FInd the moment of inertia of hollow cone of mass m,height h and apex angle 2@ about the central axis of symmetry.
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1 Answers
62σ=m/(πRs)
where s=R/sinα
R=h tanα
R/x = tan α
take a ring of radius r and width dx at a distance x from the vertex of the cone
dm=σ2πrdr/cos α
R
I=∫dm*r2
0
R
I=∫σ2πrdr/cos α*r2
0
R
I=∫σ2πrdr/cos α*r2
0
R
I=σ2π/cos α ∫rdr*r2
0
I=σ2π/cos αR4/4
m/(πR2)sin α . 2π/cos αR4/4
m/sin α . 1 /cos αR2/2
mαR2/sin2α
Check if there are mistakes.. there could be.. ;)
