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A block of mass $m$ is attached to one end of a light string which is wrapped on a disc of mass $2\,m$ and radius $R$. The total length of the slack portion of the string is $l$. The block is released from rest. The angular velocity of the disc just after the string becomes taut is:

enter image description here

Solution:

Using the principle of conservation of angular momentum, $$m\sqrt{2gl}\times R=\left[\frac{(2m)R^2}{2}+mR^2\right]\times\omega\\ \implies \omega=\sqrt{\frac{gl}{2R^2}}$$


I am getting a different answer when I tried solve it by using conservation of energy.

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  • $\begingroup$ Have you at leat tried to do it with conservation of energy? $\endgroup$ – Local Mathmatician Jun 17 at 7:07
  • $\begingroup$ yes i tried , but i am not getting the same answer $\endgroup$ – SpaceX Jun 17 at 7:11
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    $\begingroup$ Then you should write down your attempted solution in the post so that users can direct you where you've made a mistake. Use mathjax to write down equations. $\endgroup$ – Local Mathmatician Jun 17 at 7:18
  • $\begingroup$ i do feel that the given solution is wrong . Can you please help $\endgroup$ – SpaceX Jun 17 at 7:23
  • $\begingroup$ I can assure you that the given solution is not wrong and you must get the same result using conservation of energy. $\endgroup$ – Local Mathmatician Jun 17 at 7:25
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You can try to solve this problem using law of conservation of energy and using law of conservation of angular momentum and you will get different answers.

The thing is that you can't use conservation of energy law in this case. At the moment when the string became taut some kind of inelastic impact would happen and some portion of energy would lost. Imagine, that the string is made of elastic rubber so that no energy would be lost. But in this case there will be some oscillations. In case the string is rigid these oscillations will fade out quickly, but some energy will be lost in the process.

Use the law of conservation of angular momentum.

Update.

Since it's a homework I will not provide a full solution. Just main steps.

In the beginning the block just falls. You can find it's speed $V_1$ at the moment just before the string gets taut.

Total angular momentum of the system at that moment would be $-m V_1 R$ - we calculate the momentum around the center of disk.

After the string is taut the speed of the block will be some some $V_2$, the angular velocity of disk $\omega$, and $V_2 = \omega * R$.

Now you calculate the total momentum of disk and block.

These two calculated angular momentum should be equal. And this would be the first equation on your picture.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Jun 17 at 14:42
  • $\begingroup$ i am still confused why are we taking mr^2 for moment of inertia of the block about the origin. shouldn't it be m (l^2+r^2) because after it has travelled L distance its distance from the origin of the disc will be root (l^2+ r^2) . so moment of inertia will not be mr^2 please help $\endgroup$ – SpaceX Jun 17 at 16:02
  • $\begingroup$ @SpaceX There is no need to calculate the moment of inertia of the block to find out it's angular momentum. But ok, it's possible to follow this way also. Yes, the distance is $\sqrt{l^2+R^2}$. Yes, the moment of inertia is $m*(l^2+R^2)$. But now you will have to calculate the angular velocity of the block around the point O. It is not the same as the angular velocity of the disk. It is $V*cos(\theta)/\sqrt{l^2+R^2}$. Multiply the moment of inertia by angular velocity and you will get the angular momentum. But again, there is no need to do it. $\endgroup$ – lesnik Jun 17 at 16:19
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    $\begingroup$ Got it!!!! Thanks a lot for your help @ lesnik $\endgroup$ – SpaceX Jun 17 at 16:45

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