A solid disk, spinning counter-clockwise, has a mass of $6 k g$ $6 kg$ and a radius of $3 m$ $3 m$ . If a point on the edge of the disk is moving at $\frac{1}{2} \frac{m}{s}$ $1/2 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-02-22T07:55:46

The angular momentum is $= 28.3 k g m^{2} s^{- 1}$ $=28.3kgm^2s^-1$
The angular velocity $= 1.047 r a d s^{- 1}$ $=1.047rads^-1$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

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

$r = 3 m$ $r=3m$

So,

$ω = \frac{\frac{1}{2}}{3} \cdot 2 π = \frac{π}{3} = 1.047 r a d s^{- 1}$ $omega=(1/2)/3*2pi=pi/3=1.047rads^-1$

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, $I = 6 \cdot \frac{{(3)}^{2}}{2} = 27 k g m^{2}$ $I=6*(3)^2/2=27kgm^2$

$L = 1.047 \cdot 27 = 28.3 k g m^{2} s^{- 1}$ $L=1.047*27=28.3kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 6 kg6kg6 kg and a radius of 3 m3m3 m. If a point on the edge of the disk is moving at 1/2 m/s12ms1/2 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?