Problem is : enter image description here My all working is shown here : using 1 and 2 i am getting it to be 2pi√2R/g enter image description here using energy equation it will lead us to correct answer but why the torque method is not working? Here A is the COM and P is the bottom most point about which i calculated torque so as to prevent consideration of (friction for without slipping conditions). C is centre of cylinder shell


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  • $\begingroup$ Is A the center of mass, and $I_\rho$ the mass moment of inertia about it? Please explain your work. $\endgroup$ – ja72 Oct 11 at 14:52
  • $\begingroup$ Sry sir , didnt mention earlier : Ip is the moment of inertia about bottom most point , and A is the Centre of Mass of cylindrical shell. $\endgroup$ – Euler142857 Oct 11 at 15:53
  • $\begingroup$ Anyone pls help @aaron stevens $\endgroup$ – Euler142857 Oct 12 at 2:17
  • $\begingroup$ Hi, homework-like and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem? $\endgroup$ – stafusa Oct 12 at 11:25
  • $\begingroup$ Why do you have two angles, $\varphi$ and $\theta$. The problem has one degree of freedom, the swing angle $\theta$. Express the equations of motion in terms of this angle and solve them to the form $$\ddot{\theta} = -\omega^2 \theta$$ $\endgroup$ – ja72 Oct 12 at 23:19

The reason you are not getting the correct answer is because you need to sum up torques about the center of mass, and consider the MMOI at the center of mass

$$ \tau_{\rm COM} = I_{\rm COM} \ddot{\theta} $$

When it comes to the equations of motion, tracking the center of mass is key because it separates the linear and angular equations, and as a result, the angular equation needs to be about the center of mass. Similarly the linear motion has to be of the center of mass, as in $F = m \, a_{\rm COM}$.

  • $\begingroup$ Sir but i did by this method as as inspiration from this problem pls see problem " 20 " and its solutions here in the link , so their method is also not correct ? : Problems:slideshare.net/mobile/miranteogbonna/soal-latihan1mekanika Sol: slideshare.net/mobile/miranteogbonna/… $\endgroup$ – Euler142857 Oct 11 at 16:35
  • $\begingroup$ You cannot go wrong if you express things about the center of mass. There are special cases, just as in pure rolling, where the reference point has zero instantaneous velocity that could be used also (since the coupled linear terms are zero due to the velocity being zero). Note that Physics is not a do your homework for you site. Please consider the fundamentals of the situation and figure out on your own what steps you might have missed. $\endgroup$ – ja72 Oct 11 at 16:45
  • $\begingroup$ Sir i know about that , but i always consider to use bottom as to avoid other forces which r of no use , if question was not asking for that . Btw i cant able to think what i am doing wrong by considering bottom most point for torque calculations as done in for Q 20 in that link. I know COM simplifies easily , but to learn something new by doing with some other points and the benefits will make physics to be loved more . Pls consider my situtation and explain why in the Q20 in the link , why the method isnt wrong and here its doing wrong ? A small clue can also help me $\endgroup$ – Euler142857 Oct 11 at 16:58

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