KUIS FISIKA Kode: Olim-02-3-4

[tex]\displaystyle ROTASI \ BENDA \ TEGAR \ [OLIMPIADE] \\ \\ a.\ Ep \ pegas \ = \ \frac{1}{2} \ \times \ k \ \times \ x^2 \\ Ek \ total \ = \ Ek \ translasi \ + \ Ek \ rotasi \\ Ek \ total \ = \ \frac{1}{2} \ \times \ m \times \ v^2 \ + \ \frac{1}{2} \ \times \ I \ \times \omega^2 \\ Ek \ total \ = \ \frac{1}{2} \ \times \ m \times \ v^2 \ + \ \frac{1}{2} \ \times \ \frac{1}{2} \ \times \ m \ \times \ r^2 \ \times \ (\frac{v}{r})^2 \\ Ek \ total \ = \ \frac{3}{4} \ \times \ m \ \times \ v^2 \\ Karena \ roda \ berotasi \ tanpa \ slip \ maka \ tidak \ ada \ energi \ sistem \ yang \ hilang \\ Energi \ total \ = Ep \ + \ Ek \ total \ = \frac{1}{2} kx^2+ \frac{3}{4}mv^2 \xrightarrow \ Answer \\ \\ b.\ \Σt = I . \alpha \\ f.r = \frac{1}{2}.m.r^2. \frac{a}{r} \\ f = \frac{1}{2}.m.a...(1) \\ \\ \ΣF = m . a \xrightarrow \ F = k.x \ (kiri) \ serta \ f \ melawan \ (kanan) \\ kx-f = m.a...(2) \\ \\ Jumlahkan \ pers \ (1) \ dan \ (2) \ didapat \ : \\ kx = \frac{3}{2}.m.a \xrightarrow \ a = \omega^2 x \\ \omega = \sqrt{ \frac{2k}{3m} } \xrightarrow \ \omega = 2.\pi.f \\ f = \frac{1}{2 \pi} \sqrt{ \frac{2k}{3m} }\xrightarrow \ Answer [/tex]

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Pembahasan dan langkah ada di lampiran


A. Energi Total Sistem 
E tot = 3/4.m.(dx/dt)² + 1/2.kx²

B. frekuensi osilasi sistemnya adalah 

f = 1/(2π). √(2k/3m)

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