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A uniform straight rod of length $L$ is hinged at one end. It is free to oscillate in vertical plane. Time period of oscillation with small angular amplitude when a point mass of mass equal to that of rod is fixed with lower end is :

$1$. enter image description here

$2$. enter image description here

$3$. enter image description here

$4$. enter image description here

Hey,please could you help me solve this question. Actually,this seems to be a torsional pendulum and the time period in such a case is $2\pi\sqrt{I\over C}$ where C is the restoring torque.Now,my doubt is what should the restoring torque be?Should it be 2mg(mass of the rod as well as the one suspended)times the length..or something else because that is not giving me my answer.

High school student,so please keep it simple.

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closed as off-topic by John Rennie, Jim, BMS, Alfred Centauri, Kyle Kanos May 14 '14 at 14:57

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  • $\begingroup$ It is a compound pendulum, not a torsion pendulum. A torsion pendulum is a mass hanging from a wire where the mass rotates in alternating directions about the vertical axis that is the wire. For a step by step solution to a generalized version of the question see example 24.4 here: web.mit.edu/8.01t/www/materials/modules/chapter24.pdf $\endgroup$ – DavePhD May 14 '14 at 14:46
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Note that this is a compound pendulum.

You can take centre of mass of whole system(rod+particle) which has mass $2m$ to provide restoring torque. Hence, the general formula is : $$2\pi\sqrt{I\over mgy}$$

Where $y$ is distance of centre of mass from pivot. Keep in mind that $m$ is total mass.

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  • $\begingroup$ Exactly..so then the distance ,$y$ in your case should be $L\over 2$ ,right,which does not give the answer. $\endgroup$ – Yash Lekhwani May 14 '14 at 14:42
  • $\begingroup$ No. You need the centre of mass of rod + particle $\endgroup$ – evil999man May 14 '14 at 15:28

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