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PART - 1 CH - 6 System of Particles and Rotational Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 1 CH - 6 System of Particles and Rotational Motion3 Marks
Question
A sphere of radius $R$ has been cut out of $\left(\frac{R}{2}\right)$ solid sphere of radius $R$ as shown in the figure. Find the position of the center of mass of the remaining sphere.
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Answer

 If the density of the substance of the sphere is $\rho$, then the mass of the sphere of radius $R$ is $M=\frac{4}{3} \pi R^3 \rho$
The mass of the sphere of radius $\left(\frac{R}{2}\right)$ is
$\begin{aligned}
M_1 & =\frac{4}{3} \pi\left(\frac{R}{2}\right)^3 \rho \\
& =\frac{4}{3} \frac{\pi R^3 \rho}{8}=\frac{M}{8}
\end{aligned}$
The remaining sphere of mass is
$M_2=M-\frac{M}{8}=\frac{7 M}{8}$
If the distance of the center of mass of the reamining sphere from the center of the sphere is $x$, then the center of mass of the complete sphere is taken at the origin O .
$\begin{aligned}
M_1\left(\frac{R}{2}\right)+M_2 x & =0 \\
M_1\left(\frac{R}{2}\right) & =-M_2 x \\
x & =-\frac{M_1}{M_2}\left(\frac{R}{2}\right) \\
& =-\frac{M}{8} \times \frac{8}{7 M}\left(\frac{R}{2}\right) \\
x & =-\frac{R}{14}\quad \text { Ans. }\end{aligned}$ 

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