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A uniform circular disk of moment of inertia 8.0 kg m2 is rotating at 4.0 rad/s. A small lump of mass 1.0 kg is dropped on the disk and sticks to it at a distance of 1.0 m from the axis of rotation. What is the new rotational speed of the combined system?

Relax

Respuesta :

The new rotational speed of the combined system is 32 rad/s

Explanation:

Given-

Moment of Inertia, I = 8 kg mĀ²

speed of rotation, Ļ‰ = 4 rad/s

Mass of the lump = 1 kg

distance = 1 m

New rotational speed = ?

We know,

When the lump is dropped on the rotating disc then the angular momentum remains conserved.

Angular Momentum of the disc, Lā‚ = IĻ‰

Angular momentum of the lump, L ā‚“ = mvR

Since the angular momentum remains conserved,

Lā‚ = Lā‚“

8 X 4 = 1 X v X 1

New rotational speed, v = 32 rad/s

Therefore, the new rotational speed of the combined system is 32 rad/s.