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  • A rod 'l' has non-uniform linear mass density given by <img alt="\rho (x)=a+b\left ( \frac{x}{l} \right )^{2}," src="http://entrancecorner.oncodecogs.com/gif.latex?%5Crho%20%28x%29%3Da&plus;b%5Cleft%20%28%20%5Cfrac%7Bx%7D%7Bl%7D%2

A rod 'l' has non-uniform linear mass density given by \rho (x)=a+b\left ( \frac{x}{l} \right )^{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 :
Option: 1 \frac{3}{2}\left ( \frac{2a+b}{3a+b} \right )l
Option: 2  \frac{3}{2}\left ( \frac{a+b}{2a+b} \right )l
Option: 3  \frac{3}{4}\left ( \frac{2a+b}{3a+b} \right )l
Option: 4 \frac{4}{3}\left ( \frac{a+b}{2a+3b} \right )l
 

Answers (1)

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\begin{aligned} &\text { Given, } \lambda=\left(\mathrm{a}+\mathrm{b}\left(\frac{\mathrm{x}}{\mathrm{l}}\right)^{2}\right) \\ &\frac{\mathrm{d} \mathrm{N}}{\mathrm{dx}}=\lambda \\ &\mathrm{d} \mathrm{N}=\lambda \mathrm{dx}=\left(\mathrm{a}+\mathrm{b}\left(\frac{\mathrm{x}}{\mathrm{l}}\right)^{2}\right) \cdot \mathrm{dx} \\ &=\frac{\int \mathrm{x} \cdot \mathrm{dm}}{\int \mathrm{dm}}=\frac{\int \mathrm{x} \lambda \mathrm{dx}}{\int \lambda \mathrm{dx}} \\ &=\frac{\int_{0}^{1} \mathrm{x}\left(\mathrm{a}+\left(\frac{\mathrm{bx}^{2}}{1^{2}}\right)\right) \mathrm{dx}}{\int_{0}^{1}\left(\mathrm{a}+\frac{\mathrm{bx}^{2}}{1^{2}}\right) \mathrm{dx}}=\frac{\int_{0}^{1}\left(\mathrm{ax}+\frac{\mathrm{bx}^{3}}{1^{2}}\right) \mathrm{dx}}{\int_{0}^{1}\left(\mathrm{a}+\frac{\mathrm{bx}^{2}}{1^{2}}\right) \mathrm{dx}} \end{aligned}

\begin{aligned} &=\frac{\left[\frac{a x^{2}}{2}\right]_{0}^{1}+\frac{b}{1^{2}}\left[\frac{x^{4}}{4}\right]_{0}^{1}}{a[x]_{0}^{1}+\frac{b}{1^{2}}\left[\frac{x^{3}}{3}\right]_{0}^{1}}=\frac{\frac{a^{1^{2}}}{2}+\frac{b^{1^{2}}}{4}}{a l+\frac{b l}{3}} \\ &=\frac{(2 a+b) 1}{(3 a+b) 4} \times 3 \\ &=\frac{31}{4}\left(\frac{2 a+b}{3 a+b}\right) \end{aligned}

 

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