A solid disk, spinning counter-clockwise, has a mass of $3 k g$ $3 kg$ and a radius of $\frac{7}{5} m$ $7/5 m$ . If a point on the edge of the disk is moving at $\frac{4}{3} \frac{m}{s}$ $4/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2016-12-04T15:35:57

The angular momentum is $= \frac{4}{15} k g m s^{- 1}$ $=4/15kgms^(-1)$
The angular velocity is $= \frac{20}{21} H z$ $=20/21Hz$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = \frac{4}{3} m s^{- 1}$ $v=4/3ms^(-1)$

$r = \frac{7}{5} m$ $r=7/5m$

So,

$ω = \frac{4}{3} / \frac{7}{5} = \frac{20}{21} H z$ $omega=4/3/7/5=20/21Hz$

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

where $I$ $I$ is the moment of inertia

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

So, $I = 3 \cdot \frac{{(\frac{7}{5})}^{2}}{2} = \frac{147}{50} k g m^{2}$ $I=3*(7/5)^2/2=147/50kgm^2$

$L = \frac{14}{50} \cdot \frac{20}{21} = \frac{4}{15} k g m s^{- 1}$ $L=14/50*20/21=4/15kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 3kg3 kg and a radius of 75m7/5 m. If a point on the edge of the disk is moving at 43ms4/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?