A solid disk, spinning counter-clockwise, has a mass of $7 k g$ $7 kg$ and a radius of $\frac{5}{2} m$ $5/2 m$ . If a point on the edge of the disk is moving at $\frac{6}{5} \frac{m}{s}$ $6/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2018-02-22T13:09:38

The angular momentum is $= 10.5 k g m^{2} s^{- 1}$ $=10.5kgm^2s^-1$ and the angular velocity is $= 0.48 r a d s^{- 1}$ $=0.48rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=6/5ms^(-1)$

$r=5/2m$

So,

The angular velocity is positive (spinning counter clockwise)

$omega=(6/5)/(5/2)=12/25=0.48rads^-1$

The angular momentum is $L=Iomega$

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

The mass is $m= 7kg$

So, $I=7*(5/2)^2/2=175/8kgm^2$

The angular momentum is

$L=175/8*0.48=10.5kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 7 kg7kg7 kg and a radius of 5/2 m52m5/2 m. If a point on the edge of the disk is moving at 6/5 m/s65ms6/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?