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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

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43a+b2a+3bL

342a+b3a+bL

322a+b3a+bL

32a+b2a+bL

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detailed solution

Correct option is B

Given : dmdx=a+bx2L2 ⇒dm=a+bx2L2dx Now, xcm=∫xdm∫dm=∫0Lax+bx3L2dx∫0La+bx2L2dx ⇒ xcm=ax22+bx44L2ax+bx33L20L⇒xcm=aL22L+bL44L2aL+bL33L2 ⇒ xcm=2aL2+bL243aL+bL3 =342a+b3a+bL

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A rod of length L has non-uniform linear mass density given by ρx=a+bxL2, where a and b are constants and 0≤x≤L. The value of x for the centre of mass of the rod is at

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free study material

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