A solid disk, spinning counter-clockwise, has a mass of $6 kg$ and a radius of $2 m$ . If a point on the edge of the disk is moving at $1/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-08-11T06:00:39

The angular momentum is $=2kgm^2s^-1$ and the angular velocity is $=0.167rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=1/3ms^(-1)$

$r=2m$

So,

$omega=(1/3)/(2)=1/6=0.167rads^-1$

The angular momentum is $L=Iomega$

where $I$ is the [moment of inertia

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

So, $I=6*(2)^2/2=12kgm^2$

The angular momentum is

$L=12*0.167=2kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 6 kg6 kg and a radius of 2 m2 m. If a point on the edge of the disk is moving at 1/3 m/s1/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?