A solid disk, spinning counter-clockwise, has a mass of $5 k g$ $5 kg$ and a radius of $6 m$ $6 m$ . If a point on the edge of the disk is moving at $19 \frac{m}{s}$ $19 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2018-06-10T06:23:18

The angular momentum is $= 285.3 k g m^{2} s^{- 1}$ $=285.3kgm^2s^-1$ . The angular velocity is $= 3.17 r a d s^{- 1}$ $=3.17rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=19ms^(-1)$

$r=6m$

So,

The angular velocity is

$omega=(19)/(6)=3.17rads^-1$

The angular momentum is $L=Iomega$

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

The mass is $m= 5kg$

So, the moment of inertia is

$I=5*(6)^2/2=90kgm^2$

The angular momentum is

$L=Iomega=90*3.17=285.3kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 5 kg5kg5 kg and a radius of 6 m6m6 m. If a point on the edge of the disk is moving at 19 m/s19ms19 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?