Understanding the Rotation of a Freefalling Rod

[aaaa202]

This has been brought up numerous times but I don't really understand it. Consider a rod in freefall.
If you put your coordinate frame in the center of mass of the rod, there will be no torque around it and the rod as a whole will follow a straightline down. But now put a coordinate frame on one of the end points. Apart from the gravity pulling down on the rod as a whole, there will now be a net torque on the rod (because gravity acts in the center of mass).
What goes wrong with this picture, because clearly the rod doesn't rotate!

 

[chill_factor]

This has been brought up numerous times but I don't really understand it. Consider a rod in freefall.
If you put your coordinate frame in the center of mass of the rod, there will be no torque around it and the rod as a whole will follow a straightline down. But now put a coordinate frame on one of the end points. Apart from the gravity pulling down on the rod as a whole, there will now be a net torque on the rod (because gravity acts in the center of mass).
What goes wrong with this picture, because clearly the rod doesn't rotate!

the rod is a rigid body. the other side of the rod also has an equal torque, and due to rigidity, will be in the opposite direction.

 

[Doc Al]

This has been brought up numerous times but I don't really understand it. Consider a rod in freefall.
If you put your coordinate frame in the center of mass of the rod, there will be no torque around it and the rod as a whole will follow a straightline down. But now put a coordinate frame on one of the end points. Apart from the gravity pulling down on the rod as a whole, there will now be a net torque on the rod (because gravity acts in the center of mass).
What goes wrong with this picture, because clearly the rod doesn't rotate!

The problem is that you are you using an accelerating point as your 'pivot'. Torque about such an accelerating point does not simply equal the rate of change of angular momentum, unless that point happens to be the center of mass.

See my post in this thread: https://www.physicsforums.com/showthread.php?p=4097976

 

[Doc Al]

the rod is a rigid body. the other side of the rod also has an equal torque, and due to rigidity, will be in the opposite direction.

The only external force acting on the rod is gravity.

 

[A.T.]

But now put a coordinate frame on one of the end points. Apart from the gravity pulling down on the rod as a whole, there will now be a net torque on the rod (because gravity acts in the center of mass).

In an accelerated frame that falls with the rod, there is an inertial force upwards:
http://en.wikipedia.org/wiki/Fictitious_force#Acceleration_in_a_straight_line

The inertial force cancels gravity at any point of the rod. Regardless if the origin is in the center or the end: There is no net force on any part of the rod in such a frame, and thus no torque.

 

[Doc Al]

The inertial force cancels gravity at any point of the rod. Regardless if the origin is in the center or the end: There is no net force on any part of the rod in such a frame, and thus no torque.

That's a good way to look at it (and probably more straightforward).

The extra terms (beyond the torque due to external forces) you get when you calculate dL/dt about an accelerating point are equivalent to introducing that inertial force.

 

[mfb]

In the frame of one of the ends, the rod gains angular momentum - by falling linearly to the floor.
The torque is present, and required for a linear motion downwards in this frame.

 

[Doc Al]

In the frame of one of the ends, the rod gains angular momentum - by falling linearly to the floor.
The torque is present, and required for a linear motion downwards in this frame.

Viewed from an inertial frame, the rod gains angular momentum. But in the accelerating frame of one of its ends, it does not.

 

[D H]

The inertial force cancels gravity at any point of the rod. Regardless if the origin is in the center or the end: There is no net force on any part of the rod in such a frame, and thus no torque.

Yes, there is a torque. It's the same phenomenon that causes spaghettification. Taking advantage of, or otherwise dealing with, gravity gradient torque is an important concept for satellites in low Earth orbit.

 

[K^2]

Yes, there is a torque. It's the same phenomenon that causes spaghettification. Taking advantage of, or otherwise dealing with, gravity gradient torque is an important concept for satellites in low Earth orbit.

Problem assumes uniform gravitational field. There are no tidal forces. Doc Al and A.T. have it covered from both perspectives.