Contoh Soal: Momen Inersia Benda Tegar

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Sebuah tongkat homogen memiliki panjang 300 cm dan massa 3,5 kg. Batang tersebut  diputar dengan sumbu putar berjarak 20 cm dari salah satu ujungnya. Tentukan momen inersia putaran tongkat tersebut!

Penyelesaian:

 

Diketahu:

l = 300 cm = 3 m

m = 3,5 kg

Cara I:

I = I_{pm} + md^{2}

I = \frac{1}{12}(3,5)(3)^{2} + (3,5)(1,3)^{2} I = 2,625 + 5,915 I = 8,54 kgm^{2}

 

Cara II

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m_{1}=\frac{1}{15}m

l_{1}=\frac{1}{15}l

m_{2}=\frac{14}{15}m

l_{2}=\frac{14}{15}l

I = I_{1}+ I_{_{2}}

I = \frac{1}{3}m_{1}l_{1}^{2} + \frac{1}{3}m_{2}l_{2}^{2}

I = \frac{1}{3}(\frac{1}{15})m(\frac{1}{15}l)^{2}+\frac{1}{3}(\frac{14}{15})m(\frac{14}{15}l)^{2}

I = \frac{1}{10,125}ml^{2} + \frac{2,745}{10,125}ml^{2}

I = \frac{2,745}{10,125}ml^{2}

I = \frac{2,745}{10,125}(3,5)(3)^{2}

I = 8,54 kg m^{2}

 

 

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