First slide

Centre of mass(com)

Question

A rod of length L has non-uniform linear mass density given by $ρ (x) = a + b {(\frac{x}{L})}^{2}$ , where a and b are constants and $0 \leq x \leq L$ . The value of x for the centre of mass of the rod is at

Difficult

A

$\frac{4}{3} (\frac{a + b}{2 a + 3 b}) L$
B

$\frac{3}{4} (\frac{2 a + b}{3 a + b}) L$
C

$\frac{3}{2} (\frac{2 a + b}{3 a + b}) L$
D

$\frac{3}{2} (\frac{a + b}{2 a + b}) L$

Solution

$\begin{array}{l} G i v e n : \frac{d m}{d x} = a + b \frac{x^{2}}{L^{2}} \\ \Rightarrow d m = (a + b \frac{x^{2}}{L^{2}}) d x \\ N o w, x_{c m} = \frac{\int x d m}{\int d m} = \frac{\int_{0}^{L} (a x + \frac{b x^{3}}{L^{2}}) d x}{\int_{0}^{L} (a + \frac{b x^{2}}{L^{2}}) d x} \\ \Rightarrow x_{c m} = {\frac{\frac{a x^{2}}{2} + \frac{b x^{4}}{4 L^{2}}}{a x + \frac{b x^{3}}{3 L^{2}}}}_{0}^{L} \\ \Rightarrow x_{c m} = \frac{\frac{a L^{2}}{2 L} + \frac{b L^{4}}{4 L^{2}}}{a L + \frac{b L^{3}}{3 L^{2}}} \\ \Rightarrow x_{c m} = \frac{\frac{2 a L^{2} + b L^{2}}{4}}{\frac{3 a L + b L}{3}} = \frac{3}{4} (\frac{2 a + b}{3 a + b}) L \end{array}$

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