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Question Number 24644 by Tinkutara last updated on 23/Nov/17

The density of a non - uniform rod of

length 1 m is given by ρ (x) = a (1 + {b x}^{2})

where a and b are constants and

0 ⩽ x ⩽ 1 . The centre of mass of the rod

will be at

(1) \frac{3 (2 + b)}{4 (3 + b)}

(2) \frac{4 (2 + b)}{3 (3 + b)}

(3) \frac{3 (3 + b)}{4 (2 + b)}

(4) \frac{4 (3 + b)}{3 (2 + b)}

Answered by mrW1 last updated on 23/Nov/17

x_{c} = \frac{\int_{0}^{1} a (1 + {b x}^{2}) x d x}{\int_{0}^{1} a (1 + {b x}^{2}) d x}

= \frac{\frac{1}{2} + \frac{b}{4}}{1 + \frac{b}{3}} = \frac{3 (2 + b)}{4 (3 + b)}

\Rightarrow o p t i o n (1)

Commented by Tinkutara last updated on 24/Nov/17

Thank you Sir!

Answered by jota+ last updated on 23/Nov/17

x_{C} = \frac{\int_{0}^{1} x ρ (x) d x}{\int_{0}^{1} ρ (x) d x}

Commented by Tinkutara last updated on 24/Nov/17

Thank you Sir!