Question

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

Answer 1

The center of gravity from the end of the rod is 2 cm.

Explanation:

Given that,

Length = 3 m

Let us consider an elementary length dx.

Its mass acting per unit length is directly proportional to distance x from one of its end.

Its mass  is

$m=kx$

Where, $m=\dfrac{M}{L}$

$dm=kxdx$

We need to calculate the center of gravity of the rod

Using formula of center of gravity

$x_{cm}=\dfrac{\int{x dm}}{\int{dm}}$

Put the value into the formula

$x_{cm}=\dfrac{\int_{0}^{3}{x\timeskx dx}}{\int_{0}^{3}{kx dx}}$

$x_{cm}=\dfrac{(\dfrac{x^3}{3})_{0}^{3}}{(\dfrac{x^2}{2})_{0}^{3}}$

$x_{cm}=\dfrac{\dfrac{27}{3}}{\dfrac{9}{2}}$

$x_{cm}=2\ cm$

Hence, The center of gravity from the end of the rod is 2 cm.

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Topic : center of gravity

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