Timeline for Is work done by torque due to friction in pure rolling?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2022 at 1:56 | comment | added | Claudio Saspinski | What the article calls pseudowork is my first equation. But it agrees with the final conclusion that the work of the gravitational force alone equals the change of the total kinetic energy. | |
| May 17, 2022 at 0:43 | comment | added | Alpha Delta | As far as I know, what you have calculated is called pseudowork and is "... not equal to the total work done on the system, because forces have been multiplied by the centre-of-mass displacement rather than their individual displacements...". Resnik and Halliday seems to be of the same mind. | |
| May 16, 2022 at 18:53 | comment | added | Claudio Saspinski | Work is defined as $\int F.dx$. What follows from Newton's second law is the equality: work of the net force and change of kinetic energy for a pure translational movement. In this example, the work of the gravitational force (which is different from the net force) is equal to the change of total kinetic energy (rotational and translational). It seems a good convention to say that friction force here does no work. | |
| May 16, 2022 at 5:10 | comment | added | Alpha Delta | I am using Resnik and Halliday as reference and it mentions that the equation derived from Newton's Second Law although may "...look like work, they are not work in the sense we have defined it...". | |
| May 7, 2022 at 21:19 | history | answered | Claudio Saspinski | CC BY-SA 4.0 |