A solid disk, spinning counter-clockwise, has a mass of $16 k g$ $16 kg$ and a radius of $\frac{3}{7} m$ $3/7 m$ . If a point on the edge of the disk is moving at $\frac{7}{5} \frac{m}{s}$ $7/5 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Question

1361 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-30T13:44:21

The angular momentum is $= 4.81 k g m^{2} s^{- 1}$ $=4.81kgm^2s^-1$ and the angular velocity is $= 3.27 r a d s^{- 1}$ $=3.27rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

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

$omega=v/r$

where,

$v=7/5ms^(-1)$

$r=3/7m$

So,

The angular velocity is

$omega=(7/5)/(3/7)=3.27rads^-1$

The angular momentum is $L=Iomega$

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

The mass is $m= 16kg$

So, $I=16*(3/7)^2/2=1.47kgm^2$

The angular momentum is

$L=1.47*3.27=4.81kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 16 kg16kg16 kg and a radius of 3/7 m37m3/7 m. If a point on the edge of the disk is moving at 7/5 m/s75ms7/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?