A carpenter has constructed a toy as shown in | vedclass

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Question

Let $d$ be the density of the material of the cone and $12d$ that of the sphere. Then the mass of the cone will be

${m_c} = \frac{1}{3}\pi {\left(

Vedclass · Accepted Answer

at a distance $4R$ from $O$

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Let $d$ be the density of the material of the cone and $12d$ that of the sphere. Then the mass of the cone will be

${m_c} = \frac{1}{3}\pi {\left( {2R} \right)^2}\left( {4R} \right)d$

$ = \frac{{16}}{3}\pi {R^3}d$

The position of center of mass of the cone is ${C_1}$ is at a height $O{C_1} = {y_c} = \frac{{4R}}{4} = R$     (from $O$)

Also the mass of the sphere will be

${m_s} = \frac{4}{3}\pi {R^3}\left( {12d} \right)$

$ = 16\pi {R^3}d = 3{m_c}$

and its position of center of mass ${C_2}$ is

${y_s} = O{C_2} = 4R + R = 5R$      (from $O$)

Now ${y_{cm}} = \frac{{{m_c}{y_c} + {m_s}{y_s}}}{{{m_c} + {m_s}}}$

${y_{cm}} = \frac{{{m_c}\left( R \right) + 3{m_c}\left( {5R} \right)}}{{{m_c} + 3{m_c}}} = 4R$     (from $O$)

A carpenter has constructed a toy as shown in the adjoining figure. If the density of the material of the sphere is $12$ $times$ that of the cone, the position of the centre of mass of the toy is given by

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