Moment of inertia of a hollow sphere with external radius \(a\) and internal radius \(b\) is,
\(I=\frac{2}{5} M\left(\frac{a^{5}-b^{5}}{a^{3}-b^{3}}\right)\)
Radius of gyration,
\(K=\sqrt{\frac{I}{M}}\)
\(\Rightarrow K=\frac{\sqrt{\frac{2}{5} M\left(\frac{a^{5}-b^{5}}{a^{3}-b^{3}}\right)}}{M}\)
\(\Rightarrow K=\sqrt{\frac{2}{5}\left(\frac{a^{5}-b^{5}}{a^{3}-b^{3}}\right)}\)
So, the correct option is (D): \(\sqrt{\frac{2}{5}\left(\frac{a^{5}-b^{5}}{a^{3}-b^{3}}\right)}\)
If \( 2 \) is a solution of the inequality \( \frac{x-a}{a-2x}<-3 \), then \( a \) must lie in the interval:
The axis of a parabola is parallel to the y-axis and its vertex is at \((5, 0)\). If it passes through the point \((2, 3)\), then its equation is:
Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.
In general form, the moment of inertia can be expressed as,
I = m × r²
Where,
I = Moment of inertia.
m = sum of the product of the mass.
r = distance from the axis of the rotation.
M¹ L² T° is the dimensional formula of the moment of inertia.
The equation for moment of inertia is given by,
I = I = ∑mi ri²
To calculate the moment of inertia, we use two important theorems-