A solid disk, spinning counter-clockwise, has a mass of $8 k g$ $8 kg$ and a radius of $\frac{5}{2} m$ $5/2 m$ . If a point on the edge of the disk is moving at $\frac{9}{4} \frac{m}{s}$ $9/4 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2016-11-18T14:35:52

The angular velocity $= \frac{9}{40} H z$ $=9/40 Hz$
The angular momentum $= \frac{45}{8} k g m^{2} s^{- 1}$ $=45/8kgm^2s^(-1)$

First, let's calculate the angular velocity, $ω$ $omega$

$ω = \frac{v}{r}$ $omega=v/r$

$v = \frac{9}{4} m s^{- 1}$ $v=9/4 ms^(-1)$

$r = \frac{5}{2} m$ $r=5/2m$

$ω = \frac{\frac{9}{4}}{\frac{5}{2}} = \frac{9}{4} \cdot \frac{2}{5} = \frac{9}{40} H z$ $omega=(9/4)/(5/2)=9/4*2/5=9/40 Hz$

The angular momentum, $L$ $L$ is

$L = I \cdot ω$ $L=I*omega$

where, $I =$ $I=$ moment of inertia of the disc

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

$m = 8 k g$ $m=8 kg$

$r = \frac{5}{2} m$ $r=5/2 m$

$I = 8 \cdot \frac{25}{4} \cdot \frac{1}{2} = 25 k g m^{2}$ $I=8*25/4*1/2=25kgm^2$

So, $L = 25 \cdot \frac{9}{40} = \frac{45}{8} k g m^{2} s^{- 1}$ $L=25*9/40= 45/8kgm^2s^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 8kg8 kg and a radius of 52m5/2 m. If a point on the edge of the disk is moving at 94ms9/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?