One end of a uniform rod of mass m and length l is supported by a frictionless hinge which can withstand a tension of (1.75mg). the rod is free to rotate in a vertical plane. To what maximum angle should the rod be rotated from the vertical position so that when left the hinge does not break? Answer is is $\cos(\theta)$ = $\frac{1}{2}$. Why can not this question be solved using centre of mass concept?
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John Rennie
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This may help. Toppling of a cylinder on a block – mmesser314 Dec 22 '20 at 15:36
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By center of mass, do you mean by statics? I think the problem assumes the rod is going to released at an angle and swung down. At which point not only gravity is acting on the hinge, but also the dynamics of the rod apply. So statics will not do, because the center of mass of the rod is going to be accelerating upwards, and thus not static. – JAlex Dec 22 '20 at 17:34
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@JAlex sir, everything else is fine, but when I use kinetic energy of centre of mass 1/2 (mVc²) instead of rotational KE of rod to equate it with mgL at bottom, answer comes out to be wrong. If I have replaced the system once by its COM why should I again take into consideration the original system(rod) – Shikhar Chamoli Dec 22 '20 at 17:38
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[edit] the question and show your work thus far. You are missing something here and the comments is not the place to add information to your question. – JAlex Dec 22 '20 at 17:40
1 Answers
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Why not? Loss of potential energy (measured at C.M.) goes to rotational K.E.(about axle). At bottom of swing, force from the hinge – weight, causes the centripetal acceleration of the C.M.

R.W. Bird
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sir, if we have once replaced the rod by its centre of mass, shouldn't we use the kinetic energy of centre of mass 1/2 (mVc²) instead of rotational KE of rod. We can put centripetal acc. in terms of Vc instead of omega(ω) Why are we again taking the rod into consideration once it has been replaced by its equivalent COM – Shikhar Chamoli Dec 22 '20 at 17:21
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@ShikharChamoli When calculating KE you also have to consider the $\tfrac{1}{2} I \omega^2$ part also. – JAlex Dec 22 '20 at 17:37
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@R.W. Bird, If I am wrong, please correct me. We can put the P.E. at bottom either equal to the R.K.E. of rod (because Rod only rotates) or can put P.E. equal to translational K.E. of com + rotaional K.E. of COM.{ for R.KE of com I will use 1/2 (Iω²) and put I equal to m(L/2)² i.e. mR²} – Shikhar Chamoli Dec 22 '20 at 18:26
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Please show your work. There us no point in people trying to guess what you mean. – nasu Dec 22 '20 at 20:16
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