A solid disk, spinning counter-clockwise, has a mass of $3 k g$ $3 kg$ and a radius of $\frac{8}{3} m$ $8/3 m$ . If a point on the edge of the disk is moving at $\frac{11}{4} \frac{m}{s}$ $11/4 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2018-05-31T05:57:19

The angular momentum is $= 11 k g m^{2} s^{- 1}$ $=11kgm^2s^-1$ and the angular velocity is $= 1.03125 r a d s^{- 1}$ $=1.03125rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=11/4ms^(-1)$

$r=8/3m$

So,

The angular velocity is

$omega=(11/4)/(8/3)=1.03125rads^-1$

The angular momentum is $L=Iomega$

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

The mass is $m=3 kg$

So, $I=3*(8/3)^2/2=10.67kgm^2$

The angular momentum is

$L=10.67*1.03125=11kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 3 kg3kg3 kg and a radius of 8/3 m83m8/3 m. If a point on the edge of the disk is moving at 11/4 m/s114ms11/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?