A solid disk, spinning counter-clockwise, has a mass of $12 kg$ and a radius of $5/4 m$ . If a point on the edge of the disk is moving at $2/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-06-28T11:11:49

The angular momentum is $=5/3kgm^2s^-1$ and the angular velocity is $=8/45rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=2/9ms^(-1)$

$r=5/4m$

So,

$omega=(2/9)/(5/4)=8/45rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=12*(5/4)^2/2=75/8kgm^2$

The angular momentum is

$L=75/8*8/45=5/3kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 12 kg12 kg and a radius of 5/4 m5/4 m. If a point on the edge of the disk is moving at 2/9 m/s2/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?