NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
Find the x coordinate of the centre of mass of the non-uniform rod of length L given below. The origin is taken at the left end of the rod. The density of the rod as a function of its x -coordinate is ρ = ax 2 + bx + c , where a , b and c are constants.
Options
- A2 aL 2 + 3 bL 2 + 6 cL 2 ( 3 aL 2 + 4 bL + 8 c )
- B4 aL 3 + 3 bL 2 + 2 cL 2 ( 3 aL 2 + 2 bL + c )
- C3 aL 2 + 4 bL 2 + 6 cL 4 aL 2 + 6 bL + 8 c
- D3 aL 3 + 4 bL 2 + 6 cL 2 ( 2 aL 2 + 3 bL + 6 c )
Correct answer
D. 3 aL 3 + 4 bL 2 + 6 cL 2 ( 2 aL 2 + 3 bL + 6 c )
Step-by-step solution
∫ 0 L ( ax 2 + bx + c ) x dx ∫ 0 L ( ax 2 + bx + c ) dx ∫ 0 L ( ax 3 + bx 2 + cx ) dx ∫ 0 L ( ax 2 + bx + c ) dx = ax 4 4 + bx 3 3 + cx 2 2 0 L ax 3 3 + bx 2 2 + cx 0 L = 3 aL 4 + 4 bL 3 + 6 cL 2 12 2 aL 2 + 3 bL 2 + 6 cL 6 = 3 aL 3 + 4 bL 2 + 6 cL 2 2 aL 2 + 3 bL + 6 c