When a sphere rotates freely about an axis and completes a 180Β° rotation, its angular momentum reverses direction (from +IβΟβ to -IβΟβ), and applying conservation of angular momentum (IΟ + Iβ(-Οβ) = IβΟβ) yields the platform's final angular velocity as Ο = 2IβΟβ/I, demonstrating how free rotation allows angular momentum reversal unlike rigid rotation.
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Irodov Problem 1.280 π₯ Part 2 by Tyagi Sir | 180Β° Rotation TwistAdded:
So now let us see the second part of this problem.
Where see this, the second part is that this sphere is rotating about this axis.
Now the motor shaft will rotate it 180Β° instead of 90Β°. Look at this like this.
Now what will be the angular velocity of this platform when it has rotated 180Β°?
We call it Omega Double Dash and try to remove it. Let's get ready.
We have said this Omega Double Dash and I have paused it so that I can explain it to you in writing.
So now you're going to write for this angular momentum conserved will still be about this axis.
There is no external torque acting. So you would write i not omega0 j cap This was the initial angular momentum caused by the rotation of this sphere alone about this axis. And when it turns 180Β°, because its rotation is free about this axis, then the final angular momentum of the system will be i * omeg + i not omega0 * -j Cap this, understand clearly what is being said. Now when you will solve it you will get omeg = 2in0 upon i this will be your new angular velocity of this platform and you have to understand this very clearly that brother, the rotation of this sphere about this axis is free. So earlier I was rotating it in this direction. When it turned 180Β°, brother, it went downward. So that's why -i became not omeg0. And there is no rigidity about this axis of rotation of this sphere.
So it was rotating with the entire platform when it was 90Β°. Not doing it anymore.
Now in the opposite direction from our Omega 0.
Ok? So this omeg came.
If you find out this. Now you write the equation of energy that brother, initially the kinetic energy was 1/2 knot omega0Β² plus the platform's zero. Plus the work done by external forces acting on the shaft of the motor should be equal to what will be the final kinetic energy? Final kinetic energy is 1/2 I not omega0Β² + 1/2 omega' squared. This will give you the answer and I would like that if you benefited from this video and understood it then please like it.
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