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

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Conservation of Momentum

The distance of centre of mass from end A of a one dimensional rod (AB) having mass density

ρ = ρ_{0} (1 - \frac{x ^{2}}{L ^{2}}) kg / m

and length

L

(in meter) is

\frac{3 L}{α} m

. The value of

α

is ........... (where x is the distance form end A)

Answer: 8

Consider an element at distance

x

from end A with length

d x

. Its mass is

d m = λ d x = ρ_{0} (1 - \frac{x ^{2}}{L ^{2}}) d x

.

The center of mass is:

X_{cm} = \frac{\int x d m}{\int d m} = \frac{ρ _{0} \int _{0}^{L} x ( 1 - \frac{x ^{2}}{L ^{2}} ) d x}{ρ _{0} \int _{0}^{L} ( 1 - \frac{x ^{2}}{L ^{2}} ) d x}

Numerator:

\int_{0}^{L} (x - \frac{x ^{3}}{L ^{2}}) d x = \frac{L ^{2}}{2} - \frac{L ^{2}}{4} = \frac{L ^{2}}{4}

Denominator:

\int_{0}^{L} (1 - \frac{x ^{2}}{L ^{2}}) d x = L - \frac{L}{3} = \frac{2 L}{3}

X_{cm} = \frac{L ^{2} /4}{2 L /3} = \frac{3 L}{8}

Comparing with

\frac{3 L}{α}

, we get

α = 8

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer