Question 618: Centre of Mass and Rotational Motion

Question

The radius of gyration of a solid sphere of mass $$5 \mathrm{~kg}$$ about $$X Y$$ is $$5 \mathrm{~m}$$ as shown in figure. The radius of the sphere is $$\frac{5 x}{\sqrt{7}} \mathrm{~m}$$, then the value of $x$ is:

Accepted Answer

$$I_{X Y}=I_{C M}+M R^2=\frac{2}{5} M R^2+M R^2=\frac{7}{5} M R^2=\frac{7}{5} 	imes 5 R^2=7 R^2\quad 	ext{.... (1)}$$
$$\begin{aligned}
& I_{X Y}=M K^2=5 	imes 5^2 \quad \ldots(2) \
& 	herefore 5 	imes 5^2=7 	imes R^2 \quad 	ext { [From (1) and (2)] } \
& \Rightarrow R=\sqrt{\frac{5}{7}} 	imes 5=\frac{5 x}{\sqrt{7}} \quad 	ext { (Given) } \
& 	herefore x=\sqrt{5}
\end{aligned}$$

Answer

5

Answer

$$\sqrt2$$

Answer

$$\sqrt3$$

Answer

$$\sqrt5$$

Centre of Mass and Rotational Motion

The radius of gyration of a solid sphere of mass $5 \mathrm{~kg} about X Y is 5 \mathrm{~m} as shown in figure. The radius of the sphere is \frac{5 x}{\sqrt{7}} \mathrm{~m}, then the value of x$ is:

Choose the correct answer: