Timeline for Power in a uniform circular motion
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 28, 2016 at 17:31 | comment | added | sammy gerbil | If $\omega$ is constant then the wheel is not spinning faster or slowing down. Then there is no net torque on it. The torque provided by the flowing water equals the torque provided by the resistance in gears, friction, and whatever the wheel is driving. ... First you ask for power, then moment of inertia, now torque. Why don't you start by stating exactly what you are trying to do? A good question starts with a clear formulation of the problem, not your own attempt to solve it. | |
| Nov 28, 2016 at 17:00 | comment | added | Jan Eerland | When I know $\text{E}_\text{k}=\frac{\text{I}\times\omega^2}{2}$ what will the torque of the wheel be? | |
| Nov 28, 2016 at 16:55 | comment | added | sammy gerbil | The formula comes from that for gravitational PE $= mgh$. $\rho Q=\frac{dm}{dt}$ is equal to the mass rate of flow. Yes g=9.81$m/s^2$. | |
| Nov 28, 2016 at 16:50 | comment | added | Jan Eerland | Where does that formula comes from? And is $\text{g}\approx9.81$? | |
| Nov 28, 2016 at 16:43 | history | answered | sammy gerbil | CC BY-SA 3.0 |