A solid disk, spinning counter-clockwise, has a mass of $8 kg$ and a radius of $5/2 m$ . If a point on the edge of the disk is moving at $9/8 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-04-25T05:58:47

The angular momentum is $=11.25kgm^2s^-1$
The angular velocity is $=0.45rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=9/8ms^(-1)$

$r=5/2m$

So,

$omega=(9/8)/(5/2)=0.45rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=8*(5/2)^2/2=25kgm^2$

The angular velocity is

$L=25*0.45=11.25kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 8 kg8 kg and a radius of 5/2 m5/2 m. If a point on the edge of the disk is moving at 9/8 m/s9/8 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?