What is the angular momentum of a rod with a mass of $15 k g$ $15 kg$ and length of $4 m$ $4 m$ that is spinning around its center at $14 H z$ $14 Hz$ ?

Question

1485 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-13T19:22:58

The angular momentum is $= 1759.3 k g m^{2} s^{- 1}$ $=1759.3kgm^2s^-1$

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

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

$= \frac{1}{12} \cdot 15 \cdot 4^{2} = 20 k g m^{2}$ $=1/12*15*4^2= 20 kgm^2$

The angular velocity is

$ω = 2 π f = 2 π \cdot 14 = 28 π r a d s^{- 1}$ $omega=2pif=2pi*14=28pirads^-1$

So the angular momentum is

$L = 20 \cdot (28 π) = 1759.3 k g m^{2} s^{- 1}$ $L=20*(28pi) =1759.3kgm^2s^-1$

What is the angular momentum of a rod with a mass of 15kg15 kg and length of 4m4 m that is spinning around its center at 14Hz14 Hz?