Question 623: Centre of Mass and Rotational Motion

Question

The ratio of radius of gyration of a solid sphere of mass $M$ and radius $R$ about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is :-

Accepted Answer

To solve this problem, we need to find the ratio of the radius of gyration for a solid sphere (KÃ¢ÂÂ) to the radius of gyration for a thin hollow sphere (KÃ¢ÂÂ) of the same mass M and radius R.
The moment of inertia (I) of a solid sphere about its own axis is given by :
$I_{solid} = \frac{2}{5}MR^2$
The radius of gyration (K) is related to the moment of inertia (I) and mass (M) by the formula :
$I = MK^2$
So for the solid sphere, we can find KÃ¢ÂÂ using :
$K^2_{1} = \frac{I_{solid}}{M} = \frac{2}{5}R^2$
$K_{1} = R\sqrt{\frac{2}{5}}$
Now, for a thin hollow sphere, the moment of inertia about its axis is given by :
$I_{hollow} = \frac{2}{3}MR^2$
We can find KÃ¢ÂÂ using :
$K^2_{2} = \frac{I_{hollow}}{M} = \frac{2}{3}R^2$
$K_{2} = R\sqrt{\frac{2}{3}}$
Now, we need to find the ratio KÃ¢ÂÂ : KÃ¢ÂÂ :
$\frac{K_{1}}{K_{2}} = \frac{R\sqrt{\frac{2}{5}}}{R\sqrt{\frac{2}{3}}}$
The R terms cancel out, and we are left with :
$\frac{K_{1}}{K_{2}} = \frac{\sqrt{\frac{2}{5}}}{\sqrt{\frac{2}{3}}}$
Simplify by taking the square root of the fraction :
$\frac{K_{1}}{K_{2}} = \frac{\sqrt{2}\sqrt{3}}{\sqrt{5}\sqrt{2}}$
Now, the square root of 2 terms cancel out, giving us :
$\frac{K_{1}}{K_{2}} = \frac{\sqrt{3}}{\sqrt{5}}$

Answer

$5: 3$

Answer

$2: 5$

Answer

$\sqrt{5}: \sqrt{3}$

Answer

$\sqrt{3}: \sqrt{5}$

Centre of Mass and Rotational Motion

The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is :-

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