A solid disk, spinning counter-clockwise, has a mass of $12 k g$ $12 kg$ and a radius of $6 m$ $6 m$ . If a point on the edge of the disk is moving at $8 \frac{m}{s}$ $8 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-03-04T21:25:36

$ω = 1.33 r a d s^{- 1}$ $omega = 1.33rads^-1$

$p = 16 N m s$ $p = 16Nms$

Angular velocity is given by the equation

$ω = \frac{v}{r} = \frac{8 m s^{- 1}}{6 m} = 1.33 r a d s^{- 1}$ $omega = v/r = (8ms^-1)/(6m) = 1.33rads^-1$

It is in the units radians per second, or $r a d s^{- 1}$ $rads^-1$ .

Angular momentum is the product of angular velocity and mass, much like linear momentum is the product of linear velocity and mass, so

$p = m ω = 12 \times 1.33 = 16 N m s$ $p = momega = 12 xx 1.33 = 16Nms$

A solid disk, spinning counter-clockwise, has a mass of 12kg12 kg and a radius of 6m6 m. If a point on the edge of the disk is moving at 8ms8 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?