What is the angular momentum of a rod with a mass of $8 k g$ $8 kg$ and length of $4 m$ $4 m$ that is spinning around its center at $15 H z$ $15 Hz$ ?

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Answer 1 · 2016-12-15T10:56:05

The angular momentum is $= 80 π = 251.3 k g m^{2} s^{- 1}$ $=80pi=251.3kgm^2s^(-1)$

The angular momentum is

$L = I ω$ $L=Iomega$

The moment of inertia of a rod is $I = \frac{1}{12} m l^{2}$ $I=1/12ml^2$

$I = \frac{1}{12} \cdot 8 \cdot 4 = \frac{8}{3} k g m^{2}$ $I=1/12*8*4=8/3kgm^2$

The angular velocity is $ω = 15 \cdot 2 π r a d s^{- 1}$ $omega=15*2pi rads^(-1)$

So,

$L = \frac{8}{3} \cdot 30 π k g m^{2} s^{- 1} = 80 π k g m^{2} s^{- 1}$ $L=8/3*30pi kgm^2s^(-1)=80pi kgm^2s^(-1)$

What is the angular momentum of a rod with a mass of 8 kg8kg 8 kg and length of 4 m4m4 m that is spinning around its center at 15 Hz15Hz15 Hz?