Exam

JEE Main

Year

2020

Date

9 January

Shift

Evening Shift

Subject

Physics

Chapter

Center of Mass, Momentum and Collision

JEE Main 2020 · 9 January · Evening Shift · Physics · Center of Mass, Momentum and Collision

A rod of length l has non-uniform linear mass density given by ρ...

A rod of length $l$ has non-uniform linear mass density given by $ρ$\left(x\right) = a + b{\left(\dfrac{x}{l}\right)}^{2}, where $a$ and $b$ are constants and $0 \le x \le l$ The value of $x$ for the centre of mass of the rod is at:

Step-by-step solution

${X}_{cm} = \dfrac{1}{M}\mathop{∫}_{0}^{ι}x . dM$
$dM = ρ . dx = \left(a + b{\left(\dfrac{x}{ı}\right)}^{2}\right) . dx$
${x}_{cm} = \dfrac{∫ xdM}{∫ dm} = \dfrac{∫ xρdx}{∫ ρdx} = \dfrac{\mathop{∫}_{0}^{ι}x\left(a + \dfrac{b{x}^{2}}{{ι}^{2}}\right)dx}{\mathop{∫}_{0}^{ι}\left(a + \dfrac{b{x}^{2}}{{ι}^{2}}\right)dx}$
$= \dfrac{a{\left(\dfrac{{x}^{2}}{2}\right)}_{0}^{ι} + \dfrac{b}{{ι}^{2}}{\left(\dfrac{{x}^{4}}{4}\right)}_{0}^{ι}}{a{\left(x\right)}_{0}^{ι} + \dfrac{b}{{ι}^{2}}{\left(\dfrac{{x}^{3}}{3}\right)}_{0}^{ι}}$
= \dfrac{a\dfrac{{ι}^{2}}{2} + b\dfrac{{ι}^{2}}{4}}{aι + \dfrac{bι}{3}} = \dfrac{\left(2a + b\right)}{\left(3a + b\right)}\dfrac{ι}{4} \times 3$<br />$= \dfrac{3ι}{4}\left(\dfrac{2a + b}{3a + b}\right)