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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 ABAns) 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.
When a system returns back to its initial state after passing by dissimilar phase, the phenomena is said to be a cyclic process. In this process as ΔU = 0. From first law of the
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