A solid disk, spinning counter-clockwise, has a mass of $6 k g$ $6 kg$ and a radius of $\frac{8}{5} m$ $8/5 m$ . If a point on the edge of the disk is moving at $\frac{7}{4} \frac{m}{s}$ $7/4 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2018-01-18T06:32:00

The angular momentum is $= 8.4 k g m^{2} s^{- 1}$ $=8.4kgm^2s^-1$ and the angular velocity is $= 1.094 r a d s^{- 1}$ $=1.094rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=7/4ms^(-1)$

$r=8/5m$

So,

The angular velocity is

$omega=(7/4)/(8/5)=35/32=1.094rads^-1$

The angular momentum is $L=Iomega$

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

The mass is $m=6 kg$

So, $I=6*(8/5)^2/2=192/25kgm^2$

The angular momentum is

$L=192/25*1.094=8.4kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 6 kg6kg6 kg and a radius of 8/5 m85m8/5 m. If a point on the edge of the disk is moving at 7/4 m/s74ms7/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?