The distance of centre of mass from end A of a one dimensional rod (AB) having mass density $\rho \mathit{ }\mathit{=}\mathit{ }{\rho }_{\mathit{0}}\left(1- \frac{x^2}{L^2}\right)\mathit{ }$kg/m and length L. (in meter) is $\frac{3L}{\alpha }m$. The value of $\alpha$ is......... (where x is the distance form end A)

Formula for Center of Mass:

The position of the center of mass is:

x_cm = ( ∫₀ᴸ x ρ(x) dx ) / ( ∫₀ᴸ ρ(x) dx )

Here:

• ∫₀ᴸ ρ(x) dx: Total mass of the rod.
• ∫₀ᴸ x ρ(x) dx: Moment of mass about the origin.

Step 1: Total Mass (M)

The total mass of the rod is:

M = ∫₀ᴸ ρ(x) dx = ∫₀ᴸ ρ₀ (1 - x² / L²) dx

Split the integral:

M = ρ₀ ∫₀ᴸ 1 dx - ρ₀ ∫₀ᴸ (x² / L²) dx

Evaluate the integrals:

• ∫₀ᴸ 1 dx = L
• ∫₀ᴸ x² dx = (x³ / 3)₀ᴸ = L³ / 3

Substitute back:

M = ρ₀ (L - (L³ / 3L²)) = ρ₀ (L - L / 3) = ρ₀ (2L / 3)

Step 2: Moment of Mass

The moment of mass is:

Moment = ∫₀ᴸ x ρ(x) dx = ∫₀ᴸ x ρ₀ (1 - x² / L²) dx

Split the integral:

Moment = ρ₀ ∫₀ᴸ x dx - ρ₀ ∫₀ᴸ (x³ / L²) dx

Evaluate the integrals:

• ∫₀ᴸ x dx = (x² / 2)₀ᴸ = L² / 2
• ∫₀ᴸ x³ dx = (x⁴ / 4)₀ᴸ = L⁴ / 4

Substitute back:

Moment = ρ₀ ((L² / 2) - (L⁴ / 4L²)) = ρ₀ ((L² / 2) - (L² / 4)) = ρ₀ (L² / 4)

Step 3: Center of Mass

Now, calculate xcm:

x_cm = (Moment) / (M) = (ρ₀ (L² / 4)) / (ρ₀ (2L / 3))

Simplify:

x_cm = (L² / 4) / (2L / 3) = 3L / 8

Step 4: Determining α

From the problem, xcm = 3L / α. Comparing:

3L / α = 3L / 8

Thus:

α = 8

Final Answer:

The value of α is 8.