Lecture 24 - UIUC

6
10-30 - 17 r Lecture # 24 .

Transcript of Lecture 24 - UIUC

Page 1: Lecture 24 - UIUC

10-30 - 17

r

Lecture# 24

.

Page 2: Lecture 24 - UIUC

Curved Surfaces and Rolling( cont 'd )

÷; Tw

£t•cno

slip £1. C

Xx#%ftfate V En

In both these scenarios we have VI. =J

To = Exotic =rwe^+

Acceleration -1

-

= .

-

.

-

.

-

s , fr / i

inside of -j| -

1 •

'

curve .

- ;*Eec/-

know that point -oxzj'-

-

0

follows a circular path-

,\

.

of

rhdiusfr.edu#rfIen\Generally \#

I

.

;r r

\ • ai-

/theoierxe.io#en1¥ . -551p=

radius of curve

This gave us Et=( ¥2)£n at A .

Page 3: Lecture 24 - UIUC

Outsideofcurverw¥÷aftertaxaT=rxE€5¥enn=Let's derive as+

= AT + Ex # - wzrtaI×e^n=÷=rx#frw¥en+kkIxtrei)

=( rw)2

- w2( renn )Fr

e^n - rw2e^n

=fIY÷)ei= . ( If)crw÷If the surface is flat EA= - ramen

,

so the outside of a curve tends to lessenaccelerations

.

Page 4: Lecture 24 - UIUC

example :

f(×)=Io # Tw ,-

s * a. ±

TenthiO-

Given :WIIO,tyFind velocities and accelerations of points A$C .

Solution velocities NI =o

slope

:-|FKK - tax =D ffz )= - 1

TIE.

⇐ tafia );en=t'±)

vT=w-xrsI= ( 10k )x( Ft ) ( ftg )

= # G. e) B-

acceleration x= - 1

,WHO ,r= I

have at = - (Pj+w÷)enWe need p : recall the formula f- =

If "Nl

f "lx)= - ± ; -Mzs= . ,

¥45312

as staffed

Page 5: Lecture 24 - UIUC

To get AT =rx E- t TIEN Tt

examine

O.TW

area,fcxtsinkx

#c_kr:B←

to@ Given

A

/\.E=

⇐Ww←

p ¥sWhich of the other points could have velocity -8 ?

@ A

Q B IF To,=Zp twxrpaw

@ C perpendicular

@ none is possible.

to Fa

Page 6: Lecture 24 - UIUC

0Given €1 µfdatn.ie#/

Which of the following could represent the

Ea and ea ?

@ If TEA

@ Is tea

@ ←Ec TEA 15

@ Tea 9€

FEE⇐ io

@w= 10

• C\

° CzEEet

�9� 119.11 > 11*11 @ 118911 >[email protected]

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