A solid disk, spinning counter-clockwise, has a mass of $4 k g$ $4 kg$ and a radius of $3 m$ $3 m$ . If a point on the edge of the disk is moving at $\frac{8}{5} \frac{m}{s}$ $8/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-01-14T06:29:24

The angular momentum is $= 60.32 k g m s^{- 1}$ $=60.32kgms^-1$
The angular velocity is $= 3.35 r a d s^{- 1}$ $=3.35rads^-1$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = \frac{8}{5} m s^{- 1}$ $v=8/5ms^(-1)$

$r = 3 m$ $r=3m$

So,

$ω = \frac{\frac{8}{5}}{3} \cdot 2 π = \frac{16}{15} π = 3.35 r a d s^{- 1}$ $omega=(8/5)/(3)*2pi=16/15pi=3.35rads^-1$

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, $I = 4 \cdot \frac{{(3)}^{2}}{2} = 18 k g m^{2}$ $I=4*(3)^2/2=18kgm^2$

$L = \frac{16}{15} π \cdot 18 = 60.32 k g m s^{- 1}$ $L=16/15pi*18=60.32kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 4kg4 kg and a radius of 3m3 m. If a point on the edge of the disk is moving at 85ms8/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?