Moment of inertia of a hollow cone in parralelto the base
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Show that the moment of inertia of a hollow cone of mass M, radius R, and height h about its base is 1 4 M ( R 2 + 2 h 2 ) 14M(R2+2h2) 2. Relevant equations I = ∫ r 2 d m I=∫r2dm where r is the perpendicular distance from the axis Surface Area of a cone = π R ( R 2 + h 2 ) 1 / 2 =πR(R2+h2)1/2 3. The attempt at a solution Mass density σ = M S A = M π R ( R 2 + h 2 ) 1 / 2 σ=MSA=MπR(R2+h2)1/2 Now I = ∫ r 2 d m = ∫ r 2 σ d A = σ ∫ r 2 d A I=∫r2dm=∫r2σdA=σ∫r2dA Now the distance from its base is given by r = r ( x ) = h ( 1 − x / R ) r=r(x)=h(1−x/R) and from the image I made it appears the area element is simply d A = 2 π x d s dA=2πxds, and d s = √ d x 2 + d y 2 = √ 1 + y ′ ( x ) 2 d x ds=dx2+dy2=1+y′(x)2dx. Trivially y ′ ( x ) 2 = ( h / R ) 2 y′(x)2=(h/R)2 giving d A = 2 π x √ 1 + ( h / R ) 2 d x dA=2πx1+(h/R)2dx and I = σ ∫ r 2 d A = σ ∫ R 0 h 2 ( 1 − x R ) 2 2 π x √ 1 + ( h R ) 2 d x I = 2 π σ h 2 √ 1 + ( h R ) 2 ∫ R 0 ( 1 − x R ) 2 x d x I=σ∫r2dA=σ∫0Rh2(1−xR)22πx1+(hR)2dxI=2πσh21+(hR)2∫0R(1−xR)2xdx which clearly isn't going to give the answer since there's no way √ 1 + ( h / R ) 2 1+(h/R)2 will be cancelled
The moment of inertia of a point mass with respect to an axis is defined as the product of the mass times the distance from the axis squared. The moment of inertia of any extended object is built up from that basic definition