A solid disk, spinning counter-clockwise, has a mass of $9 kg$ and a radius of $6 m$ . If a point on the edge of the disk is moving at $15 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-08-09T14:24:24

The angular momentum is $=405kgm^2s^-1$ and the angular velocity is $=2.5rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=15ms^(-1)$

$r=6m$

So,

$omega=(15)/(6)=5/2=2.5rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=9*(6)^2/2=162kgm^2$

The angular momentum is

$L=162*2.5=405kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 9 kg9 kg and a radius of 6 m6 m. If a point on the edge of the disk is moving at 15 m/s15 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?