A solid disk, spinning counter-clockwise, has a mass of $9 kg$ and a radius of $3/2 m$ . If a point on the edge of the disk is moving at $5/2 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2018-02-15T07:21:59

The angular momentum is $=16.875kgm^2s^-1$ and the angular velocity is $=1.67rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=5/2ms^(-1)$

$r=3/2m$

So,

The angular velocity is positive (spinning counter clockwise)

$omega=(5/2)/(3/2)=5/3=1.67rads^-1$

The angular momentum is $L=Iomega$

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

The mass is $m= 9kg$

So, $I=9*(3/2)^2/2=81/8kgm^2$

The angular momentum is

$L=81/8*5/3=16.875kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 9 kg9 kg and a radius of 3/2 m3/2 m. If a point on the edge of the disk is moving at 5/2 m/s5/2 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?