A solid disk, spinning counter-clockwise, has a mass of $11 k g$ $11 kg$ and a radius of $\frac{4}{7} m$ $4/7 m$ . If a point on the edge of the disk is moving at $\frac{2}{3} \frac{m}{s}$ $2/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Question

1386 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-22T21:10:36

The angular velocity is $= \frac{7}{6} H z$ $=7/6Hz$
The angular momentum is $= \frac{308}{147} k g m^{2} s^{- 1}$ $=308/147kgm^2s^(-1)$

The angular velocity is given by the formula $ω = \frac{v}{r}$ $omega=v/r$

The velocity $v = \frac{2}{3} m s^{- 1}$ $v=2/3ms^(-1)$

The radius $r = \frac{4}{7} m$ $r=4/7 m$

So, $ω = \frac{\frac{2}{3}}{\frac{4}{7}} = \frac{2}{3} \cdot \frac{7}{4} = \frac{7}{6} H z$ $omega=(2/3)/(4/7)=2/3*7/4=7/6 Hz$

The angular momentum, $L = I ω$ $L=Iomega$

For a solid disc,

$I = \frac{m r^{2}}{2}$ $I=(mr^2)/2$

$I = \frac{1}{2} \cdot 11 \cdot {(\frac{4}{7})}^{2} = \frac{11 \cdot 8}{49} = \frac{88}{49} k g m^{2}$ $I=1/2*11*(4/7)^2=(11*8)/(49)=88/49kgm^2$

So,

$L = \frac{88}{49} \cdot \frac{7}{6} = \frac{308}{147} k g m^{2} s^{- 1}$ $L=88/49*7/6=308/147kgm^2s^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 11kg11 kg and a radius of 47m4/7 m. If a point on the edge of the disk is moving at 23ms2/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?