What is the angular momentum of a rod with a mass of $6 k g$ $6 kg$ and length of $9 m$ $9 m$ that is spinning around its center at $3 H z$ $3 Hz$ ?

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Answer 1 · 2017-05-25T13:15:34

The angular momentum is $= 763.4 k g m^{2} s^{- 1}$ $=763.4kgm^2s^-1$

The angular momentum is $L = I ω$ $L=Iomega$

where $I$ $I$ is the moment of inertia

The moment of inertia of a rod, rotating about the center is

$I = \frac{1}{12} \cdot m L^{2}$ $I=1/12*mL^2$

$= \frac{1}{12} \cdot 6 \cdot 9^{2} = 40.5 k g m^{2}$ $=1/12*6*9^2= 40.5 kgm^2$

The angular velocity is

$ω = (3) \cdot (2 π) = 6 π r a d s^{- 1}$ $omega=(3)*(2pi)=6pirads^-1$

The angular momentum is

$L = 40.5 \cdot 6 π = 763.4 k g m^{2} s^{- 1}$ $L=40.5*6pi=763.4kgm^2s^-1$

What is the angular momentum of a rod with a mass of 6kg 6 kg and length of 9m9 m that is spinning around its center at 3Hz3 Hz?