A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{8}{9} m$ $8/9 m$ . If a point on the edge of the disk is moving at $\frac{11}{9} \frac{m}{s}$ $11/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-04-23T07:10:39

The angular momentum is $= 1.09 k g m^{2} s^{- 1}$ $=1.09kgm^2s^-1$
The angular velocity is $= 1.375 r a d s^{- 1}$ $=1.375rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

$v=r*((Deltatheta)/(Deltat))=r omega$

$omega=v/r$

where,

$v=11/9ms^(-1)$

$r=8/9m$

So,

$omega=(11/9)/(8/9)=1.375rads^-1$

The angular momentum is $L=Iomega$

For a solid disc, $I=(mr^2)/2$

So, $I=2*(8/9)^2/2=64/81kgm^2$

The angular momentum is

$L=64/81*1.375=1.09kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 2 kg2kg2 kg and a radius of 8/9 m89m8/9 m. If a point on the edge of the disk is moving at 11/9 m/s119ms11/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?