What is the angular momentum of a rod with a mass of $8 k g$ $8 kg$ and length of $6 m$ $6 m$ that is spinning around its center at $5 H z$ $5 Hz$ ?

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Answer 1 · 2017-03-03T10:35:11

The angular momentum is $= 754 k g m^{2} s^{- 1}$ $=754kgm^2s^-1$

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

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

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

So, $I = 8 \cdot \frac{{(6)}^{2}}{12} = 24 k g m^{2}$ $I=8*(6)^2/12=24kgm^2$

The angular velocity is

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

So,

The angular momentum is

$L = 24 \cdot 10 π = 754 k g m^{2} s^{- 1}$ $L=24*10pi=754kgm^2s^-1$

What is the angular momentum of a rod with a mass of 8 kg8kg8 kg and length of 6 m6m6 m that is spinning around its center at 5 Hz5Hz5 Hz?