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 $\frac{7}{4} \frac{m}{s}$ $7/4 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-02-14T12:44:24

The angular momentum is $= 38.5 k g m^{2} s^{- 1}$ $=38.5kgm^2s^(-1)$
The angular velocity is $= 3.14 r a d s^{- 1}$ $=3.14rads^-1$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = \frac{7}{4} m s^{- 1}$ $v=7/4ms^(-1)$

$r = \frac{7}{2} m$ $r=7/2m$

So,

$ω = \frac{\frac{7}{4}}{\frac{7}{2}} \cdot 2 π = π = 3.14 r a d s^{- 1}$ $omega=(7/4)/(7/2)*2pi=pi=3.14rads^-1$

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, $I = 2 \cdot \frac{{(\frac{7}{2})}^{2}}{2} = \frac{49}{4} k g m^{2}$ $I=2*(7/2)^2/2=49/4kgm^2$

$L = 3.14 \cdot \frac{49}{4} = 38.5 k g m^{2} s^{- 1}$ $L=3.14*49/4=38.5kgm^2s^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 2kg2 kg and a radius of 72m7/2 m. If a point on the edge of the disk is moving at 74ms7/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?