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KLASIˇCNA MEHANIKA - Studentske web stranice

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356 POGLAVLJE 13. PROSTORNO GIBANJE KRUTOG TIJELA<br />

Slika 13.1: Vrtnja krutog tijela kutnom brzinom ⃗ω (t) oko nepomične točke O.<br />

Izračunajmo moment količine gibanja krutog tijela<br />

⃗L =<br />

N∑<br />

j=1<br />

⃗r j × ⃗p j =<br />

N∑<br />

j=1<br />

⃗r j × m j ⃗v j =<br />

N∑<br />

j=1<br />

⃗r j × m j (⃗ω × ⃗r j ),<br />

gdje zbrajanje ide po svim točkama krutog tijela. Primjetimo da ⃗ L ne mora biti paralelan s<br />

⃗ω . U nepomičnom (inercijskom) pravokutnom koordinatnom sustavu (x, y, z) sa ishodištem u<br />

O, moment količine gibanja tijela, kutna brzina vrtnje i radij vektor j-te čestice tijela imaju<br />

komponente:<br />

⃗L = L x ˆx + L y ŷ + L z ẑ ,<br />

⃗ω = ω x ˆx + ω y ŷ + ω z ẑ ,<br />

⃗r j = x j ˆx + y j ŷ + z j ẑ .<br />

Koristeći se vektorskim identitetom ⃗ A × ( ⃗ B × ⃗ C ) = ( ⃗ A ⃗ C ) ⃗ B − ( ⃗ A ⃗ B ) ⃗ C , izraz za ⃗ L možemo<br />

napisati kao<br />

N∑<br />

N∑<br />

⃗L = m j ⃗r j × (⃗ω × ⃗r j ) = m j [rj 2 ⃗ω − (⃗r j ⃗ω ) ⃗r j ]<br />

j=1<br />

N∑<br />

= ⃗ω m j rj 2 −<br />

j=1<br />

j=1<br />

j=1<br />

N∑<br />

m j (x j ω x + y j ω y + z j ω z ) ⃗r j ,

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