The linear mass density of a ladder of length `l` increases uniformly from one end `A` to the other end `B`,
(a) Form an expression for linear mass density as function of distance `x` from end `A` where linear mass density `lambda_(0)`. The density at one end being twice that of the other end.
(b) find the position of the centre of mass from end `A`.

Answer

(a) `lambda(x)=lambda_(0)+(lambda_(0)x)/(L)`, (b) `(5)/(9)L`

Explanation

(a) `lambda="Ax+B"`
`lambda_(A)=A(0)+B=B=lambda_(0)[x=0]`
`lambda_(B)=AL+B=2lambda_(0)rArrA=(lambda_(0))/(L)`
`rArr" "lambda=(lambda_(0))/(L)x+lambda_(0)`
(b) `x_(CM)=(int(x)lambdadx)/(intlambdadx)=(int_(0)^(L)(x)((lambda_(0))/(L)x+lambda_(0))dx)/(int_(0)^(L)((lambda_(0))/(L)x+lambda_(0))dx)=((L^(2))/(3)+L^(2)/(2))/((L)/(2)+L)=((5L^(2))/(6))/((3L)/(2))=(5)/(9)L`