A solid disk, spinning counter-clockwise, has a mass of $7 k g$ $7 kg$ and a radius of $2 m$ $2 m$ . If a point on the edge of the disk is moving at $\frac{6}{7} \frac{m}{s}$ $6/7 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-01-07T06:04:29

The angular momentum is $= 37.7 k g m^{2} s^{- 1}$ $=37.7kgm^2s^-1$
The angular velocity is $= 2.69 r a d s^{- 1}$ $=2.69rads^-1$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = \frac{6}{7} m s^{- 1}$ $v=6/7ms^(-1)$

$r = 2 m$ $r=2m$

So,

$ω = \frac{6}{7} \cdot \frac{1}{2} \cdot 2 π = \frac{6}{7} π = 2.69 r a d s^{- 1}$ $omega=6/7*1/2*2pi=6/7pi=2.69rads^-1$

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, $I = 7 \cdot \frac{{(2)}^{2}}{2} = 14 k g m^{2}$ $I=7*(2)^2/2=14kgm^2$

$L = \frac{6}{7} π \cdot 14 = 37.7 k g m^{2} s^{- 1}$ $L=6/7pi*14=37.7kgm^2s^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 7kg7 kg and a radius of 2m2 m. If a point on the edge of the disk is moving at 67ms6/7 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?