A solid disk, spinning counter-clockwise, has a mass of $4 kg$ and a radius of $2 m$ . If a point on the edge of the disk is moving at $6/7 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-09-23T06:00:13

The angular momentum is $=3.43kgm^2s^-1$ and the angular velocity is $=0.43rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

The velocity is $v=r*((Deltatheta)/(Deltat))=r omega$

Therefore, $omega=v/r$

where,

$v=6/7ms^(-1)$

$r=2m$

So,

$omega=(6/7)/(2)=3/7=0.43rads^-1$

The angular momentum is $L=Iomega$

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

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

The angular momentum is

$L=8*3/7=24/7=3.43kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 4 kg4 kg and a radius of 2 m2 m. If a point on the edge of the disk is moving at 6/7 m/s6/7 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?