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

Question

1437 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-03T15:32:06

The angular momentum is $= 9 k g m^{2} s^{- 1}$ $=9kgm^2s^-1$ and the angular velocity is $= 0.29 r a d s^{- 1}$ $=0.29rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=6/7ms^(-1)$

$r=3m$

So,

$omega=(6/7)/(3)=2/7=0.29rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=7*(3)^2/2=63/2kgm^2$

The angular momentum is

$L=2/7*63/2=9kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 7 kg7kg7 kg and a radius of 3 m3m3 m. If a point on the edge of the disk is moving at 6/7 m/s67ms6/7 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?