A solid disk, spinning counter-clockwise, has a mass of $4 k g$ $4 kg$ and a radius of $\frac{3}{7} m$ $3/7 m$ . If a point on the edge of the disk is moving at $\frac{7}{9} \frac{m}{s}$ $7/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-04-11T06:06:10

The angular momentum is $= 0.67 k g m^{2} s^{- 1}$ $=0.67kgm^2s^-1$
The angular velocity is $= 1.81 r a d s^{- 1}$ $=1.81rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=7/9ms^(-1)$

$r=3/7m$

So,

$omega=(7/9)/(3/7)=49/27=1.81rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=4*(3/7)^2/2=18/49kgm^2$

The angular momentum is

$L=18/49*1.81=0.67kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 4 kg4kg4 kg and a radius of 3/7 m37m3/7 m. If a point on the edge of the disk is moving at 7/9 m/s79ms7/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?