If the linear density of the rod of length L starting from one end x-0 varies as -A-bx- then its centre of mass of the rod will be at

The correct option is A X_cm=L(3A+2BL)/3(2A+BL) Total mass of rod is M=∫_0^LA+Bxdx=(Ax+Bx^2/2)|_0^L=(2A+BL)L/2 Now position of COM is given as X ...

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Standard XII

Physics

Area and Volume Expansion

If the linear...

Question

If the linear density of the rod of length L starting from one end (x=0) varies as $λ = A + b x$ , then its centre of mass of the rod will be at

X_{c m} = \frac{L (2 A + B L)}{3 (3 A + 2 B L)}

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X_{c m} = \frac{L (3 A + 2 B L)}{3 (2 A + B L)}

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X_{c m} = \frac{L (3 A + 2 B L)}{3}

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X_{c m} = \frac{L (2 A + 3 B L)}{3}

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Solution

The correct option is A $X_{c m} = \frac{L (3 A + 2 B L)}{3 (2 A + B L)}$
Total mass of rod is $M = \int_{0}^{L} A + B x d x = (A x + B x^{2} / 2) |_{0}^{L} = (2 A + B L) L / 2$
Now position of COM is given as $X_{c} m = \frac{\int_{0}^{L} (A + B x) x d x}{M} = \frac{(A x^{2} / 2 + B x^{3} / 3) |_{0}^{L}}{M} = \frac{(3 A + 2 B L) L^{2} / 6}{(2 A + B L) L / 2} = \frac{L (3 A + 2 B L)}{3 (2 A + B L)}$

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