What is the angular momentum of a rod with a mass of $8 k g$ $8 kg$ and length of $4 m$ $4 m$ that is spinning around its center at $12 H z$ $12 Hz$ ?

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What is the angular momentum of a rod with a mass of $8 k g$ $8 kg$ and length of $4 m$ $4 m$ that is spinning around its center at $12 H z$ $12 Hz$ ?

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Answer 1 · 2016-12-23T15:11:17

The answer is $256 π$ $256pi$ $k g \frac{m^{2}}{s}$ $kgm^2/s$ which to the nearest whole number is 804 $k g m^{/} s$ $kgm^/s$

Explanation:

Angular momentum is similar in formula to linear momentum:

$L = I \times ω$ $L=Ixxomega$ as compared to $p = m \times v$ $p =m xx v$

To find the moment of interia $I$ $I$ , most students would check a list of known expressions, as every different geometry and even a different axis of rotation will play a part in determining the nature of $I$ $I$ .

In the case of a rod being rotated about its centre, the moment of inertia is

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

$L$ $L$ being the length of the rod, and $m$ $m$ its mass.

The angular velocity $ω$ $omega$ is equal to the number of radians swept out by the rod each second. In this case, as each revolution equals $2 π$ $2pi$ radians,

$ω = 24 π$ $omega=24pi$

Put it together:

$L = (\frac{m L^{2}}{12}) \times 24 π$ $L=((mL^2)/12)xx 24pi$ = $\frac{(8) (4^{2}) (24 π)}{12}$ $((8)(4^2)(24pi))/12$ = $256 π k g \frac{m^{2}}{s}$ $256pi" " kgm^2/s$

which is approximately 804 $k g \frac{m^{2}}{s}$ $kgm^2/s$

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What is the angular momentum of a rod with a mass of $8 k g$ $8 kg$ and length of $4 m$ $4 m$ that is spinning around its center at $12 H z$ $12 Hz$ ?

1 Answer

Explanation:

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What is the angular momentum of a rod with a mass of $8 k g$ $8 kg$ and length of $4 m$ $4 m$ that is spinning around its center at $12 H z$ $12 Hz$ ?