Download - MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

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Page 1: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

MΓ‘quina de Atwood

𝐓

𝐓

π‘š1𝐠

π‘š2𝐠

π‘š1𝑔 βˆ’ 𝑇 = π‘š1π‘Ž1

𝑇 βˆ’π‘š2 𝑔 = π‘š2π‘Ž2

π‘Ž1 = π‘Ž2 = π‘Ž

π‘Ž =(π‘š1 βˆ’π‘š2)

(π‘š1 +π‘š2)𝑔

𝑇 =2π‘š1π‘š2

(π‘š1 +π‘š2)𝑔 Que forΓ§a Γ©

sentida pelo teto?

Page 2: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

π‘Ž1 m

π‘Ž =(π‘š1 βˆ’π‘š2)

(π‘š1 +π‘š2)𝑔

π‘š1 = 560 gπ‘š2 = 550 g

π‘Ž = 0,088 m/s2

1 m = 12π‘ŽΞ”π‘‘2

Δ𝑑 = 4,7 Β± 0,1 s

𝑔 = 9,8 m/s2

π‘Ž

Page 3: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

https://www.youtube.com/watch?v=4ovhEkSIqV0

Page 4: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

𝐿/2

𝐿𝐿 𝐿/3

Deslocamentos envolvendo polias mΓ³veis

Obs: cada caso Γ© um caso....

Page 5: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

Γ‰ possΓ­vel acelerar π‘š1 com π‘Ž1 > 𝑔 ?

π‘Ž1 =π‘š2

π‘š1 +π‘š2𝑔 π‘Ž1 =

π‘š2

2π‘š1 +π‘š2/2𝑔 π‘Ž1 =

π‘š2

π‘š1/2 + π‘š2𝑔

Page 6: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

O caso III em detalhe

Ξ”π‘₯2π‘š2

π‘š2

= 2Ξ”π‘₯1

Ξ”π‘₯1

Page 7: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

3a Lei de Newton

Mola Slinky

Page 8: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

3a Lei = Lei da Ação e Reação

β€œSe o corpo A faz uma forΓ§a 𝐅 no corpo B, o corpo B faz uma forΓ§a βˆ’π… no corpo A”

EvidΓͺncia experimental: conservação do momento linear total de um par de partΓ­culas interagentes (veremos mais adiante no curso)

Page 9: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

𝐅𝑂𝐢

𝐅𝐢𝑂(= 𝐍)

𝐅𝑆𝑇𝐅𝑇𝑆

𝐅𝑇𝑀

𝐅𝑀𝑇

Page 10: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

Um problema

𝐅

1

2

3

𝐅

3π‘š

𝐹 βˆ’π‘šπ‘” βˆ’ 𝑓21 = π‘šπ‘Ž

𝑓12 βˆ’π‘šπ‘” βˆ’ 𝑓32 = π‘šπ‘Ž

𝑓23 βˆ’π‘šπ‘” = π‘šπ‘Ž

π‘Ž =𝐹

3π‘šβˆ’ 𝑔 𝑓23 = 𝑓32=

𝐹

3𝑓12 = 𝑓21=

2𝐹

3

Qual é a aceleração do conjunto? Quais são as forças entre os elos?

𝐹 βˆ’ 3π‘š 𝑔 = 3π‘š π‘Ž

Page 11: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

Corda massiva suspensa

TensΓ£o nΓ£o Γ© uniforme!

E se a corda estiver em queda livre?

𝐿 βˆ’ π‘₯

π‘₯

πœ‡π‘₯ 𝑔

𝑇 π‘₯ = πœ‡π‘₯ 𝑔

𝑇

𝑇

𝑇

𝑇

Page 12: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

Mola Slinky suspensa

demonstração

Page 13: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

https://www.youtube.com/watch?v=8UimHnsWSBc

Page 14: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

Corda massiva deslizando por um prego

A massa do pequeno trechocircular é desprezível ele nãopode estar sujeito a nenhumaforça nem a nenhum torque

[πœ‡π‘₯]𝐠

[πœ‡ 𝐿 βˆ’ π‘₯ ]𝐠 π‘₯𝐿 βˆ’ π‘₯

𝐓𝐷𝐓𝐸

βˆ’π“π·βˆ’π“πΈ

|𝐓𝐸| = |𝐓𝐷|

Page 15: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

[πœ‡π‘₯]𝐠

[πœ‡ 𝐿 βˆ’ π‘₯ ]𝐠 π‘₯𝐿 βˆ’ π‘₯

π“π“πœ‡π‘₯ π‘Ž = [πœ‡π‘₯]𝑔 βˆ’ 𝑇

πœ‡ 𝐿 βˆ’ π‘₯ π‘Ž = 𝑇 βˆ’ [πœ‡ 𝐿 βˆ’ π‘₯ ]𝑔

π‘Ž =2π‘₯ βˆ’ 𝐿

𝐿𝑔

π‘Ž = 0 quando π‘₯ = 𝐿/2; π‘Ž = 𝑔 quando π‘₯ = 𝐿.

Page 16: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

Desafio: em quanto tempo o fio abaixo, solto em repouso, vai passar pelo prego?

2𝐿/3𝐿/3

Obs: Despreze o pequeno trecho sobre o prego

Page 17: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

Globo da Morte

Page 18: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

https://www.youtube.com/watch?v=qn1jWi_Po9c

Page 19: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

π―π‘‘π‘œπ‘π‘œ

π‘šπ π

Qual Γ© a menor velocidade no topo que permite executar o loop vertical?

π‘šπ‘£π‘‘π‘œπ‘π‘œ2

𝑅= 𝑁 +π‘šπ‘”

𝑅 π‘šπ‘£π‘‘π‘œπ‘π‘œ2

𝑅β‰₯ π‘šπ‘”

π‘£π‘‘π‘œπ‘π‘œ β‰₯ 𝑔𝑅

𝑅 ~ 2m ⟢ π‘£π‘‘π‘œπ‘π‘œ > 16 km/h

Page 20: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

https://www.youtube.com/watch?v=oYUfkpyvdOg

Page 21: MΓ‘quina de AtwoodΒ‘quina de Atwood 𝐓 𝐓 π‘š1𝐠 π‘š2𝐠 π‘š1 βˆ’π‘‡=π‘š1π‘Ž1 π‘‡βˆ’π‘š2 =π‘š2π‘Ž2 π‘Ž1=π‘Ž2=π‘Ž π‘Ž= (π‘š1βˆ’π‘š2) (π‘š1+π‘š2) 𝑇= ... 3a

Os copos nΓ£o executam um MCU na vertical

π‘šπ π“

π―π―π‘‘π‘œπ‘π‘œ

|π―π‘‘π‘œπ‘π‘œ| β‰₯ 𝑔𝑅

Para a Γ‘gua permanecer no copo no topo:

Obs: no caso da Γ‘gua, 𝐓 β†’ 𝐍

π…π‘‘π‘œπ‘‘

π―π‘π‘Žπ‘–π‘₯π‘œ Quem tem que ser |π―π‘π‘Žπ‘–π‘₯π‘œ|

para isso acontecer?


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