Question

Derivation of moment of inertia of solid and hollow cone

Answer 1

Answer:

See the proof below

Explanation:

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The mass of the elemental disc is

d

m

=

ρ

⋅

π

r

2

d

z

The density of the cone is

ρ

=

M

V

=

M

1

3

π

R

2

h

Therefore,

d

m

=

M

1

3

π

R

2

h

π

r

2

d

z

d

m

=

3

M

R

2

h

r

2

d

z

But

R

r

=

h

z

r

=

R

z

h

d

m

=

3

M

R

2

h

⋅

R

2

h

2

⋅

z

2

d

z

=

3

M

h

3

z

2

d

z

The moment of inertia of the elemental disc about the

z

−

axis is

d

I

=

1

2

d

m

r

2

d

I

=

1

2

⋅

3

M

h

3

z

2

⋅

z

2

R

2

h

2

d

z

d

I

=

3

2

⋅

M

R

2

h

5

z

4

d

z

Integrating both sides,

I

=

3

2

⋅

M

R

2

h

5

∫

h

0

z

4

d

z

I

=

3

2

⋅

M

R

2

h

5

[

z

5

5

]

h

0

I

=

3

2

⋅

M

R

2

h

5

⋅

h

5

5

=

3

10

M

R

2

Answer 2

Rotational inertia is important in almost all physics problems that involve mass in rotational motion. It is used to calculate angular momentum and allows us to explain (via conservation of angular momentum) how rotational motion changes when the distribution of mass changes.