Question

A uniform thin rod ab of length l has linear mass density ¥=a +bx/l if centre of mass lies at a distance of 7l/12 from a then a and b are related as

Answer 1

The length of the rod is given as

Equation of linear mass density

D($x$) $= a \ + b*\frac{x}{l}$ .

where x is the distance measured from the point A.

Also given that the centre of mass (CM) lies at a distance,

$x_{cm}$ = $\frac{7*l}{12}$ cm .

we know that the Centre of mass of a body can be given by,

$x_{cm} = \frac{\int\limits^l_0 { D (x)*x} \, dx }{\int\limits^l_0 {D(x)} \, dx }$

⇒ $x_{cm } = \frac{\int\limits^l_0 {[(a+b*\frac{x}{l})*x] } .\, dx }{\int\limits^l_0 {(a+b*\frac{x}{l}). } \, dx }$

⇒  $\frac{7*l}{12} = \frac{(a\frac{l^2}{2} +b*\frac{l^2}{3}) }{(al\ +\ b *\frac{l}{2} )}$

⇒ $\frac{7}{12} = \frac{\frac{a}{2} +\frac{b}{3} }{a + \frac{b}{2} } = \frac{(\frac{3a+2b}{6} )}{(\frac{2a+b}{2}) }$

⇒ 14a + 7b = 12a + 8b

       ∴ 2a = b

HOPE THIS HELPS YOU !!    : )