The distance of the centre of mass of a solid | vedclass

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(A) Consider a solid cone of height $h$ and base radius $R$. Let the density of the material be $\rho$.We place the vertex at the origin $(0,0)$ and t

Vedclass · Accepted Answer

$\frac{3h}{4}$

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(A) Consider a solid cone of height $h$ and base radius $R$. Let the density of the material be $\rho$.We place the vertex at the origin $(0,0)$ and the axis of the cone along the $y$-axis.At a distance $y$ from the vertex,consider a thin elemental disk of thickness $dy$ and radius $r$.By similar triangles,$\frac{r}{y} = \frac{R}{h}$,which implies $r = \frac{R}{h}y$.The mass of this elemental disk is $dm = \rho \cdot \pi r^2 dy = \rho \pi \left(\frac{R}{h}y\right)^2 dy$.The centre of mass $z_0$ (or $y_{cm}$) is given by:$z_0 = \frac{\int y dm}{\int dm} = \frac{\int_0^h y \cdot \rho \pi \frac{R^2}{h^2} y^2 dy}{\int_0^h \rho \pi \frac{R^2}{h^2} y^2 dy}$$z_0 = \frac{\int_0^h y^3 dy}{\int_0^h y^2 dy} = \frac{[y^4/4]_0^h}{[y^3/3]_0^h} = \frac{h^4/4}{h^3/3} = \frac{3h}{4}$.

The distance of the centre of mass of a solid uniform cone from its vertex is $z_0$. If the radius of its base is $R$ and its height is $h$,then $z_0$ is equal to:

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