Timeline for Contradiction in work-energy theorem for rigid bodies?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2021 at 4:43 | comment | added | Kashmiri | Yes I'm studying your comment dear. | |
| Jan 10, 2021 at 4:42 | comment | added | user65081 | I added two comments in your former answer, I hope it helps. But you can keep asking me there | |
| Jan 10, 2021 at 4:41 | comment | added | Kashmiri | Yes dear, I don't have any teacher to go to. I have only prayers to give you | |
| Jan 10, 2021 at 4:40 | vote | accept | Kashmiri | ||
| Jan 10, 2021 at 4:15 | comment | added | user65081 | I have no problem on giving it a look though | |
| Jan 10, 2021 at 4:15 | comment | added | user65081 | the question I answered is correct, now you are asking me a new and different question. | |
| Jan 10, 2021 at 4:07 | history | edited | Sandejo | CC BY-SA 4.0 |
improved formatting
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| Jan 10, 2021 at 3:22 | comment | added | Kashmiri | But this is a contradiction, could you please have a look here physics.stackexchange.com/questions/606440/… | |
| Jan 10, 2021 at 3:21 | comment | added | Kashmiri | Thank you, say we've a spinning body spinning at $\omega_0$ set down on a horizontal plane and it skids and then begins to roll .The work done by friction is $W_{f}=\frac{1}{2} m \cdot V_{c m}^{2} - 0=\frac{1}{2} m \cdot \omega^{2} R^{2}$ But work done by friction will also be change in total kinetic energy which is $W_f=\Delta K\Rightarrow \frac{1}{2} \operatorname{m} \omega^{2} R^{2}=\left(\frac{1}{2} m \omega^{2} R^{2}+\frac{1}{2} I \omega^{2}\right)-\frac{1}{2} I \omega_{0}^{2}$. | |
| Jan 9, 2021 at 23:10 | history | edited | user65081 | CC BY-SA 4.0 |
deleted 1 character in body
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| Jan 9, 2021 at 20:36 | history | edited | user65081 | CC BY-SA 4.0 |
added 8 characters in body
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| Jan 9, 2021 at 20:28 | history | answered | user65081 | CC BY-SA 4.0 |