Why this seesaw is working against physics laws

[vikram_gupta11]

But you can also play better than me.

[sm0ky2]

Try rotating your degrees of freedom more into the vertical domain.
At no point should the center of gravity drop below the axis.


If the arm is 90-degrees facing the right side:
Draw a vertical line where the arm is.


Now raise the arm until the center of mass
Crosses that line near the 45-deg or so
The weight will fall left.
Balancing the arm to either side of this line
Should be the operation of the device.
The ball falls twice.
Left then right

[vikram_gupta11]

Dear Sir,
As per sketch there will be need of tilting the device just few degree more so that the ball could fall down.
There will be a lock mechanism which will work to prevent the counterweight up to that time when ball hits the upper part of tube .
The lock mechanism will work in this way that it will be unlocked due to impact of counterweight.

[sm0ky2]

(bumped for upcoming experiments)

[vikram_gupta11]

Dear Sm0ky2 sir,
As per my calculations it must work.if there is something wrong then please clarify it.
The scale length is 2 meter.
There is a 2 or 4 meter long tube.
Let's take tube length 2 meter.
This 2 meter long tube is mounted on left arm of seesaw in middle so it's 1 meter part is above fulcrum and 1 meter is below fulcrum.
A 10 kg.ball is resting at the bottom of the tube.
Now counterweight.
After calculating distance from fulcrum the counterweight will be 14.4 kg.to balance the seesaw as the distance of ball is 1.44 meter from fulcrum as the ball is in rest position in the tube.
Now the seesaw is in balanced position.
But I will take counterweight 20 kg.due to this excess counterweight the seesaw will be in tilting position at 90 degree angle.
Now when I lift up the seesaw then I will not have to calculate the input energy of  the 1 meter distance of tube which is mounted below fulcrum as I have added mass in counterweight.
Now calculate input using mgh formula
Mgh=10*10*1=100 Joule
But calculate output as ball will fall down from 2 meter height
Mgh= 10*10*2=200 Joule
Ball will again fall down from 2 meter height at the time of reversing
Mgh=10*10*2=200 Joule
So total output is 400 Joule but input is 100 Joule.
The very interesting point is that it is not important that the device must complete one cycle as ball has completed one cycle.
See the link.you can see momentum in the video.
https://youtu.be/TQr-MR2yj_U