A carpenter has constructed a toy as shown in | vedclass

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Question

If the density of cone be $\rho,$ then its mass will be,

$m_{1}=\frac{1}{3} \pi(2 R)^{2}(4 R) \rho=\frac{16}{3} \pi R^{3} \rho$

and its centre o

Vedclass · Accepted Answer

at a distance of $4R$ from $O$

Vedclass · Answer

If the density of cone be $ho,$ then its mass will be,

$m_{1}=\frac{1}{3} \pi(2 R)^{2}(4 R) ho=\frac{16}{3} \pi R^{3} ho$

and its centre of mass will be at a height

$\frac{\mathrm{h}}{4}=\frac{4 \mathrm{R}}{4}=\mathrm{R}$

from $0$ on the line of symmetry,

i.e., $y_{1}=R$

Similarly, the mass of the sphere,

$\mathrm{m}_{2}=\frac{4}{3} \pi \mathrm{R}^{3}(12 ho)=16 \pi \mathrm{R}^{3} ho=3 \mathrm{m}_{1}$

and its centre of mass will be at its centre $\mathrm{O}_{2}$

i.e., $\left.y_{2}=4 R+R=5 R 	ext { (from } Oight)$ Now, treating the sphere and cone as point masses with their masses concentrated at their centres of mass respectively and taking the line of symmetry as $y$ $-$axis with origin at $O,$ the centre of mass of the toy is given by$:$

$Y_{C M}=\frac{m_{1} y_{1}+m_{2} y_{2}}{m_{1}+m_{2}}=\frac{m_{1} 	imes R+3 m_{1} 	imes 5 R}{m_{1}+3 m_{1}}=4 R$

i.e., centre of mass of the toy is at a distance of

$4R$ from $0$ on the line of symmetry, i.e., at the apex of the cone.

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 cone, the position of the centre of mass of the toy is given by

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