Question 611: Centre of Mass and Rotational Motion

Question

If radius of the solid sphere is doubled by keeping its mass constant, the ratio of their moment of inertia about any of its diameter is

Accepted Answer

Key Idea Moment of inertia of a solid sphere about its diameter is given as, $I=\frac{2}{5} M R^2$, where $M$ and $R$ are the mass and radius of the sphere, respectively.
Moment of inertia of a solid sphere about its diameter is $I=\frac{2}{5} M R^2\quad 	ext{.... (i)}$
As the radius of a solid sphere is doubled then the new moment of inertia of the sphere will be
$$I^{\prime}=\frac{2}{5} M(2 R)^2=\frac{8}{5} M R^2\quad	ext{.... (ii)}$$
From Eqs. (i) and (ii), we get
$$\frac{I}{I^{\prime}}=\frac{\frac{2}{5} M R^2}{\frac{8}{5} M R^2}=\frac{1}{4}=1: 4$$
The ratio of their moment of inertia about any of its diameter is $1: 4$.

Answer

$1: 8$

Answer

$2: 5$

Answer

$2: 3$

Answer

$1: 4$

Centre of Mass and Rotational Motion

If radius of the solid sphere is doubled by keeping its mass constant, the ratio of their moment of inertia about any of its diameter is

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