Consider a rod AB of length L Whose mass per length is given by Lambda...
Solution
Finding the moment of inertia of the rod
The moment of inertia of the rod about an axis passing through a point P can be calculated as:
I = ∫r² dm
where r is the distance of an infinitesimal element of the rod from the axis of rotation and dm is the mass of the element.
Expressing the mass per unit length in terms of x
Given the mass per unit length of the rod, lambda = (1 + kx), we can express the mass of an infinitesimal element dx as:
dm = lambda dx = (1 + kx) dx
Finding the distance of an infinitesimal element from the axis of rotation
Let the distance of an infinitesimal element dx from point P be y. Then, the distance of the element from point A is (L-x) and the distance of the element from point P is (y+AP). Using the Pythagorean theorem, we get:
(L-x)² = (y+AP)² + y²
Simplifying the above equation, we get:
y = (L-2x-AP²)/(2AP)
Substituting the expressions for dm and r in the moment of inertia equation
Substituting the expression for dm and r in the moment of inertia equation, we get:
I = ∫r² dm
I = ∫(y+AP)² (1+kx) dx
I = ∫((L-2x-AP²)/(2AP) + AP)² (1+kx) dx
Simplifying the above equation, we get:
I = (1/3)(L² + AP²)(1+kL) + (1/2)k(AP^4)/3
Finding the value of AP that minimizes the moment of inertia
To find the value of AP that minimizes the moment of inertia, we need to differentiate I with respect to AP and set it to zero.
dI/d(AP) = 2AP(1+kL) - (2/3)k(AP³) = 0
Solving the above equation, we get:
AP = L/(3k)^(1/2)
Therefore, AP = L/√3k is the value that minimizes the moment of inertia of the rod about an axis passing through point P.