Question Number 22956 by Physics lover last updated on 24/Oct/17
Commented by Physics lover last updated on 24/Oct/17
$${find}\:{angular}\:{velocity}\:{after}\: \\ $$$${collision}\:{and}\:{loss}\:{in}\:{kinetic} \\ $$$${energy}? \\ $$
Commented by ajfour last updated on 24/Oct/17
Commented by ajfour last updated on 24/Oct/17
$$\:\:\:\:\omega{l}/\mathrm{2}\:={v}_{\mathrm{1}} −{V}\:\:\:\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\:\:\:\:{conserving}\:{linear}\:{momentum} \\ $$$$\:\:\:\:\:\left(\mathrm{2}{m}\right){V}+{mv}_{\mathrm{1}} ={mv} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{2}{V}+{v}_{\mathrm{1}} ={v}\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$${conserving}\:{angular}\:{momentum} \\ $$$${about}\:{a}\:{point}\:{on}\:{table}\:{just}\:{below} \\ $$$${the}\:{right}\:{end}\:{of}\:{stick}\:\left(\bullet\right). \\ $$$$\:\:\:\left(\frac{\mathrm{2}{ml}^{\mathrm{2}} }{\mathrm{12}}\right)\omega−\left(\mathrm{2}{m}\right){V}\left(\frac{{l}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\omega{l}=\mathrm{6}{V}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$${using}\:{this}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:{v}_{\mathrm{1}} −{V}=\mathrm{3}{V}\:\:\:{or}\:\:\:\:\boldsymbol{{v}}_{\mathrm{1}} =\mathrm{4}\boldsymbol{{V}}\:\: \\ $$$${using}\:{this}\:{in}\:\left({ii}\right) \\ $$$$\:\:\:\:\mathrm{6}{V}={v}\:\:\:\:\:{or}\:\:{V}=\frac{\boldsymbol{{v}}}{\mathrm{6}}\: \\ $$$${then}\:\:\:\:{v}_{\mathrm{1}} =\mathrm{4}{V}\:=\:\frac{\mathrm{2}\boldsymbol{{v}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\omega{l}}=\mathrm{6}\boldsymbol{{V}}\:=\boldsymbol{{v}} \\ $$$$\:\:\:\boldsymbol{\omega}=\boldsymbol{{v}}/\boldsymbol{{l}}\:\:. \\ $$$$\boldsymbol{{K}}_{\boldsymbol{{i}}} =\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{mv}}^{\mathrm{2}} \\ $$$$\boldsymbol{{K}}_{\boldsymbol{{f}}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\boldsymbol{{m}}\right)\boldsymbol{{V}}^{\:\:\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}\boldsymbol{{ml}}^{\mathrm{2}} }{\mathrm{12}}\right)\boldsymbol{\omega}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{mv}}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{36}}+\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{2}\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{12}\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{36}}\:=\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\boldsymbol{{loss}}\:\boldsymbol{{in}}\:\boldsymbol{{K}}\:=\boldsymbol{{K}}_{\boldsymbol{{i}}} −\boldsymbol{{K}}_{\boldsymbol{{f}}} \\ $$$$\:\:\:\:\:=\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{2}}−\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{3}}\:\:\:=\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$$$ \\ $$
Commented by Physics lover last updated on 25/Oct/17
$${thankd}\:{you}\:{very}\:{much},{Mr}\:{Ajfour}. \\ $$$$ \\ $$
Answered by ajfour last updated on 24/Oct/17
$$\:\omega\:=\frac{{v}}{{l}}\:;\:\:\:\:{loss}\:{in}\:{K}\:=\frac{{mv}^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$
Commented by Physics lover last updated on 24/Oct/17
$${yes}\:{sir}!! \\ $$