Naked Science Forum

On the Lighter Side => New Theories => Topic started by: LB7 on 03/06/2016 23:26:45

Title: The sum of energy in a rotation
Post by: LB7 on 03/06/2016 23:26:45
The study:

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fs20.postimg.org%2Fe81wyvfa1%2Fimage.jpg&hash=a3bd044106a9a44654622d718b8294e1) (http://postimg.org/image/e81wyvfa1/)

A force turns around the center O in a circle (dotted lines), the force keeps its angle with the radius (beta is constant = 30°). The radius of the circle is 1. The force is 1. An external device gives the force F. It's easy to calculate the work needed to rotate the force from 0° (vertical) to 90° it is 0.5 from vertical to horizontal (rotate of 90°).

Now, take the black arm. The black arm turns around the red point, but it takes the force F at the end and it must increase its length if I want the force F turns around the center O.

Now the work is  :

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fs20.postimg.org%2F57xbeoj6h%2Fgthy.jpg&hash=5192890841fedec5e1c8044294f87156) (http://postimg.org/image/57xbeoj6h/)

There is a difference !

Maybe I need to do that:

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fs20.postimg.org%2Ftao93scgp%2Fgyh.jpg&hash=9af27ac14bd9b49783f27207ccb40f35) (http://postimg.org/image/tao93scgp/)

But there is always a difference:

2*cos(x)*sin(x+pi/6)-2*sin(x)*cos(x+pi/6) dx from pi/4 to 0

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fs20.postimg.org%2Fe389jfkm1%2Ffgb.jpg&hash=3da59e2525e10b3297e790229b167e5c) (http://postimg.org/image/e389jfkm1/)

I found my error, it was the length in the integration. The sum of energy is 0.