A solid disk, spinning counter-clockwise, has a mass of $3 k g$ $3 kg$ and a radius of $\frac{8}{9} m$ $8/9 m$ . If a point on the edge of the disk is moving at $\frac{11}{4} \frac{m}{s}$ $11/4 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-10-05T13:57:33

The angular momentum is $= 3.67 k g m^{2} s^{- 1}$ $=3.67kgm^2s^-1$
The angular velocity is $= 3.09 r a d s^{- 1}$ $=3.09rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=11/4ms^(-1)$

$r=8/9m$

So,

The angular velocity is $omega=(11/4)/(8/9)=99/32=3.09rads^-1$

The angular momentum is $L=Iomega$

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

The mass of the disc is $m=3kg$

The radius of the disc is $r=8/9m$

So, $I=3*(8/9)^2/2=32/27kgm^2$

The angular momentum is

$L=32/27*99/32=3.67kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 3 kg3kg3 kg and a radius of 8/9 m89m8/9 m. If a point on the edge of the disk is moving at 11/4 m/s114ms11/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?