Timeline for Why is there a horizontal acceleration if there is no horizontal force?
Current License: CC BY-SA 4.0
12 events
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| Mar 11, 2023 at 17:57 | comment | added | Chesx | @pwf I think this isn't quite right. The point of contact is moving. So shouldn't you add the pseudo for that to which we are calculating? i.e. about POC $I\alpha=mgr-maR$ where $\alpha$ is initial angular acceleration and $a$ is initial acceleration of POC and rest you know. And by this $a$ we can find acceleration of geometric center. | |
| Mar 10, 2023 at 16:49 | comment | added | pwf | @BobD Yes, horizontal acceleration of the geometric center (not the CM, since there is no net horizontal force). | |
| Mar 10, 2023 at 13:50 | vote | accept | Chesx | ||
| Mar 10, 2023 at 13:38 | comment | added | Bob D | @pwf Horizontal acceleration of what? The centroid? As you have already pointed out, there is no horizontal acceleration of the COM. | |
| Mar 9, 2023 at 19:19 | comment | added | pwf | Yes, the problem can be solved using torques about the point of contact with the surface. It turns out to be a lot like a physical pendulum, except that the moment of inertia is not constant but is changing with angle. Sorry I don't have time to write up the details just now. At the moment of release, the horizontal acceleration is $-rRmg/I$, where $r$ is the radius to the CM, $R$ is the radius of the ball, and $I$ is the (instantaneous) moment of inertia of the ball about the point of contact. | |
| Mar 9, 2023 at 19:05 | comment | added | Chesx | @pwf So is there any criteria we can found horizontal acceleration (or velocity maybe). Just after releasing? | |
| Mar 9, 2023 at 19:04 | comment | added | pwf | @Chesx Yes, the geometric center will. | |
| Mar 9, 2023 at 19:02 | comment | added | Chesx | @pwf do you mean sphere will oscillate about initial position horizontally? | |
| Mar 9, 2023 at 17:49 | comment | added | gandalf61 | @pwf Thank you. Fixed. | |
| Mar 9, 2023 at 17:49 | history | edited | gandalf61 | CC BY-SA 4.0 |
added 9 characters in body
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| Mar 9, 2023 at 16:55 | comment | added | pwf | This isn't quite right. You are correct that there is no horizontal acceleration for the CM, and so the CM will only move up and down, but in order for the CM to maintain its horizontal position the ball will shift left and right. The point of contact will not remain constant - if the CM starts to the right of the point of contact, later, when the CM is at its lowest point, the point of contact will be below it, and then as the CM rises the point of contact will be to the right of the CM. If the CM is interior and not on the surface, the ball will both roll and slip back and forth. | |
| Mar 8, 2023 at 18:49 | history | answered | gandalf61 | CC BY-SA 4.0 |