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 $\frac{5}{9} \frac{m}{s}$ $5/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2016-11-16T13:59:13

The angular velocity is $= \frac{20}{63} H z$ $=20/63 Hz$
The angular momentum is $= \frac{35}{36} k g m^{2} s^{- 1}$ $=35/36kgm^2s^(-1)$

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

where $v = v e l o c i t y = (\frac{5}{9}) m s^{- 1}$ $v= velocity =(5/9)ms^(-1)$

and $r = r a d i u s = \frac{7}{4} m$ $r=radius=7/4 m$

$ω = \frac{\frac{5}{9}}{\frac{7}{4}} = \frac{5}{9} \cdot \frac{4}{7} = \frac{20}{63} H z$ $omega=(5/9)/(7/4)=5/9*4/7=20/63 Hz$

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

Where $ω =$ $omega=$ angular velocity

For a solid disc, $I = \frac{m r^{2}}{2}$ $I=(mr^2)/2$

$I = 2 \cdot {(\frac{7}{4})}^{2} \cdot (\frac{1}{2}) = \frac{49}{16} k g m^{2}$ $I=2*(7/4)^2*(1/2)=49/16 kgm^2$

So, $L = (\frac{49}{16}) \cdot \frac{20}{63} = \frac{35}{36} k g m^{2} s^{- 1}$ $L=(49/16)*20/63=35/36kgm^2s^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 2kg2 kg and a radius of 74m7/4 m. If a point on the edge of the disk is moving at 59ms5/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?