Question on rotation of a thin rod.

[RGClark]

What velocity will a thin rod attain if you release it after twirling it around in a circle at high speed?
If you have a heavy bob attached to a light string rotating in a circle, then it will fly off with the speed it had in rotating when you release it.
However, if you have a rod rotating around at a fixed end, then release it, it will not fly off with the same velocity as the tip because it will continue to rotate at some speed because of angular momentum conservation.
So some of its energy will be in rotational motion and the rest in linear translational motion.
How do you calculate how much will be in linear translational energy, and so its linear velocity?


Bob Clark

 

[quasar987]

supposing you're in no-gravity space, the rod will continue to rotate in place endlessly once the constraint is removed. On the other hand, the bob will fly off

this is because the thin rod's centre of mass is at its center, and its center of mass's velocity is at all time zero when the constraint is there. It will remain zero once the constraint is removed and will continue to spin according to Newton's first law (or conservation of angular momentum).

but in the rotating bob system, the center of mass is in the center of the bob itself, and its velocity is tangeantial to the circle of rotation at all time while the constraint is on. when it is removed, the bob flys off in the tangeantial direction according to Newton's first law. (one can verify that angular momentum around the point of roration is also conserved)

 

[RGClark]

supposing you're in no-gravity space, the rod will continue to rotate in place endlessly once the constraint is removed. On the other hand, the bob will fly off

this is because the thin rod's centre of mass is at its center, and its center of mass's velocity is at all time zero when the constraint is there. It will remain zero once the constraint is removed and will continue to spin according to Newton's first law (or conservation of angular momentum).

but in the rotating bob system, the center of mass is in the center of the bob itself, and its velocity is tangeantial to the circle of rotation at all time while the constraint is on. when it is removed, the bob flys off in the tangeantial direction according to Newton's first law. (one can verify that angular momentum around the point of roration is also conserved)

I didn't make clear one end of the rod is fixed during the rotation. The rotation during this time is not around the center.


Bob Clark

 

[quasar987]

oh ok. then the rod would go like this: this is the moment that the rod is removed from its pivot point 'O':

O
|
|
|

and this is the rod a moment later:

O
...|
...|
...|

each particle in the rod behaves like the bob in the "rotating bob-system"

 

[RGClark]

oh ok. then the rod would go like this: this is the moment that the rod is removed from its pivot point 'O':

O
|
|
|

and this is the rod a moment later:

O
...|
...|
...|

each particle in the rod behaves like the bob in the "rotating bob-system"

I thought that too. But I did an experiment where I tied a string around the end of an ink pen and twirled it around. When I let it go, it continued to rotate.
This means some of the kinetic energy must still be in the form of rotational motion.
So not all the energy is transformed into linear translational motion.


- Bob Clark

 

[quasar987]

cool, I will have to try.

 

[Doc Al]

I thought that too. But I did an experiment where I tied a string around the end of an ink pen and twirled it around. When I let it go, it continued to rotate.
This means some of the kinetic energy must still be in the form of rotational motion.

The twirling pen is rotating about its center as well as translating (revolving) in a circle. When released, its center of mass will continue in a straight line (ignoring gravity) and the pen will continue to rotate about its center of mass at the same rotational speed it had while twirling.

Note that the twirling pen must make a complete rotation about its center in the same time that it makes a complete revolution about your hand.