A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{7}{4} m$ $7/4 m$ . If a point on the edge of the disk is moving at $\frac{7}{9} \frac{m}{s}$ $7/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{7}{4} m$ $7/4 m$ . If a point on the edge of the disk is moving at $\frac{7}{9} \frac{m}{s}$ $7/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2016-11-11T12:13:44

The angular momentum is $= \frac{49}{36} k g m^{2} s^{- 1}$ $=49/36 kgm^2s^(-1)$
The angular velocity is $= \frac{4}{9} H z$ $=4/9 Hz$

Explanation:

The angular velocity is $ω = \frac{v}{r}$ $omega=v/r$

Here $v = \frac{7}{9} m s^{- 1}$ $v=7/9 ms^(-1)$ and $r = \frac{7}{4} m$ $r=7/4 m$

therefore $ω = \frac{7}{9} \cdot \frac{4}{7} = \frac{4}{9} H z$ $omega = 7/9*4/7=4/9 Hz$

The angular momentum is $L = I \cdot ω$ $L=I*omega$
Where $I$ $I$ is the moment of inertia
For a solid disc $I = \frac{m r^{2}}{2} = 2 \cdot \frac{{(\frac{7}{4})}^{2}}{2} = \frac{49}{16} k g m^{2}$ $I=(mr^2)/2=2*(7/4)^2/2=49/16 kgm^2$

So, the angular momentum is $\frac{49}{16} \cdot \frac{4}{9} = \frac{49}{36} k g m^{2} s^{- 1}$ $49/16*4/9=49/36 kgm^2s^(-1)$

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A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{7}{4} m$ $7/4 m$ . If a point on the edge of the disk is moving at $\frac{7}{9} \frac{m}{s}$ $7/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Explanation:

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1 Answer

Explanation:

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A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{7}{4} m$ $7/4 m$ . If a point on the edge of the disk is moving at $\frac{7}{9} \frac{m}{s}$ $7/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?