Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

Thanks

$${Thanks} \\ $$

Commented by Tawa11 last updated on 03/Apr/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Commented by peter frank last updated on 04/Apr/22

good question.Rotation dynamics

$$\mathrm{good}\:\mathrm{question}.\mathrm{Rotation}\:\mathrm{dynamics} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com