A rod of length l is pivoted about an end. Find moment of inertia of the rod about this axis if the linear mass density of rod varies as ρ = ax² + b (x is distance from axis of rotation). If it is al⁵/[5(p)] + bl³/q, then p × q = ...
Physics · Mixed Chapter
JEE Mains

Question

A rod of length l is pivoted about an end. Find moment of inertia of the rod about this axis if the linear mass density of rod varies as ρ = ax² + b (x is distance from axis of rotation). If it is al⁵/[5(p)] + bl³/q, then p × q = ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Each small element of rod at distance x has mass dm=ρ(x)dx=(ax²+b)dx. Step 2: Moment of inertia about the pivot is I=∫₀ˡx²dm=∫₀ˡ(ax⁴+bx²)dx. Step 3: Integrating gives I=al⁵/5+bl³/3. Step 4: Compare with al⁵/[5p]+bl³/q to find p=1, q=3 and p×q=3.