A potter's wheel has the shape of a solid uniform disk of mass 7 kg and radius 0.65 m. It spins about an axis perpendicular to the disk at its center. A small 2.1 kg lump of very dense clay is dropped onto the wheel at a distance 0.41 m from the axis.
What is the moment of inertia of the system about the axis of spin?

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Vespertilio Vespertilio · Answer 1 · 2019-12-01T09:01:12+01:00

Answer:

1.832 kgm^2

Explanation:

mass of potter's wheel, M = 7 kg

radius of wheel, R = 0.65 m

mass of clay, m = 2.1 kg

distance of clay from centre, r = 0.41 m

Moment of inertia = Moment of inertia of disc + moment f inertia of the clay

I = 1/2 MR^2 + mr^2

I = 0.5 x 7 x 0.65 x 0.65 + 2.1 x 0.41 x 0.41

I = 1.47875 + 0.353

I = 1.832 kgm^2

Thus, the moment of inertia is 1.832 kgm^2.

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okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-08T12:12:42+02:00

The moment of inertia of the system about the axis of spin is mathematically given as

I = 1.832 kgm^2

Question Parameter(s):

A potter's wheel has the shape of a solid uniform disk of mass of 7 kg
and a radius of 0.65 m
A small 2.1 kg lump of very dense clay

the wheel at a distance of 0.41 m from the axis.

Generally, the equation for the moment of inertia is mathematically given as

I = 1/2 MR^2 + mr^2

I = 0.5 x 7 (0.65)^2 + 2.1 (0.41)^2

I = 1.47875 + 0.353

I = 1.832 kgm^2

In conclusion moment of inertia is

I = 1.832 kgm^2