A solid disk, spinning counter-clockwise, has a mass of $7 k g$ $7 kg$ and a radius of $\frac{1}{2} m$ $1/2 m$ . If a point on the edge of the disk is moving at $\frac{6}{5} \frac{m}{s}$ $6/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-05-03T08:24:09

The angular momentum is $= 14.7 k g m^{2} s^{- 1}$ $=14.7kgm^2s^-1$
The angular velocity is $= 2.4 r a d s^{- 1}$ $=2.4rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=6/5ms^(-1)$

$r=1/2m$

So,

$omega=(6/5)/(1/2)=2.4rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=7*(1/2)^2/2=49/8kgm^2$

The angular momentum is

$L=49/8*2.4=14.7kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 7 kg7kg7 kg and a radius of 1/2 m12m1/2 m. If a point on the edge of the disk is moving at 6/5 m/s65ms6/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?