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Physics Question on System of Particles & Rotational Motion

A rod of length LL has non-uniform linear mass density given by ρ(x)=a+b(xL)2,\rho\left(x\right)=a+b\left(\frac{x}{L}\right)^{2}, where a and b are constants and 0≤x≤L.0 \le x \le L. The value of xx for the centre of mass of the rod is a t :

A

32(2a+b3a+b)L\frac{3}{2}\left(\frac{2a+b}{3a+b}\right)L

B

32(a+b2a+b)L\frac{3}{2}\left(\frac{a+b}{2a+b}\right)L

C

34(2a+b3a+b)L\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L

D

43(a+b2a+3b)L\frac{4}{3}\left(\frac{a+b}{2a+3b}\right)L

Answer

34(2a+b3a+b)L\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L

Explanation

Solution

xcm=∫xdm∫dm=∫(λdx)x∫dmx_{cm} = \frac{\int xdm}{\int dm} =\frac{\int\left(\lambda dx\right)x}{\int dm}
=∫0L(a+bx2L2)xdx∫0L(a+bx2L2)dx= \frac{\int^{L}_{0}\left(a+\frac{bx^{2}}{L^{2}}\right)xdx}{\int ^{L}_{0}\left(a+\frac{bx^{2}}{L^{2}}\right)dx}
=aL22+bL2.L44aL+bL2.L33= \frac{\frac{aL^{2}}{2}+\frac{b}{L^{2}}. \frac{L^{4}}{4}}{aL+\frac{b}{L^{2}}. \frac{L^{3}}{3}}
=(4a+2b8)L(3a+b)3=34(2a+b)L(3a+b)= \frac{\left(\frac{4a+2b}{8}\right)L}{\frac{\left(3a+b\right)}{3}} = \frac{3}{4} \frac{\left(2a+b\right)L}{\left(3a+b\right)}