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Question

The moment of inertia of a uniform solid cone relative to its symmetry axis, if the mass of the cone is equal to M and the radius of its base is equal to R is I=3MR2y. Find the value of y.


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Solution

Step 1: Given data

Moment of inertia of a uniform solid cone relative to its symmetry axis, I=3MR2y

Where M is mass of the come and R is the radius of its base

We have to find the value of y.

Step 2: Writing equations from the figure

Finding moment of inertia of a solid cone

Small cross sections of the cone are considered to be discs.

Mass of elemental disc,

From the figure, consider triangle OAB and triangle OCD.

Since the two triangles are similarCDAB,COD=AOB, their corresponding sides are proportional

rR=xHr=RxH(1)

If dm is the mass of the small cross-section with thickness dx, then

dm=M13πR2H×πr2dxFrom(1)dm=3MπR2H×πR2x2H2dxdm=3Mx2H3dx(2)

Step 3: Finding the moment of inertia of the elemental disc

Moment of inertia of a disc is 12MR2

where M is mass and R is radius.

We consider the cone to be made up of thin discs of mass dm and radius r.

Moment of inertia of the small cross-section of the cone,

dI=12dmr2From(1)and(2),dI=12×3Mx2H3dx×R2x2H2

Step 4: Finding moment of inertia of the solid cone

Moment of inertia of the cone is found by integrating the moment of inertias of the small cross sections of the cone.

Moment of inertia of the cross sections are integrated

I=dI=3MR22H50Hx4dxI=3MR22H5x55oHI=3MR22H5×H55I=3MR210

The given equation is I=3MR2y and the equation we got is I=3MR210

Comparing the equations, we get y=10.


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