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A uniform flexible chain of length l rests on a fixed smooth sphere of radius R such that one end A of the chain is at top of the sphere while the other end B is hanging freely. The chain is held stationary by a horizontal thread PA as shown in the figure. Calculate the acceleration of the chain when the thread is burnt. enter image description here I have solved this and reached the correct answer, but I'm not sure whether the method I've used is 100% correct or not. Here it is:enter image description here enter image description here enter image description here Now I have a doubt:

Is it correct to integrate change in tension (dT) for each element this way, considering that the direction of dT is changing? Does it need to be further split into components? I was taught that integration should only be done for vectors that are unidirectional or be made unidirectional through components. Please help me with this.

Also, let me know if there is any other mistake I've made in this solution.

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  • $\begingroup$ It is correct as you are using polar coordinates(R,$\theta$). $\endgroup$ Commented Jul 15, 2019 at 7:10
  • $\begingroup$ Can you explain a bit more, please? @Unique $\endgroup$ Commented Jul 15, 2019 at 7:58
  • $\begingroup$ If you would have used cartesian coordinates then you would have to care about the vector addition. $\endgroup$ Commented Jul 15, 2019 at 8:01
  • $\begingroup$ Where did you learn that from? I've never heard about this before. Can you cite any source? @Unique $\endgroup$ Commented Jul 15, 2019 at 8:03
  • $\begingroup$ Go for Goldstein's classical mechanics. $\endgroup$ Commented Jul 15, 2019 at 8:08

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What you're doing is perfectly fine! It is legitimate to integrate a force "in the chain direction", even when the chain changes direction. That's an example of a generalized coordinate. Instead of Cartesian coordinates, you're using "chain coordinates", which measure the overall displacement of the chain.

Generalized coordinates of this kind are properly justified in Lagrangian mechanics, but they're very useful even if you just know Newtonian mechanics. I give a few more examples of this kind of reasoning in this handout.

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  • $\begingroup$ Thanks for confirming! I've roughly gone through the handout, and much of it seems to be a bit too advanced for me to completely understand. I'm still in high school. I'd read it thoroughly this weekend for sure! $\endgroup$ Commented Jul 15, 2019 at 10:10
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I'm also not sure about your $\int dT $ because the difference in tension at a point is already equal to the tangential force $(dm)gsin(\theta) $.

So I don't think you need to bother with the tension at all as it is already accounted for. The rest looks good though and maybe there is a good reason your T cancels in the end so I don't know... but to me the $\int_0^T dT $ just doesn't look right. Maybe someone else can explain if this makes sense or not.

Edit: I also found a video about a similar problem from a source which is probably not reliable, but they also don't bother with any tensions even if their reasoning is weird to me:

"There is tension but as we are calculating the total tangential force on complete chain, tension will be an internal force so it not needed here..."

If you calculate the difference in tension at a point it is already equal to $(dm)gsin(\theta) $, which is the expression you want to integrate.

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  • $\begingroup$ I have put limits as 0 to T considering tensions on the length of chain as a function of theta. At the topmost point, tension is zero when the thread is burnt, and at the lowermost point of the chain (the one-fourth circle chain), the tension is taken as T which is because of the force with which the freely hanging part of the chain pulls the rest of the chain. $\endgroup$ Commented Jul 15, 2019 at 16:43
  • $\begingroup$ P.S., the source that you have mentioned is actually quite reliable, but I don't understand the reasoning for neglecting tension, too. $\endgroup$ Commented Jul 15, 2019 at 16:44
  • $\begingroup$ I would just integrate $ (dm)g\sin(\theta) $ from $0$ to $ pi/2$ and then add the other force as you did but I would leave out all the T terms because it seems redundant to me. But idk you should ask your physics teacher or something I'm also just trying to learn physics :). $\endgroup$
    – ctsmd
    Commented Jul 15, 2019 at 16:50
  • $\begingroup$ Yeah, I'd ask my teacher tomorrow, and I'd share it here, whatever he has to say. Thanks for your time :) $\endgroup$ Commented Jul 15, 2019 at 16:51

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