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Question-22956




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?
$${find}\:{angular}\:{velocity}\:{after}\: \\ $$$${collision}\:{and}\:{loss}\:{in}\:{kinetic} \\ $$$${energy}? \\ $$
Commented by ajfour last updated on 24/Oct/17
Commented by ajfour last updated on 24/Oct/17
    ωl/2 =v_1 −V           ...(i)      conserving linear momentum       (2m)V+mv_1 =mv  ⇒     2V+v_1 =v       ...(ii)  conserving angular momentum  about a point on table just below  the right end of stick (•).     (((2ml^2 )/(12)))ω−(2m)V((l/2))=0  ⇒   ωl=6V                         ...(iii)  using this in (i)       v_1 −V=3V   or    v_1 =4V    using this in (ii)      6V=v     or  V=(v/6)   then    v_1 =4V = ((2v)/3)        𝛚l=6V =v     𝛚=v/l  .  K_i =(1/2)mv^2   K_f =(1/2)(2m)V^(  2) +(1/2)(((2ml^2 )/(12)))𝛚^2 +                          +(1/2)mv_1 ^2          =((mv^2 )/(36))+((mv^2 )/(12))+((2mv^2 )/9)        =((12mv^2 )/(36)) =((mv^2 )/3)  loss in K =K_i −K_f        =((mv^2 )/2)−((mv^2 )/3)   =((mv^2 )/6) .
$$\:\:\:\:\omega{l}/\mathrm{2}\:={v}_{\mathrm{1}} −{V}\:\:\:\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\:\:\:\:{conserving}\:{linear}\:{momentum} \\ $$$$\:\:\:\:\:\left(\mathrm{2}{m}\right){V}+{mv}_{\mathrm{1}} ={mv} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{2}{V}+{v}_{\mathrm{1}} ={v}\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$${conserving}\:{angular}\:{momentum} \\ $$$${about}\:{a}\:{point}\:{on}\:{table}\:{just}\:{below} \\ $$$${the}\:{right}\:{end}\:{of}\:{stick}\:\left(\bullet\right). \\ $$$$\:\:\:\left(\frac{\mathrm{2}{ml}^{\mathrm{2}} }{\mathrm{12}}\right)\omega−\left(\mathrm{2}{m}\right){V}\left(\frac{{l}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\omega{l}=\mathrm{6}{V}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$${using}\:{this}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:{v}_{\mathrm{1}} −{V}=\mathrm{3}{V}\:\:\:{or}\:\:\:\:\boldsymbol{{v}}_{\mathrm{1}} =\mathrm{4}\boldsymbol{{V}}\:\: \\ $$$${using}\:{this}\:{in}\:\left({ii}\right) \\ $$$$\:\:\:\:\mathrm{6}{V}={v}\:\:\:\:\:{or}\:\:{V}=\frac{\boldsymbol{{v}}}{\mathrm{6}}\: \\ $$$${then}\:\:\:\:{v}_{\mathrm{1}} =\mathrm{4}{V}\:=\:\frac{\mathrm{2}\boldsymbol{{v}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\omega{l}}=\mathrm{6}\boldsymbol{{V}}\:=\boldsymbol{{v}} \\ $$$$\:\:\:\boldsymbol{\omega}=\boldsymbol{{v}}/\boldsymbol{{l}}\:\:. \\ $$$$\boldsymbol{{K}}_{\boldsymbol{{i}}} =\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{mv}}^{\mathrm{2}} \\ $$$$\boldsymbol{{K}}_{\boldsymbol{{f}}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\boldsymbol{{m}}\right)\boldsymbol{{V}}^{\:\:\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}\boldsymbol{{ml}}^{\mathrm{2}} }{\mathrm{12}}\right)\boldsymbol{\omega}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{mv}}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{36}}+\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{2}\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{12}\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{36}}\:=\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\boldsymbol{{loss}}\:\boldsymbol{{in}}\:\boldsymbol{{K}}\:=\boldsymbol{{K}}_{\boldsymbol{{i}}} −\boldsymbol{{K}}_{\boldsymbol{{f}}} \\ $$$$\:\:\:\:\:=\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{2}}−\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{3}}\:\:\:=\frac{\boldsymbol{{mv}}^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$$$ \\ $$
Commented by Physics lover last updated on 25/Oct/17
thankd you very much,Mr Ajfour.
$${thankd}\:{you}\:{very}\:{much},{Mr}\:{Ajfour}. \\ $$$$ \\ $$
Answered by ajfour last updated on 24/Oct/17
 ω =(v/l) ;    loss in K =((mv^2 )/6) .
$$\:\omega\:=\frac{{v}}{{l}}\:;\:\:\:\:{loss}\:{in}\:{K}\:=\frac{{mv}^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$
Commented by Physics lover last updated on 24/Oct/17
yes sir!!
$${yes}\:{sir}!! \\ $$

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