Question

A non-uniform rod AB has a mass M ad length 2l. The mass 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 (a) through A (b) through the mid-point of AB.

Answer 1

Given :

• mass of rod = M
• length of rod = 2l
• mass per unit length of rod = mx of point at distance x from A

To find :

• moment of inertia of rod about axis perpendicular to the rod :

1. through A
2. through midpoint of AB

Solution :

Part A:

• Given that the mass of rod is M and length is 2l.
• Consider a small mass of length dx and mass dm from point A, then mass per unit length can be given as:

• dM= mxdx

• On integrating we will get total mass of rod on LHS (i.e M) and RHS can be integrated from x=0 to x=2l. Equation becomes,

• $M =\int\limits^{2l}_0 {mx} \, dx$

• On integrating and applying limits,

• M = 2ml²

• $m = \frac{M}{2L^{2} }$ .............(1)

• Moment of inertia of rod is

• dI = mxx²dx
• For inertia at A, we integrate from 0 to 2l.

• $I= \int\limits^{2l}_ 0{mx^{3} } \, dx$

• $I =m [ \frac{x^{4} }{4} ]\left \{ {{x=2l} \atop {x=0}} \right.$

• Substituting value of m from equation (1),

• I = 2ML².

For part B,

• Consider mid-point as a origin and length of the rod is from -L to +L and element of mass is dm and length dx at (L-x)

• Mass of the element is,
• dM= m(L-x)dx
• Moment of inertia of rod can be given by,
• dI= m(L-x)x² dx.

• Integrating the equation for limits -L to +L we get,

• $I = m[\frac{Lx^{3} }{3} - \frac{x^{4} }{4} ]\left \{ {{x=+L} \atop {x=-L}} \right.$

• We substitute value of m from (1).

• On applying limits and simplifying,

• $I = \frac{Ml^{3} }{3}$.

Answer :

At point A, the moment of inertia = 2ML²

At mid-point through AB, the moment of inertia= ML³/3.