Resolução Moyses volume 2 cap. 3

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𝐶𝑜𝑚𝑜 𝑜𝑠 𝑒𝑓𝑒𝑖𝑡𝑜𝑠 𝑑𝑒 𝑑𝑖𝑠𝑠𝑖𝑝𝑎çã𝑜 𝑑𝑒 𝑒𝑛𝑒𝑟𝑔𝑖𝑎 𝑠ã𝑜 𝑑𝑒𝑠𝑝𝑟𝑒𝑠í𝑣𝑒𝑖𝑠 𝑒𝑛ã𝑜 ℎá 𝑎çã𝑜 𝑑𝑒 𝑎𝑙𝑔𝑢𝑚𝑎 
𝑓𝑜𝑟ç𝑎 𝑒𝑥𝑡𝑒𝑟𝑛𝑎, 𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑙𝑖𝑛𝑒𝑎𝑟 𝑡𝑜𝑡𝑎𝑙 �⃗� 𝑠𝑒 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎: 
 �⃗� 𝑖 = �⃗� 𝑓 
𝑚𝑣 = 𝑣𝑓(𝑚 + 𝑀) 
𝑣𝑓 =
𝑚𝑣
(𝑚 + 𝑀)
 
 
𝐴𝑔𝑜𝑟𝑎 𝑣𝑎𝑚𝑜𝑠 𝑎𝑛𝑎𝑙𝑖𝑠𝑎𝑟 𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑝𝑒𝑙𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑙𝑒𝑖 𝑑𝑒 𝑁𝑒𝑤𝑡𝑜𝑛: 
 
 𝐹 𝑒𝑙 𝐹 𝑒𝑙 
 
𝑜𝑏𝑠. : 𝐴 𝑓𝑜𝑟ç𝑎 𝑓𝑜𝑟ç𝑎 𝑎𝑝𝑎𝑟𝑒𝑐𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑎 𝑝𝑜𝑟𝑞𝑢𝑒 𝑠𝑒𝑚𝑝𝑟𝑒 𝑠𝑒 𝑜𝑝õ𝑒 𝑎𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 
𝑚𝑎 = −𝐹𝑒𝑙 
(𝑚 + 𝑀)�̈� = −𝑘𝑥 
�̈� +
𝑘
(𝑚 + 𝑀)
𝑥 = 0 
 
𝑝𝑜𝑟 𝑐𝑜𝑛𝑣𝑒𝑛𝑖ê𝑛𝑐𝑖𝑎, 𝑐ℎ𝑎𝑚𝑜: 
𝑘
(𝑚+𝑀)
= 𝜔2 
�̈� + 𝜔2 𝑥 = 0 
 
𝐴 𝑠𝑜𝑙𝑢çã𝑜 𝑑𝑒𝑠𝑠𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑗á é 𝑠𝑎𝑏𝑖𝑑𝑎 𝑐𝑜𝑚𝑜: 
 
𝑜𝑛𝑑𝑒 𝑎 𝑒 𝑏 𝑠ã𝑜 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑠 𝑞𝑢𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑟ã𝑜 𝑜 𝑐𝑜𝑚𝑝𝑜𝑟𝑡𝑎𝑚𝑒𝑛𝑡𝑜 𝑑𝑒𝑠𝑠𝑎 𝑜𝑠𝑐𝑖𝑙𝑎çã𝑜. 
 
𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒, 𝑒𝑚 𝑡 = 0, 𝑜 𝑏𝑙𝑜𝑐𝑜 𝑒𝑠𝑡á 𝑝𝑎𝑟𝑎𝑑𝑜 𝑛𝑎 𝑝𝑜𝑠𝑖çã𝑜: 𝑥(0) = 0 
𝑒𝑛𝑡ã𝑜 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑡 = 0 𝑛𝑎 𝑒𝑞. 3.2.15: 
 
𝑥(0) = acos0 + 𝑏 sin0 = 𝑎 → 𝑎 = 0 
 
𝑎𝑔𝑜𝑟𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑚𝑜𝑠 𝑎 𝑒𝑞. 3.2.15 𝑝𝑎𝑟𝑎 𝑠𝑎𝑏𝑒𝑟𝑚𝑜𝑠 𝑎 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑𝑒: 
 
�̇�(𝑡) = −ωa sinωt + ω𝑏 cosωt 
 
𝑒 𝑎 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑𝑒 𝑖𝑛𝑖𝑐𝑖𝑎𝑙 𝑑𝑜 𝑏𝑙𝑜𝑐𝑜 𝑣𝑓 𝑗á 𝑠𝑎𝑏𝑒𝑚𝑜𝑠: �̇�(0) = 𝑣𝑓 =
𝑚𝑣
(𝑚+𝑀)
 
𝑙𝑜𝑔𝑜: 
�̇�(0) = −ωa sin 0 + ω𝑏 cos 0 = ω𝑏 =
𝑚𝑣
(𝑚 + 𝑀)
 
𝑏 =
𝑚𝑣
ω(𝑚 + 𝑀)
 
𝑠𝑎𝑏𝑒𝑛𝑑𝑜 𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑠, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑎𝑠 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑟 𝑛𝑎 𝑒𝑞. 3.2.15: 
 
𝑥(𝑡) =
𝑚𝑣
ω(𝑚 + 𝑀)
sinωt 
 
 
 
 
 𝑙0 → 𝑠𝑜𝑙𝑡𝑎𝑚𝑜𝑠 → 𝑧(𝑡) 
 𝐹 𝑒𝑙 𝐹 𝑒𝑙 
 �⃗� 
𝑉𝑎𝑚𝑜𝑠 𝑎𝑛𝑎𝑙𝑖𝑠𝑎𝑟 𝑎𝑠 𝑓𝑜𝑟ç𝑎𝑠 𝑞𝑢𝑒 𝑎𝑡𝑢𝑎𝑚 𝑛𝑜 𝑠𝑖𝑠𝑡𝑒𝑚𝑎: 
�⃗� − 𝐹 𝑒𝑙 = 𝑚𝑎 
𝑚𝑔 − 𝑘𝑧 = 𝑚�̈� 
𝑔 =
𝑘
𝑚
𝑧 + �̈� 
 
𝑃𝑜𝑟 𝑐𝑜𝑛𝑣𝑒𝑛𝑖ê𝑛𝑐𝑖𝑎, 𝑐ℎ𝑎𝑚𝑜: 
𝑘
𝑚
= 𝜔2 
 
𝑔 = 𝜔2𝑧 + �̈� 
 
𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑑𝑎 𝑝𝑎𝑟𝑡𝑒 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎 𝑑𝑒𝑠𝑠𝑎 𝐸𝐷𝑂 (𝜔2𝑧 + �̈� = 0): 
 
 
𝐶𝑜𝑚𝑜 𝑎 𝑝𝑎𝑟𝑡𝑒 𝑛ã𝑜 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎 é 𝑢𝑚𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒: 
𝑧(𝑡) = 𝑎 cos𝜔𝑡 + 𝑏 sin𝜔𝑡 + 𝐶 
�̇�(𝑡) = −𝜔𝑎 sin𝜔𝑡 + 𝜔𝑏 cos𝜔𝑡 
�̈�(𝑡) = −𝜔2𝑎 cos𝜔𝑡 − 𝜔2𝑏 sin𝜔𝑡 
 
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑛𝑎 𝑒𝑞 𝑑𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
𝑔 = 𝜔2(𝑎 cos𝜔𝑡 + 𝑏 sin𝜔𝑡 + 𝐶) − 𝜔2𝑎 cos𝜔𝑡 − 𝜔2𝑏 sin𝜔𝑡 
𝑔 = 𝜔2𝑎 cos𝜔𝑡 + 𝜔2𝑏 sin𝜔𝑡 + 𝜔2𝐶 − 𝜔2𝑎 cos𝜔𝑡 − 𝜔2𝑏 sin𝜔𝑡 
𝑔 = 𝜔2𝐶 
𝐶 =
𝑔
𝜔2
 
𝑙𝑜𝑔𝑜: 
𝑧(𝑡) = 𝑎 cos𝜔𝑡 + 𝑏 sin𝜔𝑡 +
𝑔
𝜔2
 
�̇�(𝑡) = −𝑎𝜔 sin𝜔𝑡 + 𝑏𝜔 cos𝜔𝑡 
𝐴𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑑𝑒 𝑐𝑜𝑛𝑡𝑜𝑟𝑛𝑜 𝑠𝑒𝑟ã𝑜: 
𝑧(0) = 𝑙0 
�̇�(0) = 0 
𝑎𝑠 𝑢𝑠𝑎𝑚𝑜𝑠 𝑒𝑚 𝑧(𝑡): 
𝑧(0) = 𝑎 cos 0 + 𝑏 sin0 +
𝑔
𝜔2
= 𝑙0 
𝑧(0) = 𝑎 +
𝑔
𝜔2
= 𝑙0 
𝑎 = 𝑙0 −
𝑔
𝜔2
 
 
�̇�(0) = −𝑎𝜔 sin0 + 𝑏𝜔 cos 0 = 0 
�̇�(0) = 𝑏𝜔 = 0 
𝑐𝑜𝑚𝑜 𝜔 𝑛ã𝑜 𝑝𝑜𝑑𝑒 𝑠𝑒𝑟 𝑧𝑒𝑟𝑜: 𝑏 = 0 
 
𝑠𝑎𝑏𝑒𝑛𝑑𝑜 𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑠, 𝑎𝑠 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑒𝑚 𝑧(𝑡): 
 
𝑧(𝑡) = (𝑙0 −
𝑔
𝜔2
) cos𝜔𝑡 +
𝑔
𝜔2
 
 
 
 
 
 1𝑐𝑚 = 10−2𝑚 
𝑑𝑎𝑑𝑜𝑠: 
𝑚 = 10−2𝑘𝑔 
𝑘 = 100
𝑁
𝑚
 
𝑥(0) = 10−2𝑚 
�̇�(0) = √3𝑚/𝑠 
 
𝑉𝑎𝑚𝑜𝑠 𝑎𝑛𝑎𝑙𝑖𝑠𝑎𝑟 𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑎𝑝𝑒𝑛𝑎𝑠 𝑢𝑚𝑎 𝑝𝑎𝑟𝑡í𝑐𝑢𝑙𝑎 𝑎𝑡é 𝑖𝑚𝑒𝑑𝑖𝑎𝑡𝑎𝑚𝑒𝑛𝑡𝑒 𝑎𝑛𝑡𝑒𝑠 𝑞𝑢𝑒 
𝑒𝑙𝑎𝑠 𝑠𝑒 𝑡𝑜𝑞𝑢𝑒𝑚, 𝑗á 𝑞𝑢𝑒 𝑟𝑒𝑎𝑙𝑖𝑧𝑎𝑟ã𝑜 𝑜 𝑚𝑒𝑠𝑚𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜. 
 
−𝑘𝑥 = 𝑚𝑎 
�̈� +
𝑘
𝑚
𝑥 = 0 → �̈� + 𝜔2𝑥 = 0 
𝑜𝑛𝑑𝑒: 𝜔 = √
𝑘
𝑚
= √
100
10−2
= 102𝑠−1 
 
𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑑𝑒𝑠𝑠𝑎 𝐸𝐷𝑂 é: 
𝑥(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜑) 
�̇�(𝑡) = −𝐴𝜔 sin(𝜔𝑡 + 𝜑) 
 
𝑢𝑠𝑎𝑛𝑑𝑜 𝑜𝑠 𝑑𝑎𝑑𝑜𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑖𝑠: 
𝑥(0) = 𝐴 cos(𝜑) = 10−2 
𝐴 =
10−2
cos(𝜑)
 
 
�̇�(0) = −𝐴𝜔 sin(𝜑) = √3 
−(
10−2
cos(𝜑)
)𝜔 sin(𝜑) = √3 
−10−2𝜔
sin(𝜑)
cos(𝜑)
= √3 
−10−2𝜔 tan (𝜑) = √3 
𝑠𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒𝑚 é 𝜔: 
−10−2 × 102 tan (𝜑) = √3 
tan (𝜑) = −√3 
𝑙𝑜𝑔𝑜: 𝜑 = −
𝜋
3
 
 
𝑃𝑎𝑟𝑎 𝑎𝑐ℎ𝑎𝑟𝑚𝑜𝑠 𝑎 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒, 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑒𝑠𝑠𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜 𝑒𝑚 𝑥(0): 
𝐴 =
10−2
cos (−
𝜋
3)
=
10−2
1
2
= 0,02𝑚 
 
𝑜 𝑑𝑒𝑠𝑙𝑜𝑐𝑎𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑎𝑚𝑏𝑎𝑠 𝑎𝑠 𝑝𝑎𝑟𝑡í𝑐𝑢𝑙𝑎𝑠 𝑠𝑒𝑟á: 
 
𝑥(𝑡) = 0,02 cos (102𝑡 −
𝜋
3
) 
 
𝑏) 𝑎𝑠 𝑝𝑎𝑟𝑡í𝑐𝑢𝑙𝑎𝑠 𝑐𝑜𝑙𝑖𝑑𝑖𝑟ã𝑜 𝑛𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒 𝑎 𝑝𝑜𝑠𝑖çã𝑜 (𝑥(𝑡))𝑣𝑜𝑙𝑡𝑎𝑟 𝑎 𝑧𝑒𝑟𝑜: 
 
𝑥(𝑡) = 0,02 cos (102𝑡 −
𝜋
3
) = 0 
cos (102𝑡 −
𝜋
3
) = 0 
∴ 
102𝑡 −
𝜋
3
=
𝜋
2
 
102𝑡 =
5𝜋
6
 
𝑡 =
𝜋
120
 𝑠 
 
 
 
 
𝑃𝑜𝑟 𝑠𝑒 𝑡𝑟𝑎𝑡𝑎𝑟 𝑑𝑒 𝑢𝑚 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟, 𝑣𝑎𝑚𝑜𝑠 𝑢𝑠𝑎𝑟 𝑢𝑚𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑑𝑒 𝑡𝑜𝑟𝑞𝑢𝑒: 
 
𝜏 = 𝑟 × 𝐹 
 
 𝑟 
 
 �⃗� 
 
 
 
𝑗𝑢𝑛𝑡𝑎𝑛𝑑𝑜 𝑎𝑠 𝑒𝑞𝑢𝑎çõ𝑒𝑠 𝑎𝑐𝑖𝑚𝑎: 
𝐼𝛼 = −𝑚𝑔𝑟 sin𝜃 
𝐼�̈� + 𝑚𝑔𝑟 sin 𝜃 = 0 
𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 𝐼 𝑑𝑒 𝑢𝑚𝑎 𝑝𝑎𝑟𝑡í𝑐𝑢𝑙𝑎 é 𝑑𝑎𝑑𝑜 𝑝𝑜𝑟: 
𝐼 = 𝑚𝑟2 
𝑒𝑛𝑡ã𝑜: 
𝑚𝑟2�̈� + 𝑚𝑔𝑟 sin𝜃 = 0 
�̈� +
𝑔
𝑟
sin𝜃 = 0 
𝑐𝑜𝑚𝑜 𝑒𝑠𝑡𝑎𝑚𝑜𝑠 𝑡𝑟𝑎𝑡𝑎𝑛𝑑𝑜 𝑑𝑒 â𝑛𝑔𝑢𝑙𝑜𝑠 𝑚𝑢𝑖𝑡𝑜 𝑝𝑒𝑞𝑢𝑒𝑛𝑜𝑠: 
sin𝜃 ≅ 𝜃 
 
�̈� +
𝑔
𝑟
𝜃 = 0; 
𝑔
𝑟
= 𝜔2 
�̈� + 𝜔2𝜃 = 0 
𝑆𝑒𝑛𝑑𝑜 𝑒𝑠𝑡𝑎, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑢𝑚𝑎 𝐸𝐷𝑂 𝑐𝑢𝑗𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑗á 𝑐𝑜𝑛ℎ𝑒𝑐𝑒𝑚𝑜𝑠: 
𝜃(𝑡) = 𝑎 cos𝜔𝑡 + 𝑏 sin𝜔𝑡 
𝑒, 𝑎𝑠𝑠𝑖𝑚, 𝑠𝑒 𝑡𝑟𝑎𝑡𝑎𝑛𝑑𝑜 𝑑𝑒 𝑢𝑚 𝑀𝐻𝑆 (𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 ℎ𝑎𝑟𝑚ô𝑛𝑖𝑐𝑜 𝑠𝑖𝑚𝑝𝑙𝑒𝑠). 
 
𝑃𝑎𝑟𝑎 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜, 𝑢𝑠𝑎𝑟𝑒𝑚𝑜𝑠 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑒𝑞𝑢𝑎çã𝑜: 
 
𝑇 =
2𝜋
𝜔
=
2𝜋
√
𝑔
𝑟
= 2𝜋√
𝑟
𝑔
 
 
 
 
 
 𝐹 𝑒𝑙 ℎ 
 
 �⃗� 𝑘 
 
 
𝑉𝑎𝑚𝑜𝑠 𝑢𝑠𝑎𝑟 𝑎 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎çã𝑜 𝑑𝑎 𝑒𝑛𝑒𝑟𝑔𝑖𝑎 𝑚𝑒𝑐â𝑛𝑖𝑐𝑎 𝑝𝑎𝑟𝑎 𝑠𝑎𝑏𝑒𝑟 𝑐𝑜𝑚 𝑞𝑢𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑𝑒 𝑎 
𝑚𝑎𝑠𝑠𝑎 𝑐ℎ𝑒𝑔𝑎 𝑛𝑎 𝑏𝑎𝑙𝑎ç𝑎: 
𝐸𝑖 = 𝐸𝑓 
𝑈𝑖 + 𝑇𝑖 = 𝑈𝑓 + 𝑇𝑓 
𝑐𝑜𝑚𝑜 𝑇𝑖 = 𝑈𝑓 = 0: 
𝑈𝑖 = 𝑇𝑓 
𝑚𝑔ℎ =
1
2
𝑚𝑣2 
𝑣 = √2𝑔ℎ 
 
𝐷𝑒𝑐𝑜𝑚𝑝𝑜𝑛𝑑𝑜 𝑎𝑠 𝑓𝑜𝑟ç𝑎𝑠: 
𝑃 − 𝐹𝑒𝑙 = 𝑚𝑎 
𝑚𝑔 − 𝑘𝑧 = 𝑚�̈� 
�̈� + 𝜔2𝑧 = 𝑔 
𝑜𝑛𝑑𝑒: 𝜔2 =
𝑘
𝑚
 
𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑑𝑎 𝑝𝑎𝑟𝑡𝑒 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎 (�̈� + 𝜔2𝑧 = 0): 𝑧(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜑) 
𝐶𝑜𝑚𝑜 𝑎 𝑝𝑎𝑟𝑡𝑒 𝑛ã𝑜 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎 é 𝑢𝑚𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒: 
𝑧(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜑) + 𝐶 
�̇�(𝑡)= −𝜔𝐴sin(𝜔𝑡 + 𝜑) 
�̈�(𝑡) = −𝜔2𝐴cos(𝜔𝑡 + 𝜑) 
 
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑛𝑎 𝑒𝑞 𝑑𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
𝑔 = 𝜔2(𝐴 cos(𝜔𝑡 + 𝜑) + 𝐶) − 𝜔2𝐴cos(𝜔𝑡 + 𝜑) 
𝑔 = 𝜔2𝐴cos(𝜔𝑡 + 𝜑) + 𝜔2𝐶 − 𝜔2𝐴cos(𝜔𝑡 + 𝜑) 
𝑔 = 𝜔2𝐶 
𝐶 =
𝑔
𝜔2
 
𝑙𝑜𝑔𝑜: 
𝑧(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜑) +
𝑔
𝜔2
 
�̇�(𝑡) = −𝜔𝐴sin(𝜔𝑡 + 𝜑) 
𝐴𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑖𝑠 𝑠𝑒𝑟ã𝑜: 
𝑧(0) = 0 
�̇�(0) = √2𝑔ℎ 
𝑙𝑜𝑔𝑜: 
𝑧(0) = 𝐴 cos(𝜑) +
𝑔
𝜔2
= 0 
𝐴 cos(𝜑) = −
𝑔
𝜔2
 (1) 
 
�̇�(0) = −𝜔𝐴sin(𝜑) = √2𝑔ℎ 
𝐴 sin(𝜑) = −
√2𝑔ℎ
𝜔
 (2) 
 
𝑡𝑒𝑚𝑜𝑠 𝑢𝑚 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑐𝑜𝑚 𝑎𝑠 𝑒𝑞. 𝑠 (1) 𝑒 (2): 
{
𝐴 cos(𝜑) = −
𝑔
𝜔2
𝐴 sin(𝜑) = −
√2𝑔ℎ
𝜔
 
𝑒𝑙𝑒𝑣𝑎𝑚𝑜𝑠 𝑎𝑚𝑏𝑎𝑠 𝑎𝑜 𝑞𝑢𝑎𝑑𝑟𝑎𝑑𝑜: 
{
𝐴2 cos2(𝜑) =
𝑔2
𝜔4
𝐴2 sin2(𝜑) =
2𝑔ℎ
𝜔2
 
𝑎𝑔𝑜𝑟𝑎 𝑎𝑠 𝑠𝑜𝑚𝑎𝑚𝑜𝑠: 
𝐴2(cos2(𝜑) + sin2(𝜑)) =
𝑔2
𝜔4
+
2𝑔ℎ
𝜔2
 
𝐴2 =
𝑔2
𝜔4
+
2𝑔ℎ
𝜔2
 
𝑜𝑛𝑑𝑒: 𝜔2 =
𝑘
𝑚
 
𝐴2 =
𝑚2𝑔2
𝑘2
+
2𝑔ℎ𝑚
𝑘
 
𝑝𝑟𝑜 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜 𝑓𝑖𝑐𝑎𝑟 𝑏𝑜𝑛𝑖𝑡𝑜: 
𝐴2 =
𝑚2𝑔2
𝑘2
+
2𝑔ℎ𝑚
𝑘
(
𝑘
𝑘
𝑔
𝑔
𝑚
𝑚
) 
𝐴2 =
𝑚2𝑔2
𝑘2
+
2ℎ𝑘𝑚2𝑔2
𝑚𝑔𝑘2
 
𝐴 =
𝑚𝑔
𝑘
√1 +
2ℎ𝑘
𝑚𝑔
 
𝑏) 𝐴 𝑒𝑛𝑒𝑟𝑔𝑖𝑎 𝑚𝑒𝑐â𝑛𝑖𝑐𝑎 𝑠𝑒𝑟á 𝑎 𝑠𝑜𝑚𝑎 𝑑𝑎 𝑐𝑖𝑛é𝑡𝑖𝑐𝑎 𝑐𝑜𝑚 𝑎 𝑝𝑜𝑡𝑒𝑛𝑐𝑖𝑎𝑙 𝑒𝑙á𝑠𝑡𝑖𝑐𝑎: 
𝐸 = 𝑇 + 𝑈 
𝐸 =
1
2
𝑚�̇�2 +
1
2
𝑘𝑧2 
𝑜𝑛𝑑𝑒 𝑐𝑜𝑛ℎ𝑒𝑐𝑒𝑚𝑜𝑠 𝑧(𝑡)𝑒 �̇�(𝑡): 
𝐸 =
1
2
𝑚(−𝜔𝐴sin(𝜔𝑡 + 𝜑))2 +
1
2
𝑘(𝐴 cos(𝜔𝑡 + 𝜑))2 
𝐸 =
1
2
𝑚𝜔2𝐴2 sin2(𝜔𝑡 + 𝜑) +
1
2
𝑘𝐴2 cos2(𝜔𝑡 + 𝜑) 
𝑜𝑛𝑑𝑒: 𝜔2 =
𝑘
𝑚
 
𝐸 =
1
2
𝑘𝐴2 sin2(𝜔𝑡 + 𝜑) +
1
2
𝑘𝐴2 cos2(𝜔𝑡 + 𝜑) 
𝐸 =
1
2
𝑘𝐴2(sin2(𝜔𝑡 + 𝜑) + cos2(𝜔𝑡 + 𝜑) 
𝐸 =
1
2
𝑘𝐴2 
𝑜𝑛𝑑𝑒: 𝐴 =
𝑚𝑔
𝑘
√1 +
2ℎ𝑘
𝑚𝑔
 
𝐸 =
1
2
𝑘
𝑚2𝑔2
𝑘2
(1 +
2ℎ𝑘
𝑚𝑔
) 
𝐸 =
1
2
𝑚2𝑔2
𝑘
+ 𝑚𝑔ℎ 
 
 
 
𝑂 𝑡𝑜𝑟𝑞𝑢𝑒 𝑝𝑎𝑟𝑎 𝑜 𝑝ê𝑛𝑑𝑢𝑙𝑜 𝑑𝑒 𝑡𝑜𝑟çã𝑜 é 𝑑𝑒𝑓𝑖𝑛𝑖𝑑𝑜 𝑝𝑜𝑟: 
 
𝜏 = −𝐾𝜑 = 𝐼�̈� 
𝐾
𝐼
𝜑 + �̈� = 0 
𝜔2𝜑 + �̈� = 0 
𝑜𝑛𝑑𝑒: 𝜔2 =
𝐾
𝐼
 
𝑒 𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 𝑑𝑜 𝑑𝑖𝑠𝑐𝑜 (𝑎) 𝑒 𝑑𝑜 (𝑏) 𝑠ã𝑜: 
𝐼𝑎 =
1
2
𝑀𝑅2 
 
𝐼𝑏 =
1
4
𝑀𝑅2 
 
𝑠𝑒𝑛𝑑𝑜 𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜 𝑑𝑎𝑑𝑜 𝑝𝑜𝑟: 
𝑇 =
2𝜋
𝜔
 
 
𝑇𝑎 =
2𝜋
√
𝐾
𝐼𝑎
 → 𝑇𝑎 =
2𝜋
√
𝐾
1
2𝑀𝑅
2
= 𝜋𝑅√
2𝑀
𝑘
 
 
𝑇𝑏 =
2𝜋
√
𝐾
𝐼𝑏
 → 𝑇𝑎 =
2𝜋
√
𝐾
1
4𝑀𝑅
2
= 𝜋𝑅√
𝑀
𝑘
 
 
 
 𝑙 = 1𝑚 𝜃 
 
 
 10−2𝑘𝑔 10𝑘𝑔 
𝑃𝑎𝑟𝑎 𝑎𝑐ℎ𝑎𝑟 𝑎 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑𝑒 𝑞𝑢𝑒 𝑜 𝑝ê𝑛𝑑𝑢𝑙𝑜 𝑎𝑑𝑞𝑢𝑖𝑟𝑖, 𝑢𝑠𝑎𝑚𝑜𝑠 𝑎 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎çã𝑜 𝑑𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑡𝑜𝑡𝑎𝑙: 
 �⃗� 𝑖 = �⃗� 𝑓 
𝑚𝑣 = 𝑣𝑓(𝑚 + 𝑀) 
𝑣𝑓 =
𝑚𝑣
(𝑚 + 𝑀)
 
𝑣𝑓 =
10−2 × 300
(10−2 + 10)
= 0,3𝑚/𝑠 
 
𝐴𝑔𝑜𝑟𝑎 𝑢𝑠𝑎𝑚𝑜𝑠 𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑐𝑜𝑛ℎ𝑒𝑐𝑖𝑑𝑎 𝑑𝑜 𝑡𝑜𝑟𝑞𝑢𝑒: 
𝐼�̈� + (𝑀 + 𝑚)𝑔𝑙 sin𝜃 = 0 
𝑒 𝑐𝑜𝑛ℎ𝑒𝑐𝑖𝑑𝑜 𝑡𝑎𝑚𝑏é𝑚, 𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 𝑑𝑜 𝑝ê𝑛𝑑𝑢𝑙𝑜: 
𝐼 = 𝑀𝑅2 
𝑛𝑜 𝑐𝑎𝑠𝑜: 𝐼 = (𝑀 + 𝑚)𝑙2 
𝑙𝑜𝑔𝑜: 
(𝑀 + 𝑚)𝑙2�̈� + (𝑀 + 𝑚)𝑔𝑙 sin 𝜃 = 0 
𝑝𝑎𝑟𝑎 𝑜𝑠𝑐𝑖𝑙𝑎çõ𝑒𝑠 𝑚𝑢𝑖𝑡𝑜 𝑝𝑒𝑞𝑢𝑒𝑛𝑎𝑠: sin 𝜃 ≈ 𝜃 
�̈� +
𝑔
𝑙
𝜃 = 0 
�̈� + 𝜔2𝜃 = 0 
𝐶𝑜𝑛ℎ𝑒𝑐𝑒𝑚𝑜𝑠 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑑𝑒𝑠𝑠𝑎 𝐸𝐷𝑂: 
𝜃(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜑) 
�̇�(𝑡) = −𝜔𝐴sin(𝜔𝑡 + 𝜑) 
 
𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑖𝑠: 
𝜃(0) = 0 
�̇�(0) = 0,3 𝑚/𝑠 
 
𝐴𝑝𝑙𝑖𝑐𝑎𝑚𝑜𝑠 𝑖𝑠𝑠𝑜 𝑒𝑚 𝜃(𝑡): 
𝜃(0) = 𝐴 cos(𝜑) = 0 
𝐶𝑜𝑚𝑜 𝐴 ≠ 0: 
 cos𝜑 = 0 ∴ 𝜑 =
𝜋
2
 
 
�̇�(0) = −𝜔𝐴 sin(𝜑) = 0,3 
�̇�(0) = −𝜔𝐴sin (
𝜋
2
) = 0,3 
�̇�(0) = −𝜔𝐴 = 0,3 
𝑉𝑎𝑚𝑜𝑠, 𝑎𝑔𝑜𝑟𝑎 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝜔2 =
𝑔
𝑙
: 
𝜔 = √
9,8
1
≅ 3,13 𝑠−1 
𝑉𝑜𝑙𝑡𝑎𝑛𝑑𝑜… 
�̇�(0) = −3,13 × 𝐴 = 0,3 
𝐴 = −0,096 𝑚 
𝑁𝑎 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒, 𝑜 𝑠𝑖𝑛𝑎𝑙 𝑛ã𝑜 𝑓𝑎𝑧 𝑑𝑖𝑓𝑒𝑟𝑒𝑛ç𝑎, 𝑗á 𝑞𝑢𝑒 𝑣𝑎𝑙𝑒 𝑡𝑎𝑛𝑡𝑜 𝑝𝑟𝑎 𝑐𝑖𝑚𝑎(+)𝑞𝑢𝑎𝑛𝑡𝑜 𝑝𝑎𝑟𝑎 𝑏𝑎𝑖𝑥𝑜 (−) 
𝑙𝑜𝑔𝑜: 
𝜃(𝑡) = 0,096 cos (3,13𝑡 +
𝜋
2
 ) 
𝑞𝑢𝑒 𝑡𝑎𝑚𝑏é𝑚 𝑝𝑜𝑑𝑒 𝑠𝑒𝑟 𝑒𝑠𝑐𝑟𝑖𝑡𝑜 𝑎𝑠𝑠𝑖𝑚: 
𝜃(𝑡) = 0,096 sin(3,13𝑡 ) 
 
 
𝑉𝑎𝑚𝑜𝑠 𝑎𝑛𝑎𝑙𝑖𝑠𝑎𝑟 𝑐𝑎𝑑𝑎 𝑏𝑙𝑜𝑐𝑜 𝑠𝑒𝑝𝑎𝑟𝑎𝑑𝑎𝑚𝑒𝑛𝑡𝑒, 𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑛𝑡𝑜 𝑎𝑠 𝑓𝑜𝑟ç𝑎𝑠 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑎 𝑙𝑒𝑔𝑒𝑛𝑑𝑎: 
 
𝑁𝑂𝑅𝑀𝐴𝐿 𝐴𝑇𝑅𝐼𝑇𝑂 𝑃𝐸𝑆𝑂 𝐸𝐿Á𝑆𝑇𝐼𝐶𝐴 
 
1º 𝐵𝑙𝑜𝑐𝑜 𝑚: 
 
 
 
2º 𝐵𝑙𝑜𝑐𝑜 𝑀: 
 
 
 
 
𝑝𝑒𝑟𝑐𝑒𝑏𝑒𝑚𝑜𝑠, 𝑒𝑛𝑡ã𝑜 𝑞𝑢𝑒, 𝑜𝑛𝑑𝑒 𝑛𝑜 𝑏𝑙𝑜𝑐𝑜 𝑀 𝑎 𝑓𝑜𝑟ç𝑎 𝑑𝑒 𝑟𝑒𝑠𝑡𝑎𝑢𝑟𝑎çã𝑜 é 𝑎 𝑎𝑙á𝑠𝑡𝑖𝑐𝑎, 
𝑛𝑜 𝑏𝑙𝑜𝑐𝑜 𝑚, é 𝑎 𝑑𝑒 𝑎𝑡𝑟𝑖𝑡𝑜. 
𝐶𝑜𝑚𝑜 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑚𝑜𝑙𝑎 𝑛𝑜 𝑏𝑙𝑜𝑐𝑜 𝑚, 𝑒𝑙𝑒 𝑜𝑠𝑐𝑖𝑙𝑎 𝑝𝑜𝑟𝑞𝑢𝑒 𝑜 𝑏𝑙𝑜𝑐𝑜 𝑀 𝑜 segura. 
 
𝐹𝑜𝑟ç𝑎𝑠 𝑑𝑜 𝑏𝑙𝑜𝑐𝑜 𝑚 𝑛𝑎 𝑑𝑖𝑟𝑒çã𝑜 𝑥: 
−𝐹𝑎𝑡 = 𝑚𝑎 
−𝜇𝑒𝑁 = 𝑚�̈� 
𝑃𝑎𝑟𝑎 𝑑𝑒𝑠𝑐𝑜𝑏𝑟𝑖𝑟 𝑁, 𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑚𝑜𝑠 𝑎𝑠 𝑓𝑜𝑟ç𝑎𝑠 𝑑𝑜 𝑏𝑙𝑜𝑐𝑜 𝑚 𝑛𝑎 𝑑𝑖𝑟𝑒çã𝑜 𝑦: 
𝑁 − 𝑃 = 0 
𝑁 = 𝑚𝑔 
𝑣𝑜𝑙𝑡𝑎𝑛𝑑𝑜… 
−𝜇𝑒𝑚𝑔 = 𝑚�̈� 
−𝜇𝑒𝑔 = �̈� (∗∗) 
 
𝐹𝑜𝑟ç𝑎𝑠 𝑑𝑜 𝑏𝑙𝑜𝑐𝑜 𝑀 𝑛𝑎 𝑑𝑖𝑟𝑒çã𝑜 𝑥: 
𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑟𝑒𝑚𝑜𝑠 𝑚 + 𝑀 𝑐𝑜𝑚𝑜 𝑢𝑚 𝑠ó 𝑐𝑜𝑟𝑝𝑜 𝑝𝑜𝑟𝑞𝑢𝑒 𝑎 𝑚𝑜𝑙𝑎 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑎 𝑎 𝑚𝑎𝑠𝑠𝑎 𝑑𝑒 
𝑎𝑚𝑏𝑜𝑠 𝑎𝑜 𝑚𝑒𝑠𝑚𝑜 𝑡𝑒𝑚𝑝𝑜: 
−𝐹𝑒𝑙 = (𝑀 + 𝑚)𝑎 
−𝑘𝑥 = (𝑀 + 𝑚)�̈� 
𝑘
(𝑀 + 𝑚)
𝑥 + �̈� = 0 
𝑜𝑛𝑑𝑒
𝑘
𝑀+𝑚
= 𝜔2 
𝜔2𝑥 + �̈� = 0 
 
𝐸 𝑗á 𝑐𝑜𝑛ℎ𝑒𝑐𝑒𝑚𝑜𝑠 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑑𝑒𝑠𝑠𝑎 𝐸𝐷𝑂: 
𝑥(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜑) 
𝑆𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑜 𝑒𝑚 𝑟𝑒𝑙𝑎çã𝑜 𝑎 𝑡: 
�̇�(𝑡) = −𝜔𝐴sin(𝜔𝑡 + 𝜑) 
�̈�(𝑡) = −𝜔2𝐴cos(𝜔𝑡 + 𝜑) 
 
É 𝑓á𝑐𝑖𝑙 𝑝𝑒𝑟𝑐𝑒𝑏𝑒𝑟 𝑞𝑢𝑒 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑏𝑙𝑜𝑐𝑜𝑠 𝑝𝑜𝑠𝑠𝑢𝑒𝑚 𝑎 𝑚𝑒𝑠𝑚𝑎 𝑝𝑜𝑠𝑖çã𝑜, 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑𝑒 𝑒 𝑎𝑐𝑒𝑙𝑒𝑟𝑎çã𝑜 
(𝑥, �̇� 𝑒 �̈�). 𝐸𝑛𝑡ã𝑜, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑟 �̈� 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 (∗∗): 
 
�̈�(𝑡) = −𝜔2𝐴cos(𝜔𝑡 + 𝜑) = −𝜇𝑒𝑔 
𝐴 cos(𝜔𝑡 + 𝜑) =
𝜇𝑒𝑔
𝜔2
 (∗∗∗) 
 
𝑝𝑒𝑟𝑐𝑒𝑏𝑎 𝑞𝑢𝑒, 𝑒𝑠𝑡𝑎 ú𝑙𝑡𝑖𝑚𝑎 𝑒𝑞𝑢𝑎çã𝑜, 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎 𝑓𝑜𝑟𝑚𝑎 𝑖𝑑ê𝑛𝑡𝑖𝑐𝑎 𝑎 𝑥(𝑡): 
𝑥(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜑) =
𝜇𝑒𝑔
𝜔2
 
𝑀𝑎𝑠 𝑎𝑖𝑛𝑑𝑎 𝑎𝑠𝑠𝑖𝑚, é 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 ? ! 
𝐼𝑠𝑠𝑜 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎 𝑞𝑢𝑒, 𝑠𝑒 𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑟𝑚𝑜𝑠 (�̇�(𝑡)), 𝑠𝑒𝑟á 𝑖𝑔𝑢𝑎𝑙 𝑎 𝑧𝑒𝑟𝑜: 
�̇�(𝑡) =
𝑑
𝑑𝑡
(𝐴 cos(𝜔𝑡 + 𝜑)) =
𝑑
𝑑𝑡
(
𝜇𝑒𝑔
𝜔2
) = 0 
𝐸𝑛𝑡ã𝑜, 𝑠𝑖𝑚, 𝑜𝑠 𝑏𝑙𝑜𝑐𝑜𝑠 𝑒𝑠𝑡ã𝑜 𝑝𝑎𝑟𝑎𝑑𝑜𝑠‼ 
𝐼𝑠𝑠𝑜 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎 𝑞𝑢𝑒 𝑜 𝑡 𝑑𝑎 𝑒𝑞𝑢𝑎çã𝑜 (∗∗∗), 𝑠𝑒 𝑡𝑟𝑎𝑡𝑎 𝑑𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒 𝑜𝑠 𝑏𝑙𝑜𝑐𝑜𝑠 
𝑎𝑡𝑖𝑛𝑔𝑒𝑚 𝑜 𝑑𝑒𝑠𝑙𝑜𝑐𝑎𝑚𝑒𝑛𝑡𝑜 𝑚á𝑥𝑖𝑚𝑜, 𝑜𝑛𝑑𝑒 cos(𝜔𝑡 + 𝜑) = 1: 
𝐴 =
𝜇𝑒𝑔
𝜔2
= 𝜇𝑒𝑔 (
𝑀 + 𝑚
𝑘
) 
 
 
 
 5 𝑐𝑚 
 
 
 
 5 𝑐𝑚 
 
 
𝐶𝑜𝑚𝑜 𝑖𝑙𝑢𝑠𝑡𝑟𝑎𝑑𝑜, 𝑓𝑎𝑟𝑒𝑚𝑜𝑠 𝑢𝑚𝑎 𝑝𝑒𝑞𝑢𝑒𝑛𝑎 𝑎𝑛𝑎𝑙𝑜𝑔𝑖𝑎: 
𝑁𝑜 𝑙𝑢𝑔𝑎𝑟 𝑑𝑒 𝑢𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑚𝑜𝑠 𝑎 𝑟𝑎𝑚𝑝𝑎 𝑐𝑜𝑚 𝑎 𝑐𝑎𝑝𝑎𝑐𝑖𝑑𝑎𝑑𝑒 𝑑𝑒 𝑠𝑒 𝑐𝑢𝑟𝑣𝑎𝑟, 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑒𝑚𝑜𝑠 𝑢𝑚𝑎𝑟𝑎𝑚𝑝𝑎 𝑠ó𝑙𝑖𝑑𝑎 𝑐𝑜𝑚 𝑢𝑚𝑎 𝑚𝑜𝑙𝑎, 𝑞𝑢𝑒 𝑓𝑎𝑟á 𝑜 𝑝𝑎𝑝𝑒𝑙 𝑑𝑎 𝑓𝑜𝑟ç𝑎 𝑟𝑒𝑠𝑡𝑎𝑢𝑟𝑎𝑑𝑜𝑟𝑎: 
 
𝐶𝑜𝑚𝑜 𝑛ã𝑜 ℎá 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
𝑃 = 𝐹𝑒𝑙 
𝑚𝑔 = 𝑘𝑧 
𝑔 =
𝑘
𝑚
𝑧 
𝑜𝑛𝑑𝑒, 𝑗á 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒:
𝑘
𝑚
= 𝜔2 
𝑔 = 𝜔2𝑧 
𝜔 = √
𝑔
𝑧
 
𝜔 = 14 𝑠−1 
 
 
 
 
𝑉𝑎𝑚𝑜𝑠 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑟 𝑞𝑢𝑒 𝑡𝑜𝑑𝑎 𝑎 𝑚𝑎𝑠𝑠𝑎 (𝑛𝑜𝑠 𝑑𝑜𝑖𝑠 𝑐𝑎𝑠𝑜𝑠), 𝑒𝑠𝑡á 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑑𝑎 𝑛𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 
𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎. 
𝑄𝑢𝑒, 𝑛𝑜 𝑐𝑎𝑠𝑜, 𝑐𝑜𝑖𝑛𝑐𝑖𝑑𝑒 𝑐𝑜𝑚 𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑜 𝑎𝑟𝑜 𝑒 𝑑𝑜 𝑑𝑖𝑠𝑐𝑜: 𝐶𝑀 =
𝑙
2
 
 
𝑙 
 
 
𝐸, 𝑛𝑜𝑣𝑎𝑚𝑒𝑛𝑡𝑒, 𝑐𝑜𝑚𝑜 𝑛𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜 4, 𝑢𝑠𝑎𝑟𝑒𝑚𝑜𝑠 𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑑𝑒 𝑡𝑜𝑟𝑞𝑢𝑒: 
𝜏𝑎 = 𝐼𝑎�̈� = −𝑀𝑔
𝑙
2
sin𝜃 
𝑝𝑎𝑟𝑎 𝑜𝑠𝑐𝑖𝑙𝑎çõ𝑒𝑠 𝑝𝑒𝑞𝑢𝑒𝑛𝑎𝑠: sin𝜃 ≅ 𝜃 
�̈� +
𝑀𝑔𝑙
2𝐼𝑎
𝜃 = 0 
𝑙𝑜𝑔𝑜: 
𝑀𝑔𝑙
2𝐼𝑎
= 𝜔𝑎
2 
𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 𝑑𝑜 𝑎𝑛𝑒𝑙 𝑞𝑢𝑒 𝑔𝑖𝑟𝑎 𝑒𝑚 𝑡𝑜𝑟𝑛𝑜 𝑑𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎 é: 
𝐼𝐶𝑀 = 𝑀𝑅
2 → 𝐼𝐶𝑀 = 𝑀(
𝑙
2
)
2
 
𝑃𝑜𝑟é𝑚, 𝑐𝑜𝑚𝑜 𝑒𝑙𝑒 𝑔𝑖𝑟𝑎 𝑒𝑚 𝑡𝑜𝑟𝑛𝑜 𝑑𝑎 𝑝𝑎𝑟𝑡𝑒 𝑠𝑢𝑠𝑝𝑒𝑛𝑠𝑎 𝑂𝑎: 
𝑣𝑜𝑙𝑢𝑚𝑒 1 
 𝐼𝑎 = 𝑀(
𝑙
2
)
2
+ 𝑀 (
𝑙
2
)
2
 
 𝐼𝑎 =
𝑀𝑙2
2
 
𝑣𝑜𝑙𝑡𝑎𝑛𝑑𝑜 𝑎 𝑓𝑟𝑒𝑞𝑢ê𝑛𝑐𝑖𝑎: 
𝜔𝑎
2 =
𝑀𝑔𝑙
2
𝑀𝑙2
2
 
𝜔𝑎
2 =
𝑔
𝑙
 
𝐶𝑜𝑛ℎ𝑒𝑐𝑒𝑚𝑜𝑠 𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑎𝑏𝑎𝑖𝑥𝑜 𝑝𝑎𝑟𝑎 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜: 
𝑇𝑎 =
2𝜋
𝜔𝑎
 
𝑇𝑎 = 2𝜋√
𝑙
𝑔
 
 
𝑃𝑎𝑟𝑎 𝑜 𝑑𝑖𝑠𝑐𝑜, 𝑓𝑎𝑟𝑒𝑚𝑜𝑠 𝑜𝑠 𝑚𝑒𝑠𝑚𝑜𝑠 𝑝𝑟𝑜𝑐𝑒𝑑𝑖𝑚𝑒𝑛𝑡𝑜𝑠, 𝑜𝑛𝑑𝑒 𝑎𝑝𝑒𝑛𝑎𝑠 𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 
𝑖𝑛é𝑟𝑐𝑖𝑎 𝑠𝑒𝑟á 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒: 
 
𝜏𝑏 = 𝐼𝑏�̈� = −𝑀𝑔
𝑙
2
sin 𝜃 
𝑝𝑎𝑟𝑎 𝑜𝑠𝑐𝑖𝑙𝑎çõ𝑒𝑠 𝑝𝑒𝑞𝑢𝑒𝑛𝑎𝑠: sin𝜃 ≅ 𝜃 
�̈� +
𝑀𝑔𝑙
2𝐼𝑏
𝜃 = 0 
𝑙𝑜𝑔𝑜: 
𝑀𝑔𝑙
2𝐼𝑏
= 𝜔𝑏
2 
𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 𝑑𝑜 𝑑𝑖𝑠𝑐𝑜 𝑞𝑢𝑒 𝑔𝑖𝑟𝑎 𝑒𝑚 𝑡𝑜𝑟𝑛𝑜 𝑑𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎 é: 
𝐼𝐶𝑀 =
1
2
𝑀𝑅2 → 𝐼𝐶𝑀 =
1
2
𝑀 (
𝑙
2
)
2
 
𝑃𝑜𝑟é𝑚, 𝑐𝑜𝑚𝑜 𝑒𝑙𝑒 𝑔𝑖𝑟𝑎 𝑒𝑚 𝑡𝑜𝑟𝑛𝑜 𝑑𝑎 𝑝𝑎𝑟𝑡𝑒 𝑠𝑢𝑠𝑝𝑒𝑛𝑠𝑎 𝑂𝑏: 
𝑣𝑜𝑙𝑢𝑚𝑒 1 
 𝐼𝑏 =
1
2
𝑀 (
𝑙
2
)
2
+ 𝑀(
𝑙
2
)
2
 
 𝐼𝑏 =
3𝑀𝑙2
8
 
𝑣𝑜𝑙𝑡𝑎𝑛𝑑𝑜 𝑎 𝑓𝑟𝑒𝑞𝑢ê𝑛𝑐𝑖𝑎: 
𝜔𝑏
2 =
𝑀𝑔𝑙
2
3𝑀𝑙2
8
 
𝜔𝑏
2 =
4𝑔
3𝑙
 
 
𝐶𝑜𝑛ℎ𝑒𝑐𝑒𝑚𝑜𝑠 𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑎𝑏𝑎𝑖𝑥𝑜 𝑝𝑎𝑟𝑎 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜: 
 
𝑇𝑏 =
2𝜋
𝜔𝑏
 
𝑇𝑏 = 2𝜋√
3𝑙
4𝑔
 → 𝑇𝑏 = 𝜋√
3𝑙
𝑔
 
 
𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜 𝑑𝑒 𝑢𝑚 𝑝ê𝑛𝑑𝑢𝑙𝑜 𝑠𝑖𝑚𝑝𝑙𝑒𝑠(𝑣𝑒𝑟 𝑞𝑢𝑒𝑠𝑡ã𝑜 4)é: 
𝑇 = 2𝜋√
𝑙
𝑔
 
𝑞𝑢𝑒 é 𝑖𝑔𝑢𝑎𝑙 𝑎𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜 𝑞𝑢𝑒 𝑒𝑚 𝑐𝑜𝑛𝑡𝑟𝑎𝑚𝑜𝑠 𝑑𝑜 𝑎𝑛𝑒𝑙: 
𝑇𝑎 = 𝑇 = 2𝜋√
𝑙
𝑔
 
𝑝𝑎𝑟𝑎 𝑎𝑐ℎ𝑎𝑟 𝑢𝑚𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑒𝑛𝑡𝑟𝑒 𝑇𝑏 𝑒 𝑇: 
 
𝑇𝑏
𝑇
=
𝜋√
3𝑙
𝑔
2𝜋√
𝑙
𝑔
=
√3
2
 
𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜: 
𝑇𝑏 =
√3
2
𝑇 
 
 
 
 
 
 𝑠 𝑙 
𝜃 
�⃗� 
 
 
 
𝑁𝑜𝑣𝑎𝑚𝑒𝑛𝑡𝑒, 𝑐𝑜𝑚𝑜 𝑛𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜 4, 𝑢𝑠𝑎𝑟𝑒𝑚𝑜𝑠 𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑑𝑒 𝑡𝑜𝑟𝑞𝑢𝑒, 𝑞𝑢𝑒 𝑛𝑜 𝑐𝑎𝑠𝑜, 𝑠𝑒𝑟á: 
𝜏 = 𝑠 × �⃗� 
𝜏 = −𝑀𝑔𝑠 sin𝜃 = 𝐼�̈� 
𝑃𝑎𝑟𝑎 𝑝𝑒𝑞𝑢𝑒𝑛𝑎𝑠 𝑜𝑠𝑐𝑖𝑙𝑎çõ𝑒𝑠: 
𝑀𝑔𝑠
𝐼
𝜃 + �̈� = 0 
𝜔2𝜃 + �̈� = 0 
𝑐𝑜𝑚 
𝑀𝑔𝑠
𝐼
= 𝜔2 
𝑒𝑛𝑡ã𝑜 𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜 𝑠𝑒𝑟á: 
𝑇 =
2𝜋
𝜔
 
𝑇 = 2𝜋√
𝐼
𝑀𝑔𝑠
 
 
𝐶𝑜𝑚𝑜 𝑜 𝑒𝑖𝑥𝑜 𝑑𝑒 𝑟𝑜𝑡𝑎çã𝑜 𝑑𝑎 𝑏𝑎𝑟𝑟𝑎 𝑛ã𝑜 𝑒𝑠𝑡á 𝑛𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎: 
𝑣𝑜𝑙𝑢𝑚𝑒 1 
𝑜𝑛𝑑𝑒 𝑙 é 𝑎𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝑑𝑜 𝑎𝑛𝑡𝑖𝑔𝑜 𝑎𝑡é 𝑜 𝑛𝑜𝑣𝑜 𝑒𝑖𝑥𝑜 𝑑𝑒 𝑟𝑜𝑡𝑎çã𝑜, 𝑎𝑞𝑢𝑖 𝑠𝑒𝑟á 𝑠: 
 
𝐼 = 𝐼𝐶𝑀 + 𝑀𝑠
2 
 
𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 𝑐𝑜𝑚 𝑜 𝑒𝑖𝑥𝑜 𝑑𝑒 𝑟𝑜𝑡𝑎çã𝑜 𝑛𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎 é: 
𝐼𝐶𝑀 =
1
12
𝑀𝑙2 
𝑒𝑛𝑡ã𝑜: 
𝐼 = 𝐼𝐶𝑀 + 𝑀𝑠
2 
𝐼 =
1
12
𝑀𝑙2 + 𝑀𝑠2 
𝐴𝑔𝑜𝑟𝑎, 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑛𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜: 
𝑇 = 2𝜋√
(
1
12𝑀𝑙
2 + 𝑀𝑠2)
𝑀𝑔𝑠
 
𝑇 = 2𝜋√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 
𝑃𝑎𝑟𝑎 𝑎𝑐ℎ𝑎𝑟𝑚𝑜𝑠 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑚í𝑛𝑖𝑚𝑜 𝑑𝑒 𝑇, 𝑑𝑒𝑟𝑖𝑣𝑎𝑚𝑜𝑠 𝑒 𝑖𝑔𝑢𝑎𝑙𝑎𝑚𝑜𝑠 𝑎 𝑧𝑒𝑟𝑜: 
𝑑𝑇
𝑑𝑠
=
𝑑
𝑑𝑠
(2𝜋√
(𝑙2 + 12𝑠2)
12𝑔𝑠
) = 0 
𝑑𝑇
𝑑𝑠
= 2𝜋
1
2
1
√
(𝑙2 + 12𝑠2)
12𝑔𝑠
𝑑
𝑑𝑠
(
(𝑙2 + 12𝑠2)
12𝑔𝑠
) = 0 
𝑑𝑇
𝑑𝑠
=
𝜋
√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 
𝑑
𝑑𝑠
(
(𝑙2 + 12𝑠2)
12𝑔𝑠
) = 0 
𝑑𝑇
𝑑𝑠
=
𝜋
√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 (
24𝑠 × 12𝑔𝑠 − (𝑙2 + 12𝑠2)12𝑔
(12𝑔𝑠)2
) = 0 
𝑑𝑇
𝑑𝑠
=
𝜋
√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 (
24𝑠 × 12𝑔𝑠 − (𝑙2 + 12𝑠2)12𝑔
(12𝑔𝑠)2
) = 0 
𝑑𝑇
𝑑𝑠
=
𝜋
√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 (
24𝑠 × 12𝑔𝑠
(12𝑔𝑠)2
−
(𝑙2 + 12𝑠2)12𝑔
(12𝑔𝑠)2
) = 0 
𝑑𝑇
𝑑𝑠
=
𝜋
√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 (
2
𝑔
−
(𝑙2 + 12𝑠2)
12𝑔𝑠2
) = 0 
𝑑𝑇
𝑑𝑠
=
𝜋
√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 (
2
𝑔
−
𝑙2
12𝑔𝑠2
−
1
𝑔
) = 0 
𝑑𝑇
𝑑𝑠
=
𝜋
√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 
1
𝑔
 (1 −
𝑙2
12𝑠2
) = 0 
𝑐𝑜𝑚𝑜 
𝜋
√
(𝑙2+12𝑠2)
12𝑔𝑠
 
1
𝑔
≠ 0: 
(1 −
𝑙2
12𝑠2
) = 0 
𝑎𝑔𝑜𝑟𝑎 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑖𝑠𝑜𝑙𝑎𝑟 𝑠: 
1 =
𝑙2
12𝑠2
 
𝑠 = √
𝑙2
12
 
𝑠 =
𝑙
√12
 
𝑒𝑠𝑠𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑑𝑜 é 𝑜 𝑠 𝑚í𝑛𝑖𝑚𝑜, 𝑎𝑔𝑜𝑟𝑎 𝑜 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑛𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜: 
𝑇 = 2𝜋√
(𝑙2 + 12𝑠2)
12𝑔𝑠
 
𝑇 = 2𝜋√
(𝑙2 + 12(
𝑙
√12
)
2
)
12𝑔
𝑙
√12
 
𝑇 = 2𝜋√
(𝑙2 + 𝑙2)
𝑔𝑙√12
 
𝑇 = 2𝜋√
2𝑙
𝑔√12
 
𝑇 = 2𝜋√
𝑙
𝑔√3
 
 
𝐴𝑛𝑡𝑒𝑠 𝑑𝑒 𝑞𝑢𝑎𝑙𝑞𝑢𝑒𝑟 𝑐á𝑙𝑐𝑢𝑙𝑜, 𝑝𝑟𝑒𝑐𝑖𝑠𝑎𝑚𝑜𝑠 𝑠𝑎𝑏𝑒𝑟 𝑜𝑛𝑑𝑒 𝑓𝑖𝑐𝑎 𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎, 𝑝𝑜𝑖𝑠 𝑜𝑠 𝑐á𝑙𝑐𝑢𝑙𝑜𝑠 
𝑠ã𝑜 𝑓𝑒𝑖𝑡𝑜𝑠 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑛𝑑𝑜 𝑞𝑢𝑒 𝑡𝑜𝑑𝑎 𝑎 𝑚𝑎𝑠𝑠𝑎 𝑒𝑠𝑡á 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑑𝑎 𝑛𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 (𝐶𝑀). 
 
𝑂 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎 𝑑𝑒 𝑐𝑎𝑑𝑎 𝑏𝑎𝑟𝑟𝑎 (ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎) 𝑒𝑠𝑡á 𝑛𝑜 𝑚𝑒𝑖𝑜 𝑑𝑒 𝑐𝑎𝑑𝑎. 
 
 
 
 
 
 
𝐿𝑜𝑔𝑜, 𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒 𝑚𝑎𝑠𝑠𝑎 𝑑𝑒 𝑡𝑜𝑑𝑜 𝑜 𝑎𝑟𝑎𝑚𝑎 𝑓𝑖𝑐𝑎 𝑏𝑒𝑚 𝑒𝑛𝑡𝑟𝑒 (𝑜𝑠 𝑋′𝑠)𝑜𝑠 𝐶𝑀 𝑑𝑒 𝑐𝑎𝑑𝑎 𝑙𝑎𝑑𝑜 
 
 
 
 
 
𝑉𝑎𝑚𝑜𝑠 𝑑𝑖𝑣𝑖𝑑𝑖𝑟 𝑜 𝑎𝑟𝑎𝑚𝑒 𝑝𝑎𝑟𝑎 𝑎𝑐ℎ𝑎𝑟 𝑒𝑠𝑠𝑒 𝑝𝑜𝑛𝑡𝑜, 𝑙𝑜𝑔𝑜, 𝑠𝑒 𝑜 â𝑛𝑔𝑢𝑙𝑜 𝑑𝑒𝑙𝑒 𝑖𝑛𝑡𝑒𝑖𝑟𝑜 𝑣𝑎𝑙𝑒 60°, 
𝑚𝑒𝑡𝑎𝑑𝑒 𝑣𝑎𝑙𝑒 30°: 
𝑟 
 30° 
 
 
𝐴𝑔𝑜𝑟𝑎 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑎𝑐ℎ𝑎𝑟 𝑞𝑢𝑒𝑚 é 𝑟: 
cos 30° =
𝑟
𝑙
2
 → 𝑟 =
𝑙
2
cos30° 
𝑟 =
𝑙
2
√3
2
 → 𝑟 =
𝑙√3
4
 
 
𝐴𝑔𝑜𝑟𝑎, 𝑛𝑜𝑣𝑎𝑚𝑒𝑛𝑡𝑒, 𝑢𝑠𝑎𝑟𝑒𝑚𝑜𝑠 𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑑𝑒 𝑡𝑜𝑟𝑞𝑢𝑒 𝑐𝑜𝑚𝑜 𝑛𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜 4: 
 
𝜏 = −𝑚𝑔
𝑙√3
4
sin𝜃 = 𝐼�̈� 
𝑃𝑎𝑟𝑎 𝑝𝑒𝑞𝑢𝑒𝑛𝑎𝑠 𝑜𝑠𝑐𝑖𝑙𝑎çõ𝑒𝑠: sin𝜃 ≅ 𝜃 
𝑚𝑔
𝑙√3
4
𝜃 + 𝐼�̈� = 0 
𝑐𝑜𝑚𝑜 𝑒𝑠𝑡𝑎𝑚𝑜𝑠 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑛𝑑𝑜 𝑞𝑢𝑒 𝑡𝑜𝑑𝑎 𝑎 𝑚𝑎𝑠𝑠𝑎 𝑒𝑠𝑡á 𝑐𝑒𝑛𝑡𝑟𝑎𝑑𝑎 𝑛𝑜 𝐶𝑀: 
𝐼 = 𝑚𝑟2 
𝐼 = 𝑚(
𝑙√3
4
)
2
= 𝑚𝑙2
3
16
 
𝑙𝑜𝑔𝑜: 
1
𝐼
𝑚𝑔
𝑙√3
4
𝜃 + �̈� = 0 
1
(𝑚𝑙2
3
16)
𝑚𝑔
𝑙√3
4𝜃 + �̈� = 0 
𝑔
𝑙
4√3
3
𝜃 + �̈� = 0 
𝐶𝑜𝑚𝑜 𝑖𝑠𝑠𝑜: 
𝜔2 =
𝑔
𝑙
4√3
3
 
𝑒𝑛𝑡ã𝑜 𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜 𝑠𝑒𝑟á: 
𝑇 =
2𝜋
𝜔
 
𝑇 = 2𝜋√
3𝑙
4√3𝑔
 → 𝑇 = 𝜋√
√3𝑙
𝑔
 
 
𝑢𝑚 𝑜𝑠𝑐𝑖𝑙𝑎𝑑𝑜𝑟 é: 
𝑥(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜙) 
�̇�(𝑡) = −𝜔𝐴 sin(𝜔𝑡 + 𝜙) 
 
𝑜𝑛𝑑𝑒 𝜔2 =
𝑘
𝑚
 
 
𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒, 𝑎𝑠 𝑒𝑛𝑒𝑟𝑔𝑖𝑎𝑠 (𝑝𝑜𝑡𝑒𝑛𝑐𝑖𝑎𝑙 𝑒 𝑐𝑖𝑛é𝑡𝑖𝑐𝑎), 𝑠ã𝑜 𝑑𝑎𝑑𝑎𝑠 𝑝𝑜𝑟: 
 
𝐾 =
1
2
𝑚�̇�2 𝑈 =
1
2
𝑘𝑥2 
𝑙𝑜𝑔𝑜: 
𝐾 =
1
2
𝑚(−𝜔𝐴sin(𝜔𝑡 + 𝜙))2 
𝐾 =
1
2
𝑚𝜔2𝐴2 sin2(𝜔𝑡 + 𝜙) 
 
𝑈 =
1
2
𝑘(𝐴 cos(𝜔𝑡 + 𝜙))2 
𝑈 =
1
2
𝑘𝐴2 cos2(𝜔𝑡 + 𝜙) 
 
𝐴 𝑞𝑢𝑒𝑠𝑡ã𝑜 𝑠𝑒 𝑟𝑒𝑓𝑒𝑟𝑒 𝑎𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒: 
𝐾 = 3𝑈 
1
2
𝑚𝜔2𝐴2 sin2(𝜔𝑡 + 𝜙) = 3
1
2
𝑘𝐴2 cos2(𝜔𝑡 + 𝜙) 
1
2
𝑚𝜔2𝐴2 sin2(𝜔𝑡 + 𝜙) = 3
1
2
𝑚𝜔2𝐴2 cos2(𝜔𝑡 + 𝜙) 
sin2(𝜔𝑡 + 𝜙) = 3 cos2(𝜔𝑡 + 𝜙) 
 
𝑒: 𝑡 =
1
4
𝑇 
sin2 (𝜔
𝑇
4
+ 𝜙) = 3 cos2 (𝜔
𝑇
4
+ 𝜙) 
𝐶𝑜𝑚𝑜 𝑜 𝑝𝑒𝑟í𝑜𝑑𝑜 é 𝑑𝑒𝑓𝑖𝑛𝑖𝑑𝑜: 
𝑇 =
2𝜋
𝜔
 
 
sin2 (𝜔
2𝜋
𝜔
4
+ 𝜙) = 3 cos2 (𝜔
2𝜋
𝜔
4
+ 𝜙) 
sin2 (
𝜋
2
+ 𝜙) = 3 cos2 (
𝜋
2
+ 𝜙) 
𝑇𝑖𝑟𝑎𝑛𝑑𝑜 𝑎 𝑟𝑎𝑖𝑧 𝑑𝑜𝑠 𝑑𝑜𝑖𝑠 𝑙𝑎𝑑𝑜𝑠: 
sin (
𝜋
2
+ 𝜙) = √3 cos (
𝜋
2
+ 𝜙) 
𝑞𝑢𝑒 𝑝𝑜𝑑𝑒𝑚 𝑠𝑒𝑟 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑑𝑜𝑠 𝑐𝑜𝑚𝑜: 
cos(𝜙) = −√3 sin(𝜙) 
𝑉𝑜𝑙𝑡𝑎𝑛𝑑𝑜 𝑎𝑜 𝑞𝑢𝑎𝑑𝑟𝑎𝑑𝑜: 
cos2(𝜙) = 3 sin2(𝜙) 
𝑠𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒: cos2(𝑥) + sin2(𝑥) = 1 
cos2(𝜙) = 3(1 − cos2(𝜙)) 
cos2(𝜙) = 3 − 3 cos2(𝜙) 
cos2(𝜙) =
3
4
 
cos(𝜙) =
√3
2
 
𝐸𝑛𝑡ã𝑜: 
𝜙 =
𝜋
6
 
 
 
 
 
 
 𝐹1⃗⃗ ⃗ 𝐹2⃗⃗⃗⃗ 𝐹 
 
 �⃗� �⃗� 
 
 
(𝑎) 
𝑚𝑔 − 𝑧(𝑘1 + 𝑘2) = 𝑚�̈� 
𝑔 = 𝑧
(𝑘1 + 𝑘2)
𝑚
+ �̈� 
𝑙𝑜𝑔𝑜: 
(𝑘1 + 𝑘2)
𝑚
= 𝜔2 
 
(𝑏) 
𝑚𝑔 − 𝑧𝑘 = 𝑚�̈� 
𝑔 = 𝑧
𝑘
𝑚
+ �̈� 
 
𝑁𝑒𝑠𝑠𝑒 𝑐𝑎𝑠𝑜, 𝑓𝑎𝑟𝑒𝑚𝑜𝑠 𝑢𝑚𝑎 𝑎𝑛𝑎𝑙𝑜𝑔𝑖𝑎 𝑑𝑎𝑠 𝑚𝑜𝑙𝑎𝑠 𝑐𝑜𝑚 𝑟𝑒𝑠𝑖𝑠𝑡ê𝑛𝑐𝑖𝑎𝑠 𝑙𝑖𝑔𝑎𝑑𝑎𝑠 𝑒𝑚 𝑠é𝑟𝑖𝑒: 
 
1
𝑅
=
1
𝑅1
+
1
𝑅2
 
 
𝑙𝑜𝑔𝑜: 
1
𝑘
=
1
𝑘1
+
1
𝑘2
 
1
𝑘
=
𝑘1 + 𝑘2
𝑘1𝑘2
 
𝑘 =
𝑘1𝑘2
𝑘1 + 𝑘2
 
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑒 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
𝑔 = 𝑧
𝑘1𝑘2
𝑘1 + 𝑘2
𝑚
+ �̈� 
𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜: 
𝑘1𝑘2
𝑘1 + 𝑘2
𝑚
= 𝜔2 
 
 
 
 
 
 
 
 
 
 𝑥 
 
 
 
 
𝑃𝑒𝑟𝑐𝑒𝑏𝑎 𝑞𝑢𝑒 𝑡𝑎𝑛𝑡𝑜 𝑎 𝑓𝑜𝑟ç𝑎 𝑝𝑒𝑠𝑜 𝑞𝑢𝑎𝑛𝑡𝑜𝑎 𝑒𝑙á𝑠𝑡𝑖𝑐𝑎 𝑟𝑒𝑎𝑙𝑖𝑧𝑎𝑚 𝑡𝑜𝑟𝑞𝑢𝑒: 
𝜏 𝑃 = 𝑟 × �⃗� 
𝜏𝑃 = −𝑚𝑔𝑙 sin𝜃 
𝑝𝑎𝑟𝑎 𝑝𝑒𝑞𝑢𝑒𝑛𝑎𝑠 𝑜𝑠𝑐𝑖𝑙𝑎çõ𝑒𝑠: sin𝜃 ≅ 𝜃 
𝜏𝑃 = −𝑚𝑔𝑙𝜃 
 
𝜏 𝐸 = 𝑟 × 𝐹𝑒𝑙⃗⃗⃗⃗ ⃗⃗ 
𝜏𝐸 = −𝑘𝑥
𝑙
2
sin𝛼 
𝑜𝑛𝑑𝑒 𝛼 é 𝑜 â𝑛𝑔𝑢𝑙𝑜 𝑒𝑛𝑡𝑟𝑒 𝑎 𝑚𝑜𝑙𝑎 𝑒 𝑎 𝑏𝑎𝑟𝑟𝑎; 𝑒 𝑛𝑜 𝑐𝑎𝑠𝑜 𝑑𝑒 𝑝𝑒𝑞𝑢𝑒𝑛𝑎𝑠 𝑜𝑠𝑐𝑖𝑙𝑎çõ𝑒𝑠: 𝛼 ≅
𝜋
2
: 
 
𝜏𝐸 = −𝑘𝑥
𝑙
2
sin (
𝜋
2
) 
𝜏𝐸 = −𝑘𝑥
𝑙
2
 
 
𝑇𝑟𝑖𝑔𝑜𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐𝑎𝑚𝑒𝑛𝑡𝑒, 𝑝𝑒𝑙𝑎 𝑓𝑖𝑔𝑢𝑟𝑎 𝑎𝑐𝑖𝑚𝑎, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑠𝑎𝑏𝑒𝑟 𝑞𝑢𝑒𝑚 é 𝑥: 
 
sin 𝜃 =
𝑥
𝑙
2
 
𝑥 =
𝑙
2
sin 𝜃 
𝑎𝑔𝑜𝑟𝑎, 𝑣𝑜𝑙𝑡𝑎𝑛𝑑𝑜 𝑎𝑜 𝜏𝐸: 
𝜏𝐸 = −𝑘
𝑙
2
sin 𝜃
𝑙
2
 
𝜏𝐸 = −
𝑘𝑙2
4
sin 𝜃 
𝑝𝑎𝑟𝑎 𝑝𝑒𝑞𝑢𝑒𝑛𝑎𝑠 𝑜𝑠𝑐𝑖𝑙𝑎çõ𝑒𝑠: sin𝜃 ≅ 𝜃 
𝜏𝐸 = −
𝑘𝑙2
4
𝜃 
𝑂 𝑡𝑜𝑟𝑞𝑢𝑒 𝑡𝑜𝑡𝑎𝑙 𝑠𝑒𝑟á 𝑎 𝑠𝑜𝑚𝑎 𝑑𝑜𝑠 𝑑𝑜𝑖𝑠: 
𝜏 = 𝜏𝑃 + 𝜏𝐸 
𝜏 = −𝑚𝑔𝑙𝜃 −
𝑘𝑙2
4
𝜃 
𝜏 = −(𝑚𝑔𝑙 +
𝑘𝑙2
4
)𝜃 
𝑆𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒 𝑜 𝑡𝑜𝑟𝑞𝑢𝑒 𝑡𝑎𝑚𝑏é𝑚 𝑝𝑜𝑑𝑒 𝑠𝑒𝑟 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑑𝑜 𝑐𝑜𝑚𝑜: 
𝜏 = 𝐼𝛼 = 𝐼�̈� 
 
−(𝑚𝑔𝑙 +
𝑘𝑙2
4
)𝜃 = 𝐼�̈� 
1
𝐼
(𝑚𝑔𝑙 +
𝑘𝑙2
4
)𝜃 + �̈� = 0 
𝑂𝑛𝑑𝑒 𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛é𝑟𝑐𝑖𝑎 é: 
𝐼 = 𝑚𝑙2 
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑎𝑐𝑖𝑚𝑎: 
1
𝑚𝑙2
(𝑚𝑔𝑙 +
𝑘𝑙2
4
)𝜃 + �̈� = 0 
(
𝑔
𝑙
+
𝑘
4𝑚
)𝜃 + �̈� = 0 
𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜: 
(
𝑔
𝑙
+
𝑘
4𝑚
) = 𝜔2 
 
 
 
 
 
 
 
 
 
 
 
 𝑧 
 𝑧 cos𝜙 
 
 
 
 
𝐹 = −𝑚�̈� 
𝑃𝑜𝑖𝑠 𝑝𝑟𝑒𝑐𝑖𝑠𝑎𝑚𝑜𝑠 𝑑𝑒 𝑢𝑚𝑎 𝑓𝑜𝑟ç𝑎 𝑞𝑢𝑒 𝑟𝑒𝑎𝑙𝑖𝑧𝑒 𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 
𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑎 𝑝𝑜𝑖𝑠 𝑝𝑎𝑟𝑎 𝑢𝑚 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑜𝑠𝑐𝑖𝑙𝑎𝑡ó𝑟𝑖𝑜 𝑝𝑟𝑒𝑐𝑖𝑠𝑎𝑚𝑜𝑠 𝑑𝑒 𝑢𝑚𝑎 𝑓𝑜ç𝑎 𝑟𝑒𝑠𝑡𝑎𝑢𝑟𝑎𝑑𝑜𝑟𝑎. 
 
É 𝑣𝑒𝑟𝑑𝑎𝑑𝑒 𝑞𝑢𝑒: 
𝐹
𝐴
= ∆𝑃 
𝑙𝑜𝑔𝑜: 
𝐴 ∆𝑃 = −𝑚�̈� 
𝑒 𝑡𝑎𝑚𝑏é𝑚 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒: 
∆𝑃 = 𝜌𝑔ℎ 
𝑜𝑛𝑑𝑒 ℎ é 𝑎 𝑎𝑙𝑡𝑢𝑟𝑎 𝑡𝑜𝑡𝑎𝑙 𝑞𝑢𝑒 𝑜 𝑓𝑙𝑢𝑖𝑑𝑜 𝑑𝑒𝑠𝑐𝑒 (𝑜𝑢 𝑠𝑜𝑏𝑒): 
 
ℎ = 𝑧 + 𝑧 cos𝜙 
ℎ = 𝑧(1 + cos𝜙) 
𝐴𝑠𝑠𝑖𝑚, 𝑣𝑜𝑙𝑡𝑎𝑚𝑜𝑠 𝑎𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
𝐴 𝜌𝑔ℎ = −𝑚�̈� 
𝐴 𝜌𝑔𝑧(1 + cos𝜙) = −𝑚�̈� 
𝐴
𝑚
 𝜌𝑔(1 + cos𝜙)𝑧 + �̈� = 0 
 
𝐸𝑛𝑡ã𝑜: 
𝐴
𝑚
 𝜌𝑔(1 + cos𝜙) = 𝜔2