A solid disk, spinning counter-clockwise, has a mass of $16 k g$ $16 kg$ and a radius of $\frac{3}{7} m$ $3/7 m$ . If a point on the edge of the disk is moving at $\frac{8}{5} \frac{m}{s}$ $8/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-04-19T09:37:25

The angular momentum is $= 381.95 k g m^{2} s^{- 1}$ $=381.95kgm^2s^-1$
The angular velocity is $= 3.73 r a d s^{- 1}$ $=3.73rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=8/5ms^(-1)$

$r=3/7m$

So,

$omega=(8/5)/(3/7)=56/15=3.73rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=16*(8/5)^2/2=512/25=102.4kgm^2$

The angular momentum is

$L=102.4*3.73=381.95kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 16 kg16kg16 kg and a radius of 3/7 m37m3/7 m. If a point on the edge of the disk is moving at 8/5 m/s85ms8/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?