# Calculate the radius of gyration of a rod of mass 100 g and length100 cm about an axis passing through its centre of gravity andperpendicular to its length.

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Uniform rod => density = d = M / L

Take a small length dx of the rod at a distance x from the center of gravity, which is same as the geometric center. Its mass dm = d * dx

moment of inertia = 2 * Sum of dm * x² for x = 0 to L/2

I = 2 * Integral of d * dx * x²

= 2 * d * [ x³/3 ] from x = 0 to L/2

= d L³ / 12 = M L² / 12

I = M * R² = M L² /12

Radius of gyration = R = L/2√3 = 50/√3 cm

Take a small length dx of the rod at a distance x from the center of gravity, which is same as the geometric center. Its mass dm = d * dx

moment of inertia = 2 * Sum of dm * x² for x = 0 to L/2

I = 2 * Integral of d * dx * x²

= 2 * d * [ x³/3 ] from x = 0 to L/2

= d L³ / 12 = M L² / 12

I = M * R² = M L² /12

Radius of gyration = R = L/2√3 = 50/√3 cm