Assume a ball to be kept on a rough surface with initial translational velocity $v$ and no rotation. After some time it acquires pure rolling. It is acted upon by an external torque due to friction still the angular momentum of the ball does not change from that of the initial one. Why is this?
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$\begingroup$ Why do you say the angular momentum is conserved? $\endgroup$– Manish SCommented Dec 20, 2020 at 14:02
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$\begingroup$ Angular momentum of an isolated system is always conserved. $\endgroup$– my2ctsCommented Apr 5, 2021 at 12:57
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$\begingroup$ @my2cts - A ball rolling is not an isolated system per se if there is an interaction with the ground. $\endgroup$– John AlexiouCommented Dec 13, 2022 at 13:21
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2 Answers
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This kind of surprised me, as rotational momentum does not change indeed.
During the ball's release and contact with the ground, an impulse (momentum) is applied to the ball through the contact point, which changes both the translational and rotational momentum of the ball. The system is not isolated as it interacts with the ground.
Let us look at this step by step. A ball of mass $m$ and mass moment of inertia $I= \tfrac{2}{5} m r^2$ is allowed to roll after purely translating.
Initially, the ball's center of mass has speed $\boldsymbol{v}$ and zero rotational speed $\omega =0$. All quantities are shown in their positive sense below:
- The speed of the ball at the contact point is $$v_{\rm imp} = v + \omega r = v$$
- Translational momentum is $$p = m v$$
- Rotational momentum summed at the origin is $$L = I \omega - r m v = - r\, m v$$
Note the negative sign in $L$ due to the cross product of $r$ along the y-axis and $v$ along the x-axis.
When the ball is released an impulse $J$ is applied at the contact point
- The impulse magnitude needs to be $$ J =\left( \frac{1}{ \tfrac{1}{m} + \tfrac{r^2}{I} } \right) v_{\rm imp} = \frac{m I v}{I + m r^2}$$
- Such that the velocity of the ball at the contact point after the impact is zero $$ \left(v - \frac{J}{m}\right) + \left( \omega - \frac{r J}{I} \right) r = \left( v - \frac{I v}{I+m r^2}\right) + \left( 0 - \frac{m r v}{I+mr^2} \right) r =0 $$
After the impact the ball has changed its translational velocity by $\Delta v$ and acquired rotational velocity $\Delta \omega$
Translational velocity change $$\Delta v = - \frac{J}{m} = - \frac{I v}{I+m r^2}$$
Rotational velocity change $$\Delta \omega = -\frac{r J}{I} = -\frac{m r v}{I+mr^2}$$
Translational momentum after impulse $$ p + \Delta p = m ( v + \Delta v) = \tfrac{5}{7} m v $$
Rotational momentum after impulse $$ L + \Delta L = I ( \omega + \Delta \omega) - r (p + \Delta p) = - r\,m v$$
In summary, translational momentum when from $m v$ to $\frac{5}{7} m v$ but rotational momentum remains at $-r\, m v$. All due to the interaction with the ground.
answered Dec 13, 2022 at 14:27
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$\begingroup$ Concerning your first sentence: The statement is correct if you choose an origin somewhere along the surface. As was pointed out in Deep Bhowmik's answer, the contact forces exert zero torque about any such point. And ironically, that's where your diagrams seem to indicate you've chosen the origin to be. $\endgroup$ Commented Dec 13, 2022 at 14:33
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$\begingroup$ @MichaelSeifert - yes rotational momentum changes according to the summation point (the origin). But there is a unique physical location in space where it is zero (the axis of percussion) and that location is different before and after the impulse. Before the impulse, the path of the center of mass is the axis of percussion, and after the impulse, it is a location $\tfrac{2}{5}r$ above the center of mass. So rotational momentum changes physically regardless of the choice of origin. $\endgroup$ Commented Dec 13, 2022 at 14:47
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1$\begingroup$ I think I've found the problem: you've defined the direction of positive rotation inconsistently for the "translational" and "rotational" pieces of $L$. If counter-clockwise is positive (the way you've defined it for the rotation), then the translational piece of $L$ is negative for positive $v$. So you should have $L = I \omega - m r v$. From your expressions for $\Delta v$ and $\Delta \omega$, you then have $\Delta L = I \Delta \omega - m r \Delta v = 0$. $\endgroup$ Commented Dec 13, 2022 at 17:40
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$\begingroup$ (Also note that you have a sign error in your expression for $\Delta v$) $\endgroup$ Commented Dec 13, 2022 at 17:41
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$\begingroup$ @MichaelSeifert - you are right. It serves me right for doing a cross-product in my head. $\endgroup$ Commented Dec 13, 2022 at 22:42
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Angular momentum depends on the axis chosen. If the axis is chosen at the point where the friction force acts on the rolling body, then the perpendicular distance of the friction force from the chosen axis will be zero at all times and hence the torque due to the friction force shall be zero. As no other external force is mentioned in the question, the net torque about the chosen axis shall be zero and hence angular momentum about the chosen axis will remain constant throughout.
However if some other axis is chosen, then the torque due to the friction force about the axis shall be non-zero and then the angular momentum about the axis shall change. It will not be constant.
Hope this answers your question.
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$\begingroup$ Good answer. Actually for a simple calculation any point on the horizontal surface in line with the direction of motion can be taken as the reference point. Then: L = mvR + Iω. $\endgroup$ Commented Apr 5, 2021 at 15:00