What is the angular momentum of a rod with a mass of $2 k g$ $2 kg$ and length of $7 m$ $7 m$ that is spinning around its center at $5 H z$ $5 Hz$ ?

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Answer 1 · 2017-02-13T06:40:51

The angular momentum is $= 256.6 k g m^{2} s^{- 1}$ $=256.6kgm^2s^(-1)$

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

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

For a rod , $I = \frac{m L^{2}}{12}$ $I=(mL^2)/12$

So, $I = 2 \cdot \frac{{(7)}^{2}}{12} = \frac{49}{6} k g m^{2}$ $I=2*(7)^2/12=49/6kgm^2$

$ω = 5 \cdot 2 π = 10 π r a d s^{- 1}$ $omega=5*2pi=10pirads^-1$

$L = \frac{49}{6} \cdot 10 π = 256.6 k g m^{2} s^{- 1}$ $L=49/6*10pi=256.6kgm^2s^(-1)$

What is the angular momentum of a rod with a mass of 2kg2 kg and length of 7m7 m that is spinning around its center at 5Hz5 Hz?