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I saw the following problem from the USAPhO:

A uniform pool ball of radius $r$ begins at rest on a pool table. The ball is given a horizontal impulse $J$ of fixed magnitude at a distance $\beta r$ above its center, where $−1 \leq \beta \leq 1$. The coefficient of kinetic friction between the ball and the pool table is $\mu$. Find an expression for the final speed of the ball as a function of $J$, $m$, and $\beta$.

Two solutions are presented. In the first solution, we are told to "consider an axis perpendicular to the initial impulse and coplanar with the table." Then we apply conservation of angular momentum momentum with the initial state immediately after the impulse and the final state after the ball has achieved rolling without slipping (and hence has constant velocity)

From my understanding, at any given moment, the axis always passes through the point of contact between the ball and the ground. This is the part the confuses me. Since the axis is not fixed, why is angular momentum $\ell$ conserved between the initial and final states? I noticed that $\ell$ at some time when the ball is still rolling without slipping is not the same as the initial $\ell$. (If the ball has angular velocity $\omega$ at some time, then $\ell = I\omega$, which varies as $\omega$ varies.) I believe this is related to the fact that during the rolling without slipping phase, we are not in an inertial reference frame.

(In particular, the problem is is USAPhO 2008 Quarterfinal, problem 2(a). You can find the problem and its solution here: http://www.aapt.org/physicsteam/2012/exams.cfm)

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1 Answer 1

up vote 2 down vote accepted

My initial reasoning was wrong, sorry, I can't justify the use of angular momentum conservation in that non-inertial frame. One can write a balance equation for the angular momentum, but after all it should be the same as solution №2.


But ok, let's choose a single inertial frame. Let it be the one moving with the ultimate ball velocity $v_f$. This will lead the ball having some initial angular momentum $L_0$ and recalculated value for added momentum J. Then the conservation of angular momentum will be

$$(\beta+1) r (J + mvr) + L_0 = \frac{7}{5} m r^2 \omega$$

where

$$ \begin{multline} L_0 = - \int_0^{2r} \rho z v \, \pi \left[r^2 - (r-z)^2\right] dz = \\ - v \frac{m}{\frac{4}{3}\pi r^3} \pi \int_0^{2r} z \left[r^2 - (r-z)^2\right] \, dz = -mvr \end{multline} $$

Hence

$$(\beta+1) r J = \frac{7}{5} m r^2 \omega$$

Instead of direct integration one could use the expression for angular momentum in a moving frame, but I don't remember how to do it.

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Thanks for your answer! I have some questions: (1) When you say "you don't take into account usual $g$," what exactly do you mean? In gyroscopes, for example, the gravity produces a torque which causes a change in angular momentum. (2) Where does the term $L_0$ come from? I thought changing $J$ to $J+mv$ would account for the change in reference frame. –  Alan C Aug 18 '12 at 23:44
    
(2) In that new frame initially the ball is moving (without rotating) with the speed $-v$. Obviously this accounts for some initial angular momentum since the axis does not pass through the ball's center of mass. –  Yrogirg Aug 19 '12 at 6:14
    
(1) my argument was flowed, but in (2) I'm sure. –  Yrogirg Aug 19 '12 at 8:44
    
Thank you for the explanations! –  Alan C Aug 22 '12 at 4:42

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