A solid disk, spinning counter-clockwise, has a mass of $2 k g$ $2 kg$ and a radius of $\frac{7}{4} m$ $7/4 m$ . 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 · 2016-12-06T07:17:06

The angular momentum is $= \frac{49}{16} k g m s^{- 1}$ $=49/16kgms^(-1)$
THe angular velocity is $= 1 H z$ $=1Hz$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

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

$r = \frac{7}{4} m$ $r=7/4m$

So,

$ω = \frac{7}{4} / \frac{7}{4} = 1 H z$ $omega=7/4/7/4=1Hz$

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, $I = 2 \cdot \frac{{(\frac{7}{4})}^{2}}{2} = \frac{49}{16} k g m^{2}$ $I=2*(7/4)^2/2=49/16kgm^2$

$L = \frac{49}{16} \cdot 1 = \frac{49}{16} k g m s^{- 1}$ $L=49/16*1=49/16kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 2kg2 kg and a radius of 74m7/4 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?