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If a rigid rod of length $l$ rotates on one end around the origin with an angular velocity of $\omega$ and suddenly the end fixed to the origin is released allowing the rod to move freely without any external forces on it what is the new angular velocity of the rod? Around what axis would the rod now rotate? I think that the new angular velocity would still be $\omega$ but it would now be around the center of mass of the rod. This would conserve angular momentum relative to the rod's center of mass and the magnitude of velocity of the rod's center of mass would also remain unchanged (thus conserving the total kinetic energy of the rod). Is my thinking correct or does the angular momentum change because the axis of rotation is now different and the rod's moment of inertia relative to this new axis of rotation changes?

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The rod exactly before detaching (letś call $t = -\epsilon$) from the center of rotation (A) has an angulat velocity $\omega$ and angular momentum $L = I\omega$, where $I = \frac{1}{3}ml^2$.

Just after the rod being released, (letś call $t = \epsilon$) the COM is moving with a constant velocity $v = \omega \frac{l}{2}$, because that was its velocity at $t = -\epsilon$

The velocity of A is zero at $t = \epsilon$. So its relative velocity with respect to the COM is $v_A = -\omega\frac{l}{2}$. The opposite end of the rod has a velocity of $v_B = \omega l$ at the same time.

The new angular velocity with respect to the COM is: $$\omega_1 = \frac{(\omega l - \omega\frac{l}{2})}{\frac{l}{2}} = \omega$$

The new angular momentum with respect to A is the sum of $\mathbf r \times \mathbf p_{COM}$ plus the spin angular momentum with respect to the COM: $$L_1 = \frac{l}{2} m\omega \frac{l}{2} + I_1\omega$$ The new moment of inertia with respect to the center of mass is $I_1 = \frac{ml^2}{12}$

So, $$L_1 = m\frac{l^2}{4}\omega + \frac{ml^2}{12}\omega = m\frac{l^2}{3}\omega = L$$

The kinetic energy was only rotational: $$E = \frac{1}{2}I\omega^2 = \frac{1}{2}m\frac{l^2}{3}\omega^2 = \frac{1}{6}ml^2\omega^2$$

And later it is translational and rotational: $$E_1 = \frac{1}{2}mv^2 + \frac{1}{2}I_1\omega^2 = \frac{1}{2}m(\omega \frac{l}{2})^2 + \frac{1}{2}(\frac{ml^2}{12})\omega^2 = \frac{1}{6}ml^2\omega^2 = E$$

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For your situation, the total angular momentum is the same before and after the rod is released, if the angular momentum is evaluated with respect to the same point before and after release; here, taken as the point at the fixed end of the rod. The total angular momentum is ${1 \over 3} ML^2 \omega$ about that fixed point. $M$ is the rod mass, $L$ is the length of the rod, and $\omega$ is the angular speed of rotation of the rod.

If you evaluate the angular momentum about the fixed end before release, and you evaluate the angular momentum about the center of mass (a different point) after release, the angular momentum is different for the two cases. This is because here the angular momentum of the center of mass is not considered after release of the rod. The angular momentum about the center of mass is ${1 \over {12}} ML^2 \omega$ and the angular momentum of the center of mass with respect to the original location of the fixed end of the rod is ${1 \over 4} ML^2\omega$.

Details are provided in the following discussion.

General discussion: motion of a system of particles

Angular momentum (and torque) depend on the point about which they are evaluated. Let $O$ be a point fixed in space, about which the total angular momentum for a system of particles is to be evaluated. See the figure below. Relative to point $O$ $$(1) \vec J = \vec R \times M\vec V_{CM} + \sum_{i}^{} \vec r_i^{'} \times \vec p_i^{'}$$

$\vec J$ is the angular momentum with respect to $O$, $M$ is the system total mass, $\vec R$ is the location of the center of mass (CM) with respect to $O$, $\vec r_i^{'}$ is the distance from the CM to the $i^{th}$ particle, and $\vec p_i^{'}$ is the momentum of the $i^{th}$ particle with respect to the CM. [Goldstein, Classical Mechanics]. Relationship (1) states the total angular momentum is the angular momentum of the CM plus the angular momentum of motion about the CM; it depends on the point $O$.

The translational motion of the system is described by the motion of the CM: $$(2) \vec F_{ext} = M\vec a_{CM}$$ where $\vec F_{ext}$ is the total external force on the system and $\vec a_{CM}$ is the acceleration of the CM.

The motion about the CM is described as $$ (3) {d\vec J_{CM} \over dt} = \vec N_{ext}$$ where $\vec J_{CM}$ is the angular momentum with respect to the center of mass and $\vec N_{ext}$ is the total external torque on the system with respect to the CM. This relationship is valid even if the CM is accelerating with respect to $O$.

Your case

See the figure at the end of this response.

As I understand your situation, you have a rod of length $L$ rotating with angular speed $\omega$ about a fixed point at one end $O$ under no external torque, not even from gravity. Initially some force $F_{applied}$ had to be applied to the rod to provide a torque to initiate the rotation about the fixed point, and initially there was a force of constraint $F_{constraint}$ at the fixed point, but this force caused no torque about $O$. The external torque from $F_{applied}$ caused the angular momentum about $O$ to increase; after $F_{applied}$ is removed, the rod continues to rotate with constant angular velocity about $O$. There is still an external force on the rod causing centripetal acceleration of the CM, and that force is $F_{constraint}$ acting at $O$ directed inward from the CM towards $O$: $F_{constraint} = {{mL} \over 2} \omega^2$ using relationship (2). Call this state of the system case (a). Using relationship (3), the angular momentum about the CM is constant; there is no torque with respect to the CM, since $F_{constraint}$ is directed along the line connecting $O$ to the CM. With respect to the CM, the rod rotates at constant $\omega$.

Then, the constraint $F_{constraint}$ is removed. This results in no total external force, so the CM stops moving in a circle and moves in the direction it is was moving when the constraint was removed with the velocity at the time the constraint was removed. The other particles in the rod rotate about the CM. Call this state of the system case (b).

Evaluation of case (a)

Pick the left end of the constrained rod as point $O$, fixed in space, for relationship (1). It is easily shown that $J = {1 \over 3} ML^2 \omega$ about the point $O$.

Evaluation of case (b)

Once the force of constraint is released, there is no external force on the system, and the CM continues moving in the direction it was moving at the time the constraint is released. The rod continues to rotate with angular velocity $\omega$ about the moving CM, since removing the force of constraint adds no torque with respect to the CM. With respect to the CM the angular momentum is ${1 \over {12}} ML^2 \omega$ as easily shown. With respect to the original point $O$ selected as the stationary origin in case (a), the angular momentum of the CM is ${1 \over 4} ML^2\omega$, a constant (see the figure). Based on relationship (1), the total angular momentum about $O$ is ${1 \over {12}} ML^2 \omega$ + ${1 \over 4} ML^2\omega = {1 \over 3} ML^2 \omega$, the same result as for case (a).

Conclusion

The total angular momentum taken about the same point in both cases is the same; specifically ${1 \over 3} ML^2 \omega$ for the point taken to be the original fixed left end of the rod, fixed in space.

The angular momentum for case (a) taken about the left end of the rod is not the same as the angular momentum of the rod taken about the center of mass for case (b), because we are using different points about which to evaluate the angular momentum. And, for case (b) evaluating the angular momentum about the center of mass is not the complete angular momentum; the angular momentum of the center of mass must be included. enter image description here

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  • $\begingroup$ How can there be a torque on the rod in case (a) if the angular momentum of the rod doesn't change? $\endgroup$ Apr 6 at 13:17
  • $\begingroup$ This might be clearer if you added a derivation for the equivalent point particle. This simple case, i.e. a particle revolving about the center, held down by a massless string, then flies off perpendicular to its last circumferential (angle) position. $\endgroup$ Apr 6 at 13:26
  • $\begingroup$ In that scenario the angular momentum of the point particle relative to the fixed origin wouldn't change either because there is never any torque on the particle relative to the fixed origin. $\endgroup$ Apr 6 at 13:31
  • $\begingroup$ So would you agree that in this specific case the angular velocity of the rod doesn't change after the constraint force is removed and it starts rotating around its center of mass? $\endgroup$ Apr 6 at 14:04
  • $\begingroup$ I need to revise earlier answer. I forgot to address something important. Will explain in an update. Need time to explain. Coming later. Sorry! $\endgroup$
    – John Darby
    Apr 6 at 19:58
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The system (rod) is initially under the influence of a force provided by the pivot which is causing a centripetal acceleration of the centre of mass of the rod.
Since the line of action of that force is through the centre of mass of the rod there is no torque acting on the system and so its angular momentum is constant.

The angular momentum of the system can be considered in two ways.

The first as $(I_{\rm G}+ml^2)\omega$ where $I_{\rm G}+ml^2$ is the moment of inertia of the rod of mass $m$ and length $2l$ about the pivot and $\omega$ is its angular speed.

The second as $I_{\rm G}\omega$, the spin angular momentum which is independent of axis, plus $ml^2\omega$, the orbital angular momentum about the pivot.
You will note that the orbital component is also equal to $mlv$ where $v$ is the linear speed of the centre of mass because $v=l\omega$.

Now remove the constraint which provided a force on the system.

The centre of mass now moves linearly at a speed $v$ whilst the rod rotates with an angular speed $\omega$ about its centre of mass.
The angular momentum about the original pivot point would be unchanged, $I_{\rm G}\omega+mlv$.

What interests me is whether or not the OP would have asked the question for an almost equivalent system, gravity being switched off (equivalent to the pivot being removed) as the Earth orbited the Sun?

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Your assessment is correct.

The angular velocity will still be $\omega$, and kinetic energy and angular momentum conserved. The rotation will now be around the CoM axis.

Check out this simulation: https://www.desmos.com/calculator/fwiuasurzt

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  • $\begingroup$ Is there a reason why the center of mass of the rod slows down upon release? $\endgroup$ Apr 7 at 2:37
  • $\begingroup$ If the speed of the center of mass doesn't change the simulation looks like this: desmos.com/calculator/v9q1iyozab $\endgroup$ Apr 7 at 2:46
  • $\begingroup$ Angular momentum is conserved only if evaluated about the same point. For evaluation about the left end before release and evaluation about the center of mass after relief (a different point), the angular momentum is not the same; the angular momentum of the center of mass must be considered after release, also. See my answer $\endgroup$
    – John Darby
    Apr 7 at 3:09

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