A solid disk, spinning counter-clockwise, has a mass of $14 k g$ $14 kg$ and a radius of $\frac{5}{4} m$ $5/4 m$ . If a point on the edge of the disk is moving at $\frac{7}{9} \frac{m}{s}$ $7/9 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2016-12-12T07:18:48

The angular momentum is $= 170.6 k g m^{2} s^{- 1}$ $=170.6 kgm^2s^(-1)$
The angular velocity is $= 3.9 r a d s^{- 1}$ $=3.9rads^(-1)$

The angular velocity is $ω$ $omega$

$ω = \frac{v}{r} = \frac{7}{9} \cdot \frac{4}{5} \cdot 2 π r a d s^{- 1}$ $omega=v/r=7/9*4/5*2pi rads^(-1)$

$= (\frac{56}{45} π) r a d s^{- 1} = 3.9 r a d s^{- 1}$ $=(56/45pi)rads^(-1)=3.9 rads^(-1)$

The angular momentum is $L = I ω$ $L=Iomega$

For a solid disc, the moment of inertia $I = \frac{1}{2} m r^{2}$ $I=1/2mr^2$

$I = \frac{1}{2} \cdot 14 \cdot \frac{25}{4} = \frac{175}{4} k g m^{2}$ $I=1/2*14*25/4=175/4kgm^2$

The angular momentum is

$L = \frac{175}{4} \cdot \frac{56}{45} \cdot π = 170.6 k g m^{2} s^{- 1}$ $L=175/4*56/45*pi=170.6 kgm^2s^(-1)$

A solid disk, spinning counter-clockwise, has a mass of 14kg14 kg and a radius of 54m5/4 m. If a point on the edge of the disk is moving at 79ms7/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?