A solid disk, spinning counter-clockwise, has a mass of $7 kg$ and a radius of $3 m$ . If a point on the edge of the disk is moving at $16 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2016-06-02T09:42:37

For a disc rotating with its axis through the centre and perpendicular to its plane, the moment of inertia , ** $I =1/2MR^2$ **

So, the Moment of Inertia for our case, $I = 1/2MR^2 = 1/2 xx (7\ kg) xx (3\ m)^2 = 31.5\ kgm^2$

where, $M$ is the total mass of the disc and $R$ is the radius.

the angular velocity ( $omega$ ) of the disc, is given as: $omega = v/r$ where $v$ is the linear velocity at some distance $r$ from the centre.

So, the Angular velocity ( $omega$ ), in our case, = $v/r=(16ms^-1)/(3m) ~~ 5.33 \ rad"/"s$

Hence, the Angular Momentum = $I omega ~~ 31.5 xx 5.33 \ rad \ kg \ m^2\ s^-1 = 167.895 \ rad \ kg \ m^2\ s^-1$

A solid disk, spinning counter-clockwise, has a mass of 7 kg7 kg and a radius of 3 m3 m. If a point on the edge of the disk is moving at 16 m/s16 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?