A solid disk, spinning counter-clockwise, has a mass of $6 k g$ $6 kg$ and a radius of $2 m$ $2 m$ . If a point on the edge of the disk is moving at $\frac{2}{3} \frac{m}{s}$ $2/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-03-26T10:07:34

The angular momentum is $= 4 k g m^{2} s^{- 1}$ $=4kgm^2s^-1$
The angular velocity is $= 0.33 r a d s^{- 1}$ $=0.33rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=2/3ms^(-1)$

$r=2m$

So,

angular velocity, $omega=(2/3)/(2)=0.33rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=6*(2)^2/2=12kgm^2$

$L=I omega=12*0.33=4kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 6 kg6kg6 kg and a radius of 2 m2m2 m. If a point on the edge of the disk is moving at 2/3 m/s23ms2/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?