Question

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EXPLAIN PROPERLY MOMENT OF INERTIA FROM LINE 3,4,5,6⏩⏩

Answer 1

Let the  legth of rod  be l  , and  mass be  m  

Mass of  unit  length = m/l

LINE 3:- picture 1

Take  a  small  part dx  whose  distance from axis  is  x

Mass of that  part = $\frac{m}{l}dx$

Moment  of  interia of  small part  =m/l x².dx

$\text{interia of whole rod}\\\;\\I=\int\limits^l_0 {\frac{m}{l}x^{2}}\,dx\\\;\\I=\frac{m}{l}\int\limits^l_0{x^{2}} \,dx\\\;\\I=\frac{m}{l}[\frac{x^{3}}{3}}]\limits^l_0\\\;\\I=\frac{m}{3l}l^{3}\\\;\\I=\frac{ml^{2}}{3}$

Line  5: picure 2

Similary,  as  above

Mass of small part = $\frac{m}{l}dx$

Distance from axis = xsinФ

limits = 0 to l

$I=\int\limits^l_0 {\frac{m}{l}dx.(x\sin\theta)^{2} } \,\\ \\I=\frac{m\sin^{2}\theta}{l}\int\limits^l_0 {x^{2}} \,dx\\ \\I=\frac{m\sin^{2}\theta}{3l}[x^{3}]\limits^l_0\\ \\I=\frac{m\sin^{2}\theta}{3l}.l^{3}\\ \\I=\frac{ml^{2} \sin^{2}\theta}{3}$

Line 4: Picture 3

Mass of small part = $\frac{m}{l}dx$

Distance from axis = xsinФ

limits = -l/2 to +l/2

$I=\int\limits^\frac{l}{2}_\frac{-l}{2}{\frac{m}{l}(x\sin\theta)^{2}dx} \,\\\\I=\frac{m\sin^{2}\theta}{l}\int\limits^\frac{l}{2}_\frac{-l}{2}{x^{2}}dx\\\\I=\frac{m\sin^{2}\theta}{3l}[x^{3} ]\limits^\frac{l}{2}_\frac{-l}{2}\\\\I=\frac{m\sin^{2}\theta}{3l}[\frac{l^{3}}{8}+\frac{l^{3}}{8}]\\\\I=\frac{m\sin^{2}\theta}{3l}.\frac{2l^{3}}{8}\\\\I=\frac{ml^{2} \sin^{2}\theta}{12}$

Line 6:

No need  of calculus.

Mass  of rod  = m

Distance from axis = x

$I=mx^{2}$