Saved Bookmarks
| 1. |
The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3kgm^(2), find torque required to bring it to rest in 20s. |
|
Answer» Solution :Torque `TAU=Ialpha=I((omega_(2)-omega_(1)))/(t)` `:.tau=(0.3xx(0-10pi))/(20)` `=-0.15pi=-0.471Nm` |
|