A solid disk, spinning counter-clockwise, has a mass of $13 k g$ $13 kg$ and a radius of $\frac{4}{7} m$ $4/7 m$ . If a point on the edge of the disk is moving at $\frac{8}{5} \frac{m}{s}$ $8/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-03-27T10:30:41

The angular momentum is $= 5.94 k g m^{2} s^{- 1}$ $=5.94kgm^2s^-1$
The angular velocity is $= 2.8 r a d s^{- 1}$ $=2.8rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=8/5ms^(-1)$

$r=4/7m$

So,

The angular velocity is $omega=(8/5)/(4/7)=56/20=2.8rads^-1$

The angular momentum is $L=Iomega$

mass, $m=13kg$

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

So, $I=13*(4/7)^2/2=2.12kgm^2$

$L=I*omega=2.12*2.8=5.94kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 13 kg13kg13 kg and a radius of 4/7 m47m4/7 m. If a point on the edge of the disk is moving at 8/5 m/s85ms8/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?