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 January 1st, 2014, 10:11 PM #1 Newbie   Joined: Apr 2012 Posts: 18 Thanks: 0 Using moments of inertia to find the heights of objects Hi guys! Just need a little bit of help with this practice problem. Consider a solid cylinder of radius R, a cylindrical shell of outer radius R and inner radius A, and a spherical shell with outer radius R and inner radius A. All three objects have the same constant density. Find the heights of the cylinders such that all three objects have the same moments of inertia when rolling down a slope. Which cylinder has the larger height? Now i've tried random manipulations of the moments of inertia equation (the triple integral related to the density and coordinates over the volume) but I'm getting nowhere. Does any body have a solution? Thanks in advance!
 January 2nd, 2014, 01:47 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Using moments of inertia to find the heights of objects As per the techniques I've outlined in this topic, we have the moments of inertia for the three objects as follows: a) A solid cylinder of radius R: $I_a=\frac{1}{2}M_aR^2$ b) A cylindrical shell of outer radius R and inner radius A: $I_b=\frac{1}{2}M_b$$A^2+R^2$$$ c) A spherical shell with outer radius R and inner radius A: $I_c=\frac{2}{5}M_c\frac{R^5-A^5}{R^3-A^3}$ Now, let $\rho$ be the mass density common to all three objects. Hence: $M_a=\rho V_a=\rho \pi R^2h_a$ $M_b=\rho V_b=\rho \pi$$R^2-A^2$$h_b$ $M_c=\rho V_c=\rho \frac{4}{3}\pi$$R^3-A^3$$$ Hence: $I_a=\frac{1}{2}\rho \pi R^2h_aR^2=\frac{1}{2}\pi\rho R^4h_a$ $I_b=\frac{1}{2}\rho \pi$$R^2-A^2$$h_b$$A^2+R^2$$=\frac{1}{2}\pi\rho$$R^4-A^4$$h_b$ $I_c=\frac{2}{5}\rho \frac{4}{3}\pi$$R^3-A^3$$\frac{R^5-A^5}{R^3-A^3}=\frac{8}{15}\pi\rho$$R^5-A^5$$$ And so we find: The height of the solid cylinder: $I_a=I_c$ $\frac{1}{2}\pi\rho R^4h_a=\frac{8}{15}\pi\rho$$R^5-A^5$$$ $h_a=\frac{16$$R^5-A^5$$}{15R^4}$ The height of the cylindrical shell: $I_b=I_c$ $\frac{1}{2}\pi\rho$$R^4-A^4$$h_b=\frac{8}{15}\pi\rho$$R^5-A^5$$$ $h_b=\frac{16}{15}\frac{$$R^5-A^5$$}{$$R^4-A^4$$}$

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