A solid disk, spinning counter-clockwise, has a mass of $4 k g$ $4 kg$ and a radius of $\frac{3}{2} m$ $3/2 m$ . If a point on the edge of the disk is moving at $\frac{5}{7} \frac{m}{s}$ $5/7 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-02-09T12:55:08

The angular momentum is $= 13.5 k g m s^{- 1}$ $=13.5 kgms^(-1)$
The angular velocity is $= 2.99 r a d s^{- 1}$ $=2.99 rads^-1$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = \frac{5}{7} m s^{- 1}$ $v=5/7ms^(-1)$

$r = \frac{3}{2} m$ $r=3/2m$

So,

$ω = \frac{5}{7} / \frac{3}{2} \cdot 2 π = \frac{20}{21} π = 2.99 r a d s^{- 1}$ $omega=5/7/3/2*2pi=20/21pi=2.99 rads^-1$

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, $I = 4 \cdot \frac{{(\frac{3}{2})}^{2}}{2} = \frac{9}{2} k g m^{2}$ $I=4*(3/2)^2/2=9/2kgm^2$

$L = 2.99 \cdot \frac{9}{2} = 13.5 k g m s^{- 1}$ $L=2.99*9/2=13.5 kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 4kg4 kg and a radius of 32m3/2 m. If a point on the edge of the disk is moving at 57ms5/7 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?