A solid disk, spinning counter-clockwise, has a mass of $6 k g$ $6 kg$ and a radius of $\frac{7}{5} m$ $7/5 m$ . If a point on the edge of the disk is moving at $\frac{4}{3} \frac{m}{s}$ $4/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-04-19T09:13:52

The angular momentum is $= 5.6 k g m^{2} s^{- 1}$ $=5.6kgm^2s^-1$
The angular velocity is $= 0.95 r a d s^{- 1}$ $=0.95rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=4/3ms^(-1)$

$r=7/5m$

So,

$omega=(4/3)/(7/5)=20/21=0.95rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=6*(7/5)^2/2=147/25kgm^2$

The angular momentum is

$L=147/25*20/21=5.6kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 6 kg6kg6 kg and a radius of 7/5 m75m7/5 m. If a point on the edge of the disk is moving at 4/3 m/s43ms4/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?