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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

हिन्दी

A thin uniform rod of length \( L \) lies along the \( x \)-axis from \( x = 0 \) to \( x = L \). It is placed in a non-uniform gravitational field where acceleration due to gravity is directed along \( -\hat{j} \) and has magnitude \( g(x) = g_0(1 + \alpha x) \). The \( x \)-coordinate of the centre of gravity of the rod is:

  1. \( \frac{L(3 + 2\alpha L)}{3(2 + \alpha L)} \)
  2. \( \frac{L(2 + \alpha L)}{2(3 + \alpha L)} \)
  3. \( \frac{L(1 + \alpha L)}{2 + \alpha L} \)
  4. \( \frac{L(4 + 3\alpha L)}{6(1 + \alpha L)} \)

Correct answer · Option 1

\( \frac{L(3 + 2\alpha L)}{3(2 + \alpha L)} \)

Explanation

\( x_{\text{cg}} = \frac{\int_0^L x g(x) dx}{\int_0^L g(x) dx} = \frac{\int_0^L (x + \alpha x^2) dx}{\int_0^L (1 + \alpha x) dx} = \frac{L^2/2 + \alpha L^3/3}{L + \alpha L^2/2} = \frac{L(3 + 2\alpha L)}{3(2 + \alpha L)} \).
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