A solid disk, spinning counter-clockwise, has a mass of $5 k g$ $5 kg$ and a radius of $4 m$ $4 m$ . If a point on the edge of the disk is moving at $\frac{7}{3} \frac{m}{s}$ $7/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-01-03T15:36:16

The angular momentum is $= 146.6 k g m s^{- 1}$ $=146.6kgms^(-1)$
The angular velocity is $= 3.7 r a d s^{- 1}$ $=3.7rads^(-1)$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = \frac{7}{3} m s^{- 1}$ $v=7/3ms^(-1)$

$r = 4 m$ $r=4 m$

So,

$ω = \frac{7}{3} \cdot \frac{1}{4} \cdot 2 π = \frac{7}{6} \cdot π = r a d s^{- 1}$ $omega=7/3*1/4*2pi=7/6 *pi =rads^(-1)$

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, $I = 5 \cdot \frac{{(4)}^{2}}{2} = 40 k g m^{2}$ $I=5*(4)^2/2=40kgm^2$

$L = \frac{14}{12} \cdot π \cdot 40 = \frac{140}{3} \cdot π = 146.6 k g m s^{- 1}$ $L=14/12*pi*40=140/3*pi=146.6kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 5kg5 kg and a radius of 4m4 m. If a point on the edge of the disk is moving at 73ms7/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?