A solid disk, spinning counter-clockwise, has a mass of $6 k g$ $6 kg$ and a radius of $\frac{8}{5} m$ $8/5 m$ . If a point on the edge of the disk is moving at $\frac{4}{5} \frac{m}{s}$ $4/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2018-03-01T06:33:06

The angular momentum is $= 3.84 k g m^{2} s^{- 1}$ $=3.84kgm^2s^-1$ and the angular velocity is $= 0.5 r a d s^{- 1}$ $=0.5rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=4/5ms^(-1)$

$r=8/5m$

So,

The angular velocity is

$omega=(4/5)/(8/5)=0.5rads^-1$

The angular momentum is $L=Iomega$

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

The mass is $m=6 kg$

So, $I=6*(8/5)^2/2=7.68kgm^2$

The angular momentum is

$L=7.68*0.5=3.84kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 6 kg6kg6 kg and a radius of 8/5 m85m8/5 m. If a point on the edge of the disk is moving at 4/5 m/s45ms4/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?