A solid disk, spinning counter-clockwise, has a mass of $8 k g$ $8 kg$ and a radius of $\frac{3}{2} m$ $3/2 m$ . If a point on the edge of the disk is moving at $5 \frac{m}{s}$ $5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-05-17T17:18:53

The angular momentum is $= 30 k g m^{2} s^{- 1}$ $=30kgm^2s^-1$
The angular velocity is $= 3.33 r a d s^{- 1}$ $=3.33rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=5ms^(-1)$

$r=3/2m$

So,

$omega=(5)/(3/2)=3.33rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=8*(3/2)^2/2=9kgm^2$

$L=9*10/3=30kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 8 kg8kg8 kg and a radius of 3/2 m32m3/2 m. If a point on the edge of the disk is moving at 5 m/s5ms5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?