Angular motion

[dttom]

There are two masses connected by a light rod with length l, a vertical axis is at a distance x from one mass.

Suddenly the axis is removed, and the system spontaneously has the axis shifted to the centre.

 

L=mwx^2 + mw(l-x)^2

L'=2mw'(l/2)^2 = (mw'l^2)/2

L=L' (conservation of angular momentum)

w'=w[(2x/l - 1)^2 +1]

so w' must be greater than w.

 

L = L', E = Lw/2

so the system energy increases, and I think this is illogical, how come a system can have its energy increases spontaneously (without any driving force)?

 

 

Edited November 3, 2009 by dttom

[D H]

Emphasis mine:

There are two masses connected by a light rod with length l, a vertical axis is at a distance x from one mass.

Suddenly the axis is removed, and the system spontaneously has the axis shifted to the centre.

I see three possible meanings to phrase "the system spontaneously has the axis shifted to the centre."

1. The masses somehow magically teleport to some other location so that "the axis is shifted to the centre." In this case, you are dealing with a unreal situation. Conserved quantities aren't conserved, but big deal. The situation is not real.
    
    
   
2. You switch the frame of reference so that "the axis is shifted to the centre." Linear momentum, angular momentum, and energy are frame dependent quantities. Think of it this way. Consider a single particle moving at some velocity with respect to an observer. To this observer, the particle will have non-zero linear momentum and non-zero kinetic energy. To an observer who is co-moving with the particle, the particle has zero linear momentum and zero kinetic energy.
    
    
   
3. Something entirely different from the above two interpretations.

 

What exactly do you mean by the highlighted phrase?

[dttom]

Emphasis mine:

I see three possible meanings to phrase "the system spontaneously has the axis shifted to the centre."

1. The masses somehow magically teleport to some other location so that "the axis is shifted to the centre." In this case, you are dealing with a unreal situation. Conserved quantities aren't conserved, but big deal. The situation is not real.
    
    
   
2. You switch the frame of reference so that "the axis is shifted to the centre." Linear momentum, angular momentum, and energy are frame dependent quantities. Think of it this way. Consider a single particle moving at some velocity with respect to an observer. To this observer, the particle will have non-zero linear momentum and non-zero kinetic energy. To an observer who is co-moving with the particle, the particle has zero linear momentum and zero kinetic energy.
    
    
   
3. Something entirely different from the above two interpretations.

 

What exactly do you mean by the highlighted phrase?

 

I think the third option is most appropriate.

I actually mean some other things else.

I am trying to describe a phenomenon that any symmetric rotational object, without a fixed axis of rotation, would simply rotate around its centre of mass. That is, in the case discussing, the centre of the rod. I hope that makes it clear.

[insane_alien]

in order to shift the axis of rotation you need to apply a force, this means extra rotational energy can be coming from another source and hence you are trying to apply closed system equations and assumptions to an open system.

[dttom]

in order to shift the axis of rotation you need to apply a force, this means extra rotational energy can be coming from another source and hence you are trying to apply closed system equations and assumptions to an open system.

 

If an energy source is required, once the fixed axis is removed, say there was a stick pin-through a point some distance from the centre, the rod-mass system still rotate about the original axis of rotation (the pin-hole)? But what observed in daily seems to reveal that an object tends to rotate about its own centre of mass regardless of how the initial rotation was.

[D H]

L=mwx^2 + mw(l-x)^2

L'=2mw'(l/2)^2 = (mw'l^2)/2

L=L' (conservation of angular momentum)

You are switching reference frames here. In particular, the origin of the frame in which you computed L' is shifted from the origin of the frame in which you computed L by x-l/2. These are two different frames of reference, so angular momentum is not the same in the two frames.

[dttom]

You are switching reference frames here. In particular, the origin of the frame in which you computed L' is shifted from the origin of the frame in which you computed L by x-l/2. These are two different frames of reference, so angular momentum is not the same in the two frames.

 

If there is a frame shift, certain manipulation should have transform the previous frame to the latter frame, what actually the manipulation should be? Because I nearly know nothing about frame shifting (though I could understand the relative example you quoted in the first reply), could you help me a little bit.

What confuse me is, a reference frame is a set of chosen coordinate axis, but now I am describing an event that, when a rod is made to rotate about an axis other than the centre, the rod now just rotates about its centre axis when such a 'making pin' is removed.

[D H]

The thing you must do is to pick one reference frame and stick with it throughout. In terms of the diagram you recently added to the original post, I will use the following reference frame:

• The origin is at the pivot point,
  
• The [math]\hat x[/math] axis points to the left,
  
• The [math]\hat y[/math] axis points toward the top of the screen, and
  
• The [math]\hat z[/math] axis is coming out of the screen.

 

In this frame, the linear momentum, angular momentum and kinetic energy before the pin is released are

 

[math]\aligned

\vec p_- &= -m\omega_-x\hat y + m\omega_-(l-x)\hat y \\

&= m\omega_-(l-2x)\hat y \\

\vec L_- &= mx^2\omega_-\hat z + m(l-x)^2\omega_-\hat z \\

&= m(l^2-2lx+2x^2)\omega_-\hat z \\

E_- &= \frac 1 2 mx^2\omega_-^2+\frac 1 2m(l-x)^2\omega_-^2 \\

&= \frac 1 2m(l^2-2lx+2x^2)\omega_-^2

\endaligned[/math]

 

After the pin is released, the bar+masses system will move up or down the screen, depending on the location of the center of mass with respect to the pivot point. Denoting the velocity of the center of mass after the release as [math]\vec v_{cm}[/math], the linear momentum of the bar+masses system after the release is

 

[math]\vec p_+ = 2m\vec v_{cm}[/math]

 

To conserve linear momentum, [math]\vec p_+ = \vec p_-[/math] and thus

 

[math]\vec v_{cm}

= \frac{\vec p_+}{2m} = (l/2-x)\omega_-\hat y[/math]

 

The angular momentum after the release is the angular momentum due to the motion of the system as a whole plus the angular momentum due to rotation about the center of mass:

 

[math]

\vec L_+ = 2m\vec r_{cm}\times \vec v_{cm}

+ \frac 1 2 ml^2\omega_+\hat z

[/math]

 

The center of mass of the bar+masses system is

 

[math]\vec r_{cm} = \frac 1{2m}(-mx\hat x + m(l-x)\hat x) = (l/2-x)\hat x[/math]

 

Applying this to the expression for the post-release angular momentum yields

 

[math]\aligned

\vec L_+ &=

2m((l/2-x)\hat x)\times ((l/2-x)\omega_-\hat y)

+ \frac 1 2 ml^2\omega_+\hat z \\

&= \left(2m(l/2-x)^2\omega_- + \frac 1 2 ml^2\omega_+\right) \hat z

\endaligned

[/math]

 

Equating this to the pre-prelease angular momentum,

 

[math]\left(2m(l/2-x)^2\omega_- + \frac 1 2 ml^2\omega_+\right) \hat z

= m(l^2-2lx+2x^2)\omega_-\hat z[/math]

 

Solving for [math]\omega_+[/math]

 

[math]\aligned

\frac 1 2 ml^2\omega_+ &=

m\left((l^2-2lx+2x^2)-2m(l/2-x)^2\right)\omega_- \\

&= \frac 1 2 ml^2\omega_-

\endaligned[/math]

 

Thus [math]\omega_+ = \omega_-[/math]. In other words, the angular velocity does not change.

 

Is energy conserved? The post-release kinetic energy is the kinetic energy due to the translation of the center of mass plus the rotational kinetic energy. Since there is no change in angular velocity, I'll use [math]\omega[/math] sans the + and - subscripts.

 

[math]

\aligned

E_+ &= \frac 1 2

\left(2m v_{cm}^2 + \frac 1 2 ml^2 \omega^2\right) \\

&= \frac 1 2 m\left(2 (l/2-x)^2+\frac 1 2 l^2\right)\omega^2 \\

&= \frac 1 2 m(l^2-2lx+2x^2)\omega^2

\endaligned[/math]

 

This is the pre-release kinetic energy, so kinetic energy is conserved, as expected.

[dttom]

I got it eventually, thanks for your detail explanation.

 

Though the calculation of the event was settled, I still wonder why it should be like that, or why should the system rotate about its own centre of mass in the absence of a given axis of rotation (the pin in this case).

Edited November 4, 2009 by dttom

[D H]

Though the calculation of the event was settled, I still wonder why it should be like that, or why should the system rotate about its own centre of mass in the absence of a given axis of rotation (the pin in this case).

One can pick any point on an object to be the center of rotation. Given that fixed point, the combined translational and rotational motion of the object can always be described in terms of a translation of this fixed point plus a rotation about some axis passing through this fixed point. (For that matter, the fixed point does not even a point not on an object!) This is a direct consequence of Euler's rotation theorem.

 

In other words, there is nothing wrong per se with choosing to look at the system as continuing to rotate about the pre-release pivot point. There is however a very good reason for choosing the center of mass as that special fixed point. For an unconstrained, constant mass rigid body, the translational and rotational equations of motion decouple when the motion is viewed as a translation of the center of mass plus a rotation about some axis passing through the center of mass. This is not the case for any other point.

 

For example, looking at things with the pivot point viewed as the fixed point is going to lead to a mess. If no external forces act on the system, the center of mass will move in a straight line once released. The motion of the pivot point is much more complex: it circles about a straight line path. Add in external forces and things become really messy.

 

That said, there are cases where choosing the center of mass leads to a mess. When the pin is attached the mathematics are much simpler with the pivot point chosen to be the fixed point. The pin constrains the motion, so this does not qualify as an unconstrained, constant mass rigid body.

 

Another example is a launch escape system that takes a capsule away from the exploding rocket underneath the capsule. The launch escape system rockets burn at an incredible rate to make the capsule accelerate away from (translational acceleration) and out of the way (rotational acceleration) of the main rocket. The launch escape system plus capsule is not a constant mass rigid body. In fact, the location of the center of mass of system moves within the system. The inertia tensor is not constant. No matter how you slice it, the translational and rotational equations of motion are tightly coupled. Some prefer to analyze this system from the perspective of a fixed point on the system as opposed to from the perspective of the center of mass.

[dttom]

Thank you very much, that helps a lot.

Edited November 4, 2009 by dttom