Mechanical Engineering - Engineering Mechanics - Discussion

Standard formula:

Thin spherical :2/3Mr^2.
Solid sphere :2/5Mr^2.

Hence,

dI=r2dm.
Finding dm,

dm=MAdA.
Where A is the total surface area of the shell: 4√R2.
Now, dA is the area of the ring.

dA=Rdθ*2√r.
Note: 2√r is the circumference of the hoop while R dθ is the "thickness" of the hoop (its dx in the above picture). The R dθ comes from the equation for arc length: S = Rθ.

Now, we have to find a way to relate r with θ. Consider the above picture, notice that there is a right-angle triangle with angle θ at the centre of the circle. Hence,

sinθ=rR.
r=Rsinθ.
Hence, dA becomes:

dA=2*R2sinθdθ.
Substituting the equation for dA into the equation for dm, we have:

dm=Msinθ2dθ.
Substituting the equation above and the equation for r into the equation for dI, we have:

dI=MR22sin3θdθ.
Integrating with the proper limits, (from one end to the other)

I=MR22*0*sin3θdθ.

Now, we split the sin3θ into two,

I=MR22*0*sin2θ sinθdθ.
I=MR22*0*(1"cos2θ)siθvdθ.
Now, at this point, we will use the substitution: u = cos θ. Hence,

And we have,
I=23MR2.

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Please explain the derivation.

Anyone, can say the derivation part for this?