Lecture 3 Physics

8/3/2019 Lecture 3 Physics

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Phy 101 Lecture (Staggs) 10/6/2011

Week #3 Lecture Notes 1

Lecture 3

CircularMotion

Physics 1016 Oct 2011

A ball on a string is

swung in a verticalcircle. The stringhappens to break whenit is parallel to theground and the ball ismoving up. Whichtrajectory does theball follow?!

1) a2) b3) c4) d

Breaking Strings


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Kinematics with constantangular acceleration

x= v0+

1

2at

2

v = v0+ at

v2= v

0

2+ 2ax

x=1

2

v0

+ v

( )t

Remember this?

"=1

2#

0+#( )t

"=#0t+

1

2$.t

2

" ="0+#.t

"2="

0

2+ 2#$

Rotational kinematicsare analogous to one-dimensional linearkinematics!

x --> !

v --> "

a --> #

s = R!(arclength)v = R" (linear speed, tangential)a = R#(tangential acceleration)

!s

R

v

Merry-go-round: radius R=2.0 mBoy running around outside of merry-go-round: v=0.50 m/sAt t=0, favorite pony is 1/4 turn ahead of him.

At t=0, merry-go-round begins to move: # =0.010 rad/s2 !

Kinematics with constant

angular accelerationWhen doeshe catch thepony?!

What is his angular speed?!v=R"b: "b=v/R= (0.50 m/s)/(2.0m)"b=0.250 rad/s

What is his !b(t)?! !b= "bt

What is the ponys !p(t)?! !p= 1/2#t2

Boy catches pony when !b=!p+(2$)/4.!

!b= "bt = 1/2#t2+$/2 !p

!b

Solve quadratic equation for t: t = 7.4s, 43 s!


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Even if the speed arounda circle is constant, thevelocity vector is NOTconstant.

Circular Motion

r

vmmaF

2

==!

where the sum is over all forces in the radial direction

Therefore there is anacceleration, ac=v

2/r.

Newtons 2nd law:

N

mg

fs

Rotational MotionConsider the simpler ride (as in the demo) in whichthe axis of rotation remains vertical.

If the coefficient of static friction is s, and the radius isR, how large must " be before the floor can drop out?

Newtons 2nd law:

N=mac=mv2/R=m"2R=mg/s, so

that:%

Require fs=mg.

Since fs,max=sN,require N=mg/s.

" =

g

Rs

Answer

isinde

penden

tofthe

massesof

ther

iders!


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DEMO: Buckets

Ball in a cup

mg

N

Ball falls out when swingslowly so that N --> 0.

DEMO: Pendulum

g

ac=v2/R

T-mg=mv2/R

T

mgR !

v is not constant; fastest at bottom

T-mg cos !=mv2/R

T=mg+mv2/R

T=mg cos !+mv2/Rmgac=v2/R

T mg cos !

ac=v2/Rmg sin !

T


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Frames of Reference

! Galilean velocity transformation between inertialframes: Let us choose a particularly simple casewhere v1 = 0.

! Figure out the signs by imagining this case!

! Aside from negligible air resistance, there is noacceleration in the x direction in free fall.

! The ball goes straight up and straight down: Ican catch it!

v2 = velocity of anobject as measured inthe lab

v1= velocity of anobject as measured inthe moving frame

Vframe = velocity of theframe as measured inthe lab

Inertial frames are those which arenot undergoing acceleration.

Newtons Chair:

Noninertial Frames

! The launcher on the cart (moving forward at constantspeed) launches the ball straight up. Where will it land ifthe cart accelerates?

A" At the position of the launcher!

B" In front of the launcher!

C" Behind the launcher


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1. n> w.2. n= w.3. n< w.4. We cant tell about nwithout knowing v.

A car is rolling overthe top of a hill atspeed v. At thisinstant,

Apparent Weight

Definition: apparent weight is the magnitude ofthe contact force(s) that supports an object.

Heavy Balloons

1) No accelerationBalloon hangs vertical

2) Forward accelerationBalloon leans backwardso that Tx=ma.

a

!


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Helium Balloons

1)No accelerationBalloon is Vertical

2) Forward acceleration

Balloon leans Forward

a

!

Helium Balloons

a

!

1)No accelerationBalloon is Vertical

2) Forward acceleration

Balloon leans Forward


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Solid Body Rotation

Thanks to Florence Hsiao (& Bill Watterson)

Apparent Weight at the

Equator vs the Pole

Consider person standing at the equator. What is his linear speed?

The Earth rotates with T= 1 day = 24*3600 s.

Thus " = 2$f = 2$/T = 2$/(86400 s) = 7.3 x 10-5 rad/s.

Finally, v = RE " =(6.4 x106 m)(7.3 x10-5rad/s) = 470 m/s ~ 1000 mph!

What is the linear speed of a personstanding at the South Pole?

v = R" = 0*" = 0 (R measured from person to axis of rotation!)

What is the apparent weight of person at the equator?

N

mg

ac=v2/REmg-N = mac: N=mg(1-v2/(gRE))

v2/(gRE) = 0.003

You weigh 0.3% less at the equator!


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Conical Pendulum

L

R

! T

mg

Tcos !

mgTsin !

ac=v2/R

y: Tcos !=mg --> T= mg/cos !%

x: Tsin !=mv2/R = (mg/cos !)sin ! =mg tan !-->

v = Rgtan"

Given R and !, what is v of ball?%

END

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