A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{7}{2} m$ $7/2 m$ . If a point on the edge of the disk is moving at $\frac{15}{4} \frac{m}{s}$ $15/4 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{7}{2} m$ $7/2 m$ . If a point on the edge of the disk is moving at $\frac{15}{4} \frac{m}{s}$ $15/4 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2016-11-10T06:15:47

The angular velocity is $= \frac{15}{14} H z$ $=15/14 Hz$
The angular momentum $= \frac{105}{4} k g m^{2} s^{- 1}$ $=105/4 kgm^2s^-1$

Explanation:

The angular velocity $ω = \frac{v}{r}$ $omega=v/r$
$v =$ $v=$ velocity $= \frac{15}{4} m s^{- 1}$ $=15/4 ms^(-1)$
and $r =$ $r=$ radius $= \frac{7}{2} m$ $=7/2 m$
So $ω = \frac{15}{4} \cdot \frac{2}{7} = \frac{15}{14} H z$ $omega=15/4*2/7=15/14 Hz$

The angular momentum is $L = I . ω$ $L=I.omega$
$I =$ $I=$ moment of inertia $= m \frac{r^{2}}{2}$ $=mr^2/2$
$I = 2 \cdot {(\frac{7}{2})}^{2} = 2 \cdot \frac{49}{4} = \frac{49}{2} k g m^{2}$ $I=2*(7/2)^2=2*49/4=49/2 kgm^2$
$L = \frac{49}{2} \cdot \frac{15}{14} = \frac{105}{4} k g m^{2} s^{- 1}$ $L=49/2*15/14=105/4 kgm^2s^(-1)$

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