A solid disk, spinning counter-clockwise, has a mass of $12 k g$ $12 kg$ and a radius of $\frac{7}{5} m$ $7/5 m$ . If a point on the edge of the disk is moving at $\frac{8}{3} \frac{m}{s}$ $8/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-11-12T16:30:02

The angular momentum is $= 81.3 k g m^{2} s^{- 1}$ $=81.3kgm^2s^-1$ and the angular velocity is $= 1.9 r a d s^{- 1}$ $=1.9rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=8/3ms^(-1)$

$r=7/5m$

So,

The angular velocity is

$omega=(8/3)/(7/5)=40/21=1.9rads^-1$

The angular momentum is $L=Iomega$

The mass is $m=12kg$

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

So, $I=12*(8/3)^2/2=128/3kgm^2$

The angular momentum is

$L=128/3*1.9=81.3kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 12 kg12kg12 kg and a radius of 7/5 m75m7/5 m. If a point on the edge of the disk is moving at 8/3 m/s83ms8/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?