A solid disk, spinning counter-clockwise, has a mass of $6 kg$ and a radius of $3/2 m$ . If a point on the edge of the disk is moving at $3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-02-27T08:03:11

The angular momentum is $=84.8kgm^2s^-1$
The angular velocity is $=12.57rads^-1$

The angular velocity is

$omega=v/r$

where,

$v=3ms^(-1)$

$r=3/2m$

So,

$omega=(3)/(3/2)*2pi=4pi=12.57rads^-1$

The angular momentum is $L=Iomega$

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

So, $I=6*(3/2)^2/2=27/4kgm^2$

$L=12.57*27/4=84.8kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 6 kg6 kg and a radius of 3/2 m3/2 m. If a point on the edge of the disk is moving at 3 m/s3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?