Question Number 90819 by ajfour last updated on 26/Apr/20
Commented by ajfour last updated on 26/Apr/20
$${mass}\:{of}\:{rod}\:{is}\:{M},\:{and}\:\alpha=\mathrm{0},\:{but} \\ $$$${rod}\:{rotates}\:{with}\:{constant}\:{angular} \\ $$$${veloity}\:\omega_{\mathrm{0}} \:,\:{then}\:{find}\:{relative}\: \\ $$$${speed}\:{of}\:{small}\:{mass}\:{m}\:{with} \\ $$$${respect}\:{to}\:{rod},\:{as}\:{it}\:{reaches}\: \\ $$$${the}\:{extreme}\:{end}\:\left({a}={l}\right)\:{from}\:{the} \\ $$$${hinged}\:{end}\:\left({a}=\mathrm{0}\right). \\ $$
Commented by ajfour last updated on 26/Apr/20
$${this}\:{one}\:{from}\:{a}\:{renowned} \\ $$$${physics}\:{book},\:{have}\:{answer}. \\ $$
Answered by ajfour last updated on 26/Apr/20
Answered by mr W last updated on 26/Apr/20
![when sleeve is at position x: angular speed of rod =ω speed of sleeve along rod=u acc. of sleeve along rod=a=u(du/dx) Iω_0 =(I+mx^2 )ω ⇒ω=(((Ml^2 )/3)/(((Ml^2 )/3)+mx^2 ))ω_0 =((Ml^2 )/(Ml^2 +3mx^2 ))ω_0 ma−mω^2 x=0 ⇒u(du/dx)−ω^2 x=0 ⇒u(du/dx)−(((Ml^2 ω_0 )/(Ml^2 +3mx^2 )))^2 x=0 ⇒∫_0 ^u_(max) udu=∫_0 ^l (((Ml^2 ω_0 )/(Ml^2 +3mx^2 )))^2 xdx ⇒(u_(mac) ^2 /2)=(((Ml^2 ω_0 )/(3m)))^2 ∫_0 ^l ((xdx)/((((Ml^2 )/(3m))+x^2 )^2 )) ⇒u_(max) ^2 =(((Ml^2 ω_0 )/(3m)))^2 ∫_0 ^l ((d(((Ml^2 )/(3m))+x^2 ))/((((Ml^2 )/(3m))+x^2 )^2 )) ⇒u_(max) ^2 =(((Ml^2 ω_0 )/(3m)))^2 [(1/(((Ml^2 )/(3m))+x^2 ))]_l ^0 ⇒u_(max) ^2 =(((Mlω_0 )/(3m)))^2 [(1/(M/(3m)))−(1/((M/(3m))+1))] ⇒u_(max) ^2 =(((Mlω_0 )/(3m)))^2 [(1/((M/(3m))((M/(3m))+1)))] ⇒u_(max) =lω_0 (√(M/(M+3m)))](Q90901.png)
$${when}\:{sleeve}\:{is}\:{at}\:{position}\:{x}: \\ $$$${angular}\:{speed}\:{of}\:{rod}\:=\omega \\ $$$${speed}\:{of}\:{sleeve}\:{along}\:{rod}={u} \\ $$$${acc}.\:{of}\:{sleeve}\:{along}\:{rod}={a}={u}\frac{{du}}{{dx}} \\ $$$${I}\omega_{\mathrm{0}} =\left({I}+{mx}^{\mathrm{2}} \right)\omega \\ $$$$\Rightarrow\omega=\frac{\frac{{Ml}^{\mathrm{2}} }{\mathrm{3}}}{\frac{{Ml}^{\mathrm{2}} }{\mathrm{3}}+{mx}^{\mathrm{2}} }\omega_{\mathrm{0}} =\frac{{Ml}^{\mathrm{2}} }{{Ml}^{\mathrm{2}} +\mathrm{3}{mx}^{\mathrm{2}} }\omega_{\mathrm{0}} \\ $$$${ma}−{m}\omega^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$\Rightarrow{u}\frac{{du}}{{dx}}−\omega^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$\Rightarrow{u}\frac{{du}}{{dx}}−\left(\frac{{Ml}^{\mathrm{2}} \omega_{\mathrm{0}} }{{Ml}^{\mathrm{2}} +\mathrm{3}{mx}^{\mathrm{2}} }\right)^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{{u}_{{max}} } {udu}=\int_{\mathrm{0}} ^{{l}} \left(\frac{{Ml}^{\mathrm{2}} \omega_{\mathrm{0}} }{{Ml}^{\mathrm{2}} +\mathrm{3}{mx}^{\mathrm{2}} }\right)^{\mathrm{2}} {xdx} \\ $$$$\Rightarrow\frac{{u}_{{mac}} ^{\mathrm{2}} }{\mathrm{2}}=\left(\frac{{Ml}^{\mathrm{2}} \omega_{\mathrm{0}} }{\mathrm{3}{m}}\right)^{\mathrm{2}} \int_{\mathrm{0}} ^{{l}} \frac{{xdx}}{\left(\frac{{Ml}^{\mathrm{2}} }{\mathrm{3}{m}}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow{u}_{{max}} ^{\mathrm{2}} =\left(\frac{{Ml}^{\mathrm{2}} \omega_{\mathrm{0}} }{\mathrm{3}{m}}\right)^{\mathrm{2}} \int_{\mathrm{0}} ^{{l}} \frac{{d}\left(\frac{{Ml}^{\mathrm{2}} }{\mathrm{3}{m}}+{x}^{\mathrm{2}} \right)}{\left(\frac{{Ml}^{\mathrm{2}} }{\mathrm{3}{m}}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow{u}_{{max}} ^{\mathrm{2}} =\left(\frac{{Ml}^{\mathrm{2}} \omega_{\mathrm{0}} }{\mathrm{3}{m}}\right)^{\mathrm{2}} \left[\frac{\mathrm{1}}{\frac{{Ml}^{\mathrm{2}} }{\mathrm{3}{m}}+{x}^{\mathrm{2}} }\right]_{{l}} ^{\mathrm{0}} \\ $$$$\Rightarrow{u}_{{max}} ^{\mathrm{2}} =\left(\frac{{Ml}\omega_{\mathrm{0}} }{\mathrm{3}{m}}\right)^{\mathrm{2}} \left[\frac{\mathrm{1}}{\frac{{M}}{\mathrm{3}{m}}}−\frac{\mathrm{1}}{\frac{{M}}{\mathrm{3}{m}}+\mathrm{1}}\right] \\ $$$$\Rightarrow{u}_{{max}} ^{\mathrm{2}} =\left(\frac{{Ml}\omega_{\mathrm{0}} }{\mathrm{3}{m}}\right)^{\mathrm{2}} \left[\frac{\mathrm{1}}{\frac{{M}}{\mathrm{3}{m}}\left(\frac{{M}}{\mathrm{3}{m}}+\mathrm{1}\right)}\right] \\ $$$$\Rightarrow{u}_{{max}} ={l}\omega_{\mathrm{0}} \sqrt{\frac{{M}}{{M}+\mathrm{3}{m}}} \\ $$
Commented by ajfour last updated on 26/Apr/20
$${Excellent}\:{Sir}!\:{Very}\:{basic}\:{way}. \\ $$$$\:\:{thanks}\:{Sir}! \\ $$