a non uniform rod AB has a mass M and length 2L the mas per unit length of the rod is mx at a point of the rod distant x from A.find the moment of inertia of this rod about an axis perpendicular to the rod (!)through A (2)through the mid point of AB

consider a strip of the rod ''dx'' at a length ''x'' from A.mass of the small strip is mx*dx.moment of inertia of the small strip about A IS m*x*dx*x*x.on integrating we get moment of inetia about A as 4*m*L*L*L*L.

ABOUT MID POINT OF AB.consider a strip of length ''dx'' at a distance of x from mid point P.mass of it is m*(L-x)*dx.moment of inertia of half of the rod about P is m*(L-x)*dx*x*x.on integration we get moment of inertia of half of the rod.similarly do for the other half.adding them we get the answer.

Dear Park kl,

for part 1.

mass of part dx at a distance x from A=mxdx
                         2L
I= integrating... [ mx.x^2.dx]= 4mL^4
                          0

for part 2.

let the midpoint be C.then.let the part whose mass per unit length is mentioned be at a distance of x from C.then

distance from A=L-x
                          L
I= integrating... [m(L-x).x^2dx]= m(L^4)/6.
                          -L

(ANS).

Regards,
Akash.

for part 1.

mass of part dx at a distance x from A=mxdx
                         2L
I= integrating... [ mx.x^2.dx]= 4mL^4
                          0

for part 2.

let the midpoint be C.then.let the part whose mass per unit length is mentioned be at a distance of x from C.then

distance from A=L-x
                          L
I= integrating... [m(L-x).x^2dx]= m(L^4)/6.
                          -L