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 $\frac{4}{3} \frac{m}{s}$ $4/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Question

1380 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-12T10:45:06

The angular momentum is $= 40.24 k g m s^{- 1}$ $=40.24kgms^-1$
The angular velocity is $= 5.24 r a d s^{- 1}$ $=5.24rads^-1$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = \frac{4}{3} m s^{- 1}$ $v=4/3ms^(-1)$

$r = \frac{8}{5} m$ $r=8/5m$

So,

$ω = \frac{\frac{4}{3}}{\frac{8}{5}} \cdot 2 π = \frac{5}{3} π = 5.24 r a d s^{- 1}$ $omega=(4/3)/(8/5)*2pi=5/3pi=5.24 rads^-1$

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, $I = 6 \cdot \frac{{(\frac{8}{5})}^{2}}{2} = \frac{192}{25} = 7.68 k g m^{2}$ $I=6*(8/5)^2/2=192/25=7.68kgm^2$

$L = 5.24 \cdot 7.68 = 40.24 k g m s^{- 1}$ $L=5.24*7.68=40.24kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 6kg6 kg and a radius of 85m8/5 m. If a point on the edge of the disk is moving at 43ms4/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?