A solid disk, spinning counter-clockwise, has a mass of $8 k g$ $8 kg$ and a radius of $\frac{3}{2} m$ $3/2 m$ . If a point on the edge of the disk is moving at $\frac{9}{8} \frac{m}{s}$ $9/8 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2016-11-20T08:27:05

The angular momentum is $= \frac{27}{4} k g m s^{- 1}$ $=27/4kgms^(-1)$
The angular velocity is $= \frac{3}{4} H z$ $=3/4Hz$

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

where
$v = v e l o c i t y = \frac{9}{8} m s^{- 1}$ $v=velocity=9/8ms^(-1)$
$r = r a d i u s = \frac{3}{2} m$ $r=radius=3/2m$

so, $ω = \frac{9}{8} \cdot \frac{2}{3} = \frac{3}{4} H z$ $omega=9/8*2/3=3/4 Hz$

The angular momentum is

$L = I ω$ $L=Iomega$

In the case of a solid disc, $I = \frac{m r^{2}}{2}$ $I=(mr^2)/2$

$m =$ $m=$ mass of the disc, $= 8 k g$ $=8kg$

$r =$ $r=$ radius of the disc, $= \frac{3}{2} m$ $=3/2 m$

Therefore, $I = \frac{1}{2} \cdot 8 \cdot \frac{9}{4} = 9 k g m^{2}$ $I=1/2*8*9/4=9 kgm^2$

The angular momentum is $L = 9 \cdot \frac{3}{4} = \frac{27}{4} k g m s^{- 1}$ $L=9*3/4=27/4kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 8 kg8kg8 kg and a radius of 3/2 m32m3/2 m. If a point on the edge of the disk is moving at 9/8 m/s98ms9/8 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?