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 October 26th, 2019, 10:33 PM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus How do I find the time from given velocities in a circular motion? This problem has left me going in circles. No pun intended, but I'm confused at how to use the vectors given to find the time. The problem is as follows: A sphere has a uniform circular motion. This object passes through two points $A$ and $B$, when this happens its speed is $\left(-4\hat{\textrm{i}}+4\sqrt{3}\hat{\textrm{j}}\right )\frac{m}{s}$ and $\left(8\hat{\textrm{i}}\right)\frac{m}{s}$ respectively. The radius of the circle where this object is moving is $\textrm{3 meters}$. Find the time which will take the particle to move from $A$ to $B$.The alternatives given are: $\begin{array}{ll} 1.&1.57\\ 2.&3.14\\ 3.&6.28\\ 4.&0.16\\ 5.&0.31\\ \end{array}$ Typically I would try to show some effort into this. But I'm stuck at the beginning. What is it exactly that should I do to find the given time? The only thing which I could come up with is that to find the time it can be obtained from the tangential speed as: $v_{t}=\omega r = \frac{\theta}{t} r$ But apart from this I don't know how to link it with what is being asked. Can somebody give me a hand with this? Last edited by skipjack; October 27th, 2019 at 03:26 AM.
 October 27th, 2019, 04:13 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 Let's assume that "sphere" was supposed to be "particle", that the particle is moving along the circumference of a circle of radius 3 m at a constant speed, and that the particle's velocity vectors at points A and B on the circle are as given in the problem (albeit incorrectly called speeds). The magnitude of the given vectors is 8 m/s, which is the particle's constant speed. As the length of the circumference of the circle is 6$\pi$ m, the time required for a complete circuit of the circle is $\frac34\pi$ s. The distance to be covered in going from A to B may be 1/3 of the length of the circumference of the circle or 2/3 of that length (because of the angle between the given vectors). There is insufficient information to let you determine which applies unless you assume that one of the answers must be correct. You should choose the answer that is approximately $\frac14\pi$ s or $\frac12\pi$ s. Hence option 1 is the expected answer choice. Thanks from topsquark
 October 27th, 2019, 05:35 AM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 Assuming standard motion about the circle, the angular position of point B is $\theta = \dfrac{3\pi}{2}$ since its tangential velocity is purely in the positive x direction. The tangential velocity components at point A indicate its position is in quadrant I. Since tangential velocity is perpendicular to the position vector, the position vector is in the direction $\theta = \arctan\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$. The angular distance from A to B is $\dfrac{4\pi}{3}$ radians. Given the radius of 3, the linear distance from A to B is $4\pi$ meters. $t=\dfrac{d}{v} = \dfrac{4\pi}{8} = \dfrac{\pi}{2} \approx 1.57$ sec Thanks from topsquark, idontknow and Chemist116
 October 27th, 2019, 05:57 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 I assume that "standard motion" means "counterclockwise motion". For clockwise motion, one gets $\displaystyle t = \frac{\pi}{4}$, which doesn't correspond to any of the listed answer choices. Thanks from skeeter and Chemist116
October 28th, 2019, 10:01 PM   #5
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Quote:
 Originally Posted by skeeter Assuming standard motion about the circle, the angular position of point B is $\theta = \dfrac{3\pi}{2}$ since its tangential velocity is purely in the positive x direction. The tangential velocity components at point A indicate its position is in quadrant I. Since tangential velocity is perpendicular to the position vector, the position vector is in the direction $\theta = \arctan\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$. The angular distance from A to B is $\dfrac{4\pi}{3}$ radians. Given the radius of 3, the linear distance from A to B is $4\pi$ meters. $t=\dfrac{d}{v} = \dfrac{4\pi}{8} = \dfrac{\pi}{2} \approx 1.57$ sec
I believe that this answer would be benefited if it could be accompanied by a drawing or sketch. I did one. Not sure if it would help. But it seems to correspond to what you meant by your method.

If you go in one direction has half the speed from going in the other direction.

I'm not sure if this helps, but I used it as an aid to my understanding.

October 29th, 2019, 07:36 AM   #6
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No, the speeds are the same ... the CW distance is 1/2 the CCW distance, hence the time for the CCW path takes twice the time.

1st screenshot is the CCW path,2nd is the CW path
Attached Images
 CCW.png (1.5 KB, 6 views) CW.png (1.4 KB, 6 views)

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