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nomor 5
m1 = m
m2 = 2m
m3 = m
m4 = 3m
penyelesaian:
a.) poros AA'
I = I1 + I2 + I3 + I4
I = (m1 r1²) + (m2 r2²) + (m3 r3²) + (m4 r4²)
I = (m . 0²) + (2m . b²) + (m . (b + b)²) + (3m . (b + b + b)²)
I = 0 + 2mb² + (m . (2b)²) + (3m . (3b)²)
I = 0 + 2mb² + (m . 4b²) + (3m . 9b²)
I = 0 + 2mb² + 4mb² + 27mb²
I = 33 mb²
nomor 6
m = 2 kg
IA = 8 kgm²
AO = OB
IO?
penyelesaian:
Panjang batang
rAB = √IA / m
rAB = √8 / 2
rAB = √4
rAB = 2 m
Momen inersia dititk O:
IO = (1/2 m (1/2rAB)² + (1/2 m (1/2rAB)²
IO = (1/2 . 2 . (1/2 . 2)² + (1/2 . 2 . (1/2 . 2)²
IO = (1 . 1²) + (1 . 1²)
IO = 1 + 1
IO = 2 kgm²
DINAMIKA ROTASI
a. AA'
Iaa = (2m)(b)² + (m)(2b)² + (3m)(3b)²
Iaa = 2mb² + 4mb² + 27mb²
Iaa = 33mb² ← jwb
b. BB'
Ibb = (m)(2b)² + (2m)(b)² + (3m)(b)²
Ibb = 4mb² + 2mb² + 3mb²
Ibb = 9mb² ← jwb
No. 6
Jika Batang bermassa m dan panjang L
dan di ujung titik A maka
I = ⅓ ml²
disubtitusi
8=1/3 2. l²
8=⅔ l²
2l²=3.8
2l²=24
l²=24/2
l²=12
jika diputar pada pusatnya maka: