A solid disk, spinning counter-clockwise, has a mass of $9 k g$ $9 kg$ and a radius of $\frac{3}{8} m$ $3/8 m$ . If a point on the edge of the disk is moving at $\frac{7}{5} \frac{m}{s}$ $7/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-05-10T08:17:58

The angular momentum is $= 2.36 k g m^{2} s^{- 1}$ $=2.36kgm^2s^-1$
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,

$v=7/5ms^(-1)$

$r=3/8m$

So,

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

The angular momentum is $L=Iomega$

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

So, $I=9*(3/8)^2/2=0.63kgm^2$

THe angular momentum is

$L=0.63*3.73=2.36kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 9 kg9kg9 kg and a radius of 3/8 m38m3/8 m. If a point on the edge of the disk is moving at 7/5 m/s75ms7/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?