What is the angular momentum of a rod with a mass of $2 k g$ $2 kg$ 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 · 2017-04-07T07:04:13

The angular momentum is $= 254.5 k g m^{2} s^{- 1}$ $=254.5kgm^2s^-1$

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

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

and $ω$ $omega$ is the angular velocity

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

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

$= \frac{1}{12} \cdot 2 \cdot 9^{2} = 13.5 k g m^{2}$ $=1/12*2*9^2= 13.5 kgm^2$

The angular velocity is

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

The angular momentum is

$L = 13.5 \cdot 6 π = 254.5 k g m^{2} s^{- 1}$ $L=13.5*6pi=254.5kgm^2s^-1$

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