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 $19 H z$ $19 Hz$ ?

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Answer 1 · 2017-05-09T11:08:42

The angular momentum is $= 2865.1 k g m^{2} s^{- 1}$ $=2865.1kgm^2s^-1$

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

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

$= \frac{1}{12} \cdot 8 \cdot 6^{2} = 24 k g m^{2}$ $=1/12*8*6^2= 24 kgm^2$

The angular velocity is

$ω = 19 \cdot 2 π$ $omega=19*2pi$

$= (38 π) r a d s^{- 1}$ $=(38pi) rads^(-1)$

The angular momentum is

$L = 24 \cdot 38 π = 2865.1 k g m^{2} s^{- 1}$ $L=24*38pi=2865.1kgm^2s^-1$

What is the angular momentum of a rod with a mass of 8kg8 kg and length of 6m6 m that is spinning around its center at 19Hz19 Hz?