Free rotation when initially rotated about some other point.

[shanu_bhaiya]

Suppose I rotate a uniform rod about any point, except center of mass, what will happen if I leave it in an isolated space? Will it start rotation about it's own center of mass?

 

[Doc Al]

Once all external forces are removed from the rod, its motion will be a combination of:
(1) its center of mass continuing to move with constant velocity;
(2) the rod continuing rotate about its center of mass at constant angular velocity.

 

[BlackWyvern]

\tau = Fd--º- <--- rough diagram.\tau1 = 2m . a . 2d

\tau2 = m . a . d

a = \alpha . r

a₁ = \alpha . 2r
a₂ = \alpha . r

\tau₁ = 2m. \alpha . 4r²

\tau₂ = m . \alpha . r²

\tau1 = 8\tau₂

So, from my calculations, the torques are not at equilibrium, so it will not stay spinning the way you have placed it.

 

[shanu_bhaiya]

So, it keeps on rotating about same point, either it is on that rod or outside it.
Thanks to everybody for replying.

 

[Doc Al]

So, from my calculations, the torques are not at equilibrium

What torques? No external forces (or torques) act on the rod once it is released in space. (I presume that is the situation.) The rod is in equilibrium.

 

[BlackWyvern]

Isn't there a centripetal force?

 

[Doc Al]

So, it keeps on rotating about same point, either it is on that rod or outside it.

I don't understand this statement. The motion will be as I described above--it doesn't simply continue to rotate about the same non-central point.

 

[BlackWyvern]

Ah!

The centre of mass (before), was moving in a circular rotation, and you were providing the centripetal acceleration for the centre of mass. So when you release it, it will fly off at a tangent. That's the CoM sorted.

The rod will rotate about it's centre of mass uniformly.

 

[Doc Al]

Isn't there a centripetal force?

The centripetal forces that act within the rotating rod (the intermolecular tension in the rod) are internal forces that exert no torque.

The centre of mass (before), was moving in a circular rotation, and you were providing the centripetal acceleration for the centre of mass. So when you release it, it will fly off at a tangent. That's the CoM sorted.

The rod will rotate about it's centre of mass uniformly.

Now you've got it.

 

[shanu_bhaiya]

Ah!

The centre of mass (before), was moving in a circular rotation, and you were providing the centripetal acceleration for the centre of mass. So when you release it, it will fly off at a tangent. That's the CoM sorted.

The rod will rotate about it's centre of mass uniformly.

Alright, by changing the inertial frame of reference the axis of rotation also changes.

1. Just like a boy watching a tyre rolling on the road has an axis on the contact of the tyre with the road.
2. But if you're watching while sitting on the same vehicle you will see the tyre rotating about the central axis.
3. If, you're watching a rolling tyre while going opposite of the vehicle, the axis of pure rotation will be somewhere above the centre of the tyre.

It could also be understood by the combination of translation+roatation(about c.o.m.). Now, you want to say that centre of mass will do a pure translation as it was doing all before leaving the rod into space, while the whole rod will also do rotational motion about c.o.m. This rotational+translational motion can also be figured out by only considering a pure rotation about a different axis, and that must be the initial rotating point before I was leaving the rod into space. This is in exact agree with what Doc Al wanted to say.

What I am doubtful about is that while watching the rod as a pure rotational (while it's c.o.m. is still doing translation), it's axis of rotation may be anywhere even outside the rod. And, this really seems a confusing point.