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 $\frac{9}{5} \frac{m}{s}$ $9/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-04-24T06:47:24

The angular momentum is $= 3.04 k g m^{2} s^{- 1}$ $=3.04kgm^2s^-1$
The angular velocity is $= 4.8 r a d s^{- 1}$ $=4.8rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=9/5ms^(-1)$

$r=3/8m$

So,

$omega=(9/5)/(3/8)=4.8rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=9*(3/8)^2/2=81/128kgm^2$

The angular momentum is

$L=81/128*4.8=3.04kgm^2s^-1$

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 9/5 m/s95ms9/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?