A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{8}{9} m$ $8/9 m$ . If a point on the edge of the disk is moving at $\frac{11}{4} \frac{m}{s}$ $11/4 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Question

1384 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T14:59:53

The angular momentum is $= 15.4 k g m s^{- 1}$ $=15.4kgms^(-1)$
The angular velocity is $= 19.4 r a d s^{- 1}$ $=19.4 rads^(-1)$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = \frac{11}{4} m s^{- 1}$ $v=11/4ms^(-1)$

$r = \frac{8}{9} m$ $r=8/9m$

So,

$ω = \frac{11}{4} / \frac{8}{9} = \frac{99}{32} \cdot 2 π = \frac{99}{16} π = 19.4 r a d s^{- 1}$ $omega=11/4/8/9=99/32*2pi =99/16pi=19.4 rads^(-1)$

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

where $I$ $I$ is the moment of inertia

For a solid disc, $I = \frac{m r^{2}}{2}$ $I=(mr^2)/2$

So, $I = 2 \cdot \frac{{(\frac{8}{9})}^{2}}{2} = \frac{64}{81} k g m^{2}$ $I=2*(8/9)^2/2=64/81kgm^2$

$L = \frac{64}{81} \cdot \frac{99}{16} π = \frac{44}{9} π = 15.4 k g m s^{- 1}$ $L=64/81*99/16pi=44/9pi=15.4kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 2kg2 kg and a radius of 89m8/9 m. If a point on the edge of the disk is moving at 114ms11/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?