Question

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?

Answer 1

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`