The moment of inertia of a solid sphere about an axis passing through centre of gravity is , then its radius of gyration about a parallel axis at a distance 2R from first axis is

Engineering

Physics

Moment of Inertia

Question

The moment of inertia of a solid sphere about an axis passing through centre of gravity is $\frac{2}{5} {MR}^{2}$ , then its radius of gyration about a parallel axis at a distance 2R from first axis is

$\sqrt{\frac{22}{5}} R$

$\frac{5}{2} R$

$\sqrt{\frac{12}{5}} R$

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The radius of gyration K of a body about a given axis of rotation is that radial distance from the axis, the square of which on being multiplied by the total mass of the body gives the moment of inertia of the body about that axis. Thus,

I = MK2= ΣmR2

where Misthe total mass of the body.
This means that $K = \sqrt{\frac{I}{M}}$
According to the orem of parallel axis

I = I_CG + M(2R)²
is moment of inertia about an axis through centre of gravity.

therefore

I = 2/5MR²+ 4MR²= 22/5MR² or MK²= 22/5MR²
therefore

∴ K = \sqrt{\frac{22}{5}} R

The total mass of a body may be supposed to be concentrated at a radial distance K from the axis of rotation, so far as the moment of inertia of the