If an object attached to a string is being swung around in a circle, then when the string breaks, the object will continue in a straight line at a constant velocity, per Newton's First Law. If I understand correctly, the angular velocity drops to zero the instant the string breaks, correct?
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2$\begingroup$ Why do you think so? What is your definition of "angular velocity"? $\endgroup$– ACuriousMind ♦Commented Sep 27, 2022 at 16:38
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$\begingroup$ I think the angular acceleration would instantaneously drop to zero, given that the source of this acceleration is the tension in the string (F=ma). $\endgroup$– aRockStrCommented Sep 27, 2022 at 17:06
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$\begingroup$ I don't think that the definition of angular velocity is his (or hers) particularly. Its well defined in science. en.wikipedia.org/wiki/Angular_velocity $\endgroup$– AKUMA no ONICommented Sep 28, 2022 at 4:18
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$\begingroup$ In layman terms, I think what the question's author is asking is if the ball travels in a straight line when the ball is no longer attached to the stick. And by 0, I am guessing that the author means, not that there is no angular velocity to measure, but that the measurement of the angle is 0 degrees. $\endgroup$– AKUMA no ONICommented Sep 28, 2022 at 4:23
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2 Answers
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Its not that a particle has angular velocity only when it moves along a circular path. A particle moving along a straight line also has angular velocity about any point not lying on the line containing the path of the particle.
The angular velocity of a particle moving along a line, however, changes instant to instant so it is instantaneous angular velocity. Mathematically, if $\vec{v}$ is the linear velocity of the particle, its instantaneous angular velocity $\vec{\omega}$ about a point is given by $$\vec{v} = \vec{\omega} \times \vec{r}$$ where $\vec{r}$ is the position vector of the particle with respect to the point.
So, no, angular velocity does not drop to zero when the string breaks. You will keep on getting non-zero angular velocity about the center of the circle for ever assuming that particle continues its straight line motion for ever.
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$\begingroup$ This makes sense. See my comment on the answer below. $\endgroup$ Commented Sep 27, 2022 at 17:15
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1$\begingroup$ Yes, you are right; angular velocity asymptotically approaches zero. $\endgroup$– t2mCommented Sep 27, 2022 at 17:18
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No. The angular velocity would only be zero if the object's new trajectory was along a radius of the circle. But it is not - after the string breaks, the object moves along a tangent line. Its angle with respect to the centre of the circle is still changing, so it has a non-zero angular velocity.
Another way to see this is to realise that the object's angular momentum must be conserved.
answered Sep 27, 2022 at 16:47
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1$\begingroup$ So this means that when the string breaks, the angular velocity asymptotically approaches zero? I've been asked to find an equation for the angular velocity after the string breaks. Would the answer to this be the rotational angular velocity minus the tangential angular velocity? $\endgroup$ Commented Sep 27, 2022 at 17:00
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2$\begingroup$ Yes, angular velocity asymptotically approaches zero. You can find the angular velocity as a function of time by dividing the angular momentum (which is constant) by the radial distance of the object at time $t$. You will need to use trigonometry. $\endgroup$ Commented Sep 27, 2022 at 17:35