.p.m. 8.7 cm


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Centripetal acceleration (a)An object moving along a circular

path is continually changing indirection. This means that even ifit is travelling at a constant speed,vit is also continually changing itsvelocity. It is therefore undergoingan acceleration, a.

This acceleration is directedtowards the centre (centripetal) ofthe circular path and is given by:

a = v2

r

v

r

a

v

r

a


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but: v = r

combining this with: a = v2/ r

gives:

a = r2

and also:

a = v


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Complete

angular

speed

linear speed radius centripetal

acceleration

8.0 ms-1 2.0 m

2.0 rad s-1 0.50 m

9.0 rad s-1 27 ms-1

6.0 ms-1 9.0 ms-2

33r.p.m. 1.8 ms-2


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ISS Question

For the International SpaceStation in orbit about theEarth (ISS) Calculate:

(a) the centripetalacceleration and

(b) linear speed

Data:

orbital period = 90 minutes

orbital height = 400km

Earth radius = 6400km


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(a)

(b)


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Proof of: a = v2/ rNOTE: This is not required for A2 AQA Physics

Consider an object moving atconstant speed, vfrom point Atopoint Balong a circular path ofradius r.

Over a short time period, titcovers arc length, sand sweepsout angle, .

As v = s / tthen s = v t.

The velocity of the objectchanges in direction by angle as it moves from A to B.

vAA B

C

vB


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If is very small then scan be considered to be astraight line and theshape ABCto be atriangle.

Triangle ABCwill have thesame shape as the vectordiagram above.

Therefore v/ vA (or B)= s / r

-vA

vBv

The change in velocity, v

= vB- vA

Which is equivalent to:

v= vB+ (- vA)

A B

C

vBs

rr


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but s = v t

and so:

v/ v = v t / r

v/ t = v2/ r

As t approaches zero, v/ twillbecome equal to the instantaneous

acceleration, a.

Hence: a = v2/ r

In the same direction as v, towards thecentre of the circle.

-vA

vBv


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What happens when centripetal force

is removed

When the centripetalforce is removed theobject will move alonga straight linetangentially to thecircular path.


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Other examples of centripetal forces

Situation Centripetal force

Earth orbiting the Sun GRAVITY of the Sun

Car going around a bend. FRICTION on the cars tyres

Airplane banking (turning) PUSH of air on the airplanes

wings

Electron orbiting a nucleus ELECTROSTATIC attractiondue to opposite charges


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Equations for centripetal force

From Newtons 2ndlaw of motion:

F = ma

If a= centripetal acceleration

then F= centripetal force

and so:

F = m v2/ r

and F = m r 2

and F = m v


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Question 1

Calculate the centripetal tension

force in a string used to whirl amass of 200g around a horizontal

circle of radius 70cm at 4.0ms-1.


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Question 2

Calculate the maximum speed that a car of mass 800kg

can go around a curve of radius 40m if the maximumfrictional force available is 8kN.


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Question 3A mass of 300g is whirled around a

vertical circle using a piece of stringof length 20cm at 3.0 revolutionsper second.

Calculate the tension in the stringat positions:

(a) Atop

(b) Bbottom and

(c) Cstring horizontal

C

B

A


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(a) A top

(b) B bottom

(c) C horizontal string


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Question 4 meow!

Calculate the maximum speed that Pat can drive

over the bridge for Jess to stay in contact with thevans roof if the distance that Jess is from the

centre of curvature is 8.0m.


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mg

R

Forces on

Jess


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Internet LinksLadybug Revolution- PhET - Join the ladybug in an exploration ofrotational motion. Rotate the merry-go-round to change its angle, orchoose a constant angular velocity or angular acceleration. Explorehow circular motion relates to the bug's x,y position, velocity, andacceleration using vectors or graphs.

Motion in 2D- PhET - Learn about velocity and acceleration vectors.Move the ball with the mouse or let the simulation move the ball in four

types of motion (2 types of linear, simple harmonic, circle). See thevelocity and acceleration vectors change as the ball moves.

Motion produced by a force- linear & circular cases - netfirms

Uniform circular motion- Fendt

Carousel - centripetal force - Fendt

Relation between speed and centripetal force - NTNUVertical circle & force vectors- NTNU

Circular Motion & Centripetal Force- NTNU

Inertia of a lead brick & Circular motion of a water glass - 'Whys Guy'Video Clip (3 mins) (2nd of 2 clips)
http://phet.colorado.edu/new/simulations/sims.php?sim=Ladybug_Revolutionhttp://phet.colorado.edu/new/simulations/sims.php?sim=Motion_in_2Dhttp://www.ngsir.netfirms.com/englishhtm/Motion.htmhttp://www.walter-fendt.de/ph14e/circmotion.htmhttp://www.walter-fendt.de/ph11e/carousel.htmhttp://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=192.0http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=190.0http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=6.0http://web.hep.uiuc.edu/home/MATS/WCIA/wcia_030226_2.wmvhttp://web.hep.uiuc.edu/home/MATS/WCIA/wcia_030226_2.wmvhttp://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=6.0http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=190.0http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=192.0http://www.walter-fendt.de/ph11e/carousel.htmhttp://www.walter-fendt.de/ph14e/circmotion.htmhttp://www.ngsir.netfirms.com/englishhtm/Motion.htmhttp://phet.colorado.edu/new/simulations/sims.php?sim=Motion_in_2Dhttp://phet.colorado.edu/new/simulations/sims.php?sim=Ladybug_Revolution

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a2 h 41b Circularmotion