Question

9. The density of a cylinder of radius R and length / varies linearly from the central axis where p = 500 kg/m to the value p. = 3p. IfR= .05 m and I = .1 m, find: a. The average densityof the cylinder over the radius. b. The average density over its volume. c. The moment of inertia of the cylinder about its central axis.

Answer 1

Let density variation is given by

$\rho\left ( r \right ) = \rho_{o}+m r$ .....................(1)

ar r = 0 , $\rho$ (0) = = 500 kg/m³

Po

at r = R, $\rho$ (R) = 500 + m R = 1500 , hence m = 1000/R

Hence density function $\rho$ (r) = 500 + 1000 (r/R)

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average density over radius

since density varying linearly across radial direction, average is (500+1500)/2 = 1000 kg/m³

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average density over volume

$\bar{\rho } = \frac{1}{V}\int_{0}^{V}\rho (v') dv' = \frac{1}{V}\int_{o}^{R} \rho (r) 2\pi r l dr = \frac{2 \pi l}{V}\int_{0}^{R}\rho (r)dr$

$\bar{\rho } = \frac{2\pi l}{V}\int_{0}^{R}\left \{ 500 + 1000\left ( \frac{r}{R} \right ) \right \}r dr = \frac{1000\pi l R^{2}}{V}\left ( \frac{1}{2} + \frac{2}{3} \right )$

since $\pi$ R²l is volume V, then $\bar{\rho }$ = (7/6)$\times$1000 kg/m³ = 1167 kg/m³

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Moment of inertia I is given by,

$I = \int dI = \int_{0}^{R}m(r)r^{2} = \int_{0}^{R}2\pi rdr l \rho (r)r^{2}$

$I = 2\pi l \int_{0}^{R} \rho (r)r^{3} dr = 2\pi l \int_{0}^{R} \left ( 500 + 1000\left ( \frac{r}{R} \right ) \right ) r^{3} dr$

$I = 650\pi l R^{4} = 650\pi\times 0.1\times \left ( 0.05 \right )^{4} = 1.276\times 10^{-3} kg m^{2}$