A solid disk, spinning counter-clockwise, has a mass of $4 k g$ $4 kg$ and a radius of $\frac{7}{4} m$ $7/4 m$ . If a point on the edge of the disk is moving at $\frac{5}{9} \frac{m}{s}$ $5/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-04-07T06:23:23

The angular momentum is $= 1.94 k g m^{2} s^{- 1}$ $=1.94kgm^2s^-1$
The angular velocity is $= 0.32 r a d s^{- 1}$ $=0.32rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

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

$r=7/4m$

So,

$omega=(5/9)/(7/4)=20/63=0.32rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=4*(7/4)^2/2=49/8kgm^2$

The angular momentum is

$L=49/8*0.32=1.94kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 4 kg4kg4 kg and a radius of 7/4 m74m7/4 m. If a point on the edge of the disk is moving at 5/9 m/s59ms5/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?