Question

A rod of length is 3 m and its mass acting per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at :-

• A. 1.5 m
• B. 2 m
• C. 2.5 m
• D. 3.0 m

Answer 1

The Correct Option is B

To find the center of gravity of a rod whose mass per unit length is directly proportional to its distance from one end, we can use the following approach:

Step 1: Define the mass distribution.

Let the rod be placed along the x-axis from x = 0 to x = 3 \, \text{m}. Assume the mass per unit length (linear mass density) at a distance x from one end is given by \lambda(x) = kx, where k is a constant of proportionality.

Step 2: Express total mass of the rod.

The total mass M of the rod is the integral of the linear density along the length of the rod:

M = \int_0^3 \lambda(x) \, dx = \int_0^3 kx \, dx

M = \left[ \frac{kx^2}{2} \right]_0^3 = \frac{k(3)^2}{2} = \frac{9k}{2}

Step 3: Find the center of gravity.

The center of gravity (x_{\text{cg}}) is given by:

x_{\text{cg}} = \frac{\int_0^3 x \lambda(x) \, dx}{M} = \frac{\int_0^3 x (kx) \, dx}{M}

x_{\text{cg}} = \frac{\int_0^3 kx^2 \, dx}{\frac{9k}{2}} = \frac{k \left[ \frac{x^3}{3} \right]_0^3}{\frac{9k}{2}}

= \frac{k \cdot \frac{27}{3}}{\frac{9k}{2}} = \frac{9k}{\frac{9k}{2}} = 2 \, \text{m}

Thus, the center of gravity from the given end of the rod will be 2 m.