Question-168119

Question Number 168119 by Beginner last updated on 03/Apr/22

Answered by mr W last updated on 04/Apr/22

l=length of rod=24 cm I=((Ml^2 )/(12))+M((l/2)−x)^2 I(d^2 θ/dt^2 )=−Mg((l/2)−x)sin θ ≈−Mg((l/2)−x)θ (d^2 θ/dt^2 )+((Mg)/I)((l/2)−x)θ=0 ω=((Mg)/I)((l/2)−x)=((Mg((l/2)−x))/(((Ml^2 )/(12))+((l/2)−x)^2 M))=((g((l/2)−x))/((l^2 /(12))+((l/2)−x)^2 )) T=((2π)/ω)=((2π)/g)×(((l^2 /(12))+((l/2)−x)^2 )/((l/2)−x)) let u=(l/2)−x, T=((2π)/ω)=((2π)/g)×(((l^2 /(12))/u)+u) such that T is minimum, (dT/du)=0. ⇒−(l^2 /(12u^2 ))+1=0 ⇒u^2 =(l^2 /(12)) ⇒u=(l/(2(√3)))=(l/2)−x ⇒x=(l/2)(1−(1/( (√3))))=12(1−(1/( (√3))))=12(1−(1/( (√n)))) ⇒n=3 and T_(min) =((2(√3)πl)/(3g))

$${l}={length}\:{of}\:{rod}=\mathrm{24}\:{cm} \\ $$$${I}=\frac{{Ml}^{\mathrm{2}} }{\mathrm{12}}+{M}\left(\frac{{l}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} \\ $$$${I}\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=−{Mg}\left(\frac{{l}}{\mathrm{2}}−{x}\right)\mathrm{sin}\:\theta\:\approx−{Mg}\left(\frac{{l}}{\mathrm{2}}−{x}\right)\theta \\ $$$$\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }+\frac{{Mg}}{{I}}\left(\frac{{l}}{\mathrm{2}}−{x}\right)\theta=\mathrm{0} \\ $$$$\omega=\frac{{Mg}}{{I}}\left(\frac{{l}}{\mathrm{2}}−{x}\right)=\frac{{Mg}\left(\frac{{l}}{\mathrm{2}}−{x}\right)}{\frac{{Ml}^{\mathrm{2}} }{\mathrm{12}}+\left(\frac{{l}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} {M}}=\frac{{g}\left(\frac{{l}}{\mathrm{2}}−{x}\right)}{\frac{{l}^{\mathrm{2}} }{\mathrm{12}}+\left(\frac{{l}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} } \\ $$$${T}=\frac{\mathrm{2}\pi}{\omega}=\frac{\mathrm{2}\pi}{{g}}×\frac{\frac{{l}^{\mathrm{2}} }{\mathrm{12}}+\left(\frac{{l}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} }{\frac{{l}}{\mathrm{2}}−{x}} \\ $$$${let}\:{u}=\frac{{l}}{\mathrm{2}}−{x}, \\ $$$${T}=\frac{\mathrm{2}\pi}{\omega}=\frac{\mathrm{2}\pi}{{g}}×\left(\frac{\frac{{l}^{\mathrm{2}} }{\mathrm{12}}}{{u}}+{u}\right) \\ $$$${such}\:{that}\:{T}\:{is}\:{minimum},\:\frac{{dT}}{{du}}=\mathrm{0}. \\ $$$$\Rightarrow−\frac{{l}^{\mathrm{2}} }{\mathrm{12}{u}^{\mathrm{2}} }+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{2}} =\frac{{l}^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\Rightarrow{u}=\frac{{l}}{\mathrm{2}\sqrt{\mathrm{3}}}=\frac{{l}}{\mathrm{2}}−{x} \\ $$$$\Rightarrow{x}=\frac{{l}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{12}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{12}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{n}}}\right) \\ $$$$\Rightarrow{n}=\mathrm{3} \\ $$$${and}\:{T}_{{min}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}\pi{l}}{\mathrm{3}{g}} \\ $$

Commented by Beginner last updated on 03/Apr/22