A solid disk, spinning counter-clockwise, has a mass of $6 k g$ $6 kg$ and a radius of $\frac{3}{5} m$ $3/5 m$ . If a point on the edge of the disk is moving at $1 \frac{m}{s}$ $1 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-02-02T17:50:36

The angular momentum is $= 11.3 k g m s^{- 1}$ $=11.3kgms^-1$
The angular velocity is $= 10.5 r a d s^{- 1}$ $=10.5rads^-1$

The angular velocity is

$ω = \frac{v}{r}$ $omega=v/r$

where,

$v = 1 m s^{- 1}$ $v=1ms^(-1)$

$r = \frac{3}{5} m$ $r=3/5m$

So,

$ω = \frac{1}{\frac{3}{5}} \cdot 2 π = \frac{10}{3} π = 10.5 r a d s^{- 1}$ $omega=1/(3/5)*2pi=10/3pi =10.5rads^-1$

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, $I = 6 \cdot \frac{{(\frac{3}{5})}^{2}}{2} = \frac{27}{25} k g m^{2}$ $I=6*(3/5)^2/2=27/25kgm^2$

$L = 10.5 \cdot \frac{27}{25} = 11.3 k g m s^{- 1}$ $L=10.5*27/25=11.3kgms^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 6kg6 kg and a radius of 35m3/5 m. If a point on the edge of the disk is moving at 1ms1 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?