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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 exactly should I do to find the given time.

The only thing which I could come up with is that to find the time can be obtained from the tangential speed as:

$v_{t}=\omega r = \frac{\theta}{t} r$

But apart from this I dont know how to link it with what is being asked. Can somebody give me a hand with this?

With the guidance of Eric Towers I made the following sketch with do help to understand how the vectors are placed in the circle this can aid in the solution of the problem.

Sketch of the vectors in the circle

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The radius of the circle is $3 \,\mathrm{m}$. The object's speed is always $|(-4,4\sqrt{3})| = |(8,0)| = 8 \,\mathrm{m/s}$. Now we have a choice.

Recall that, in a circle, tangents (velocities) are perpendicular to radii. This means the velocity vector $8i$ can only occur at the top or bottom of the circular motion (position angles $\pi/2$ or $3\pi/2$). Similarly, the velocity vector $-4i+4\sqrt{3}j$ can only occur at position angles $\pi/6$ or $7\pi/6$. We can distinguish these because either the circulation is clockwise or it is anticlockwise.

If it is clockwise: The the velocity vector $8i$ must occur at the top of the circle, at angle $\pi/2$, and the velocity vector $-4i+4\sqrt{3}j$ occurs at angle $7\pi/6$. So the sphere travels through some number of radians and the length of the portion of the circumference is the radius times the subtended angle. Then length over speed gets you the time.

If it is anticlockwise: The velocity vector $8i$ must occur at the bottom of the circle, at position angle $3\pi/2$, and the velocity vector $-4i+4\sqrt{3}j$ must occur at position angle $\pi/6$. Now the sphere travels half as far, so it takes half the time to do so.

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  • $\begingroup$ Okay so you obtained $\left \| \left \langle -4,4\sqrt 3 \right \rangle\right \|=8$. When you mentioned that the velocity are perpendicular to the circle. Wouldn't it mean that is also occurs at $0$ and $\pi$?. But the part which I'm confused is that if it mentions the vector $\left \langle 8,0 \right \rangle$ this would be in the right side of the circle as it is positive, not in the left side. I don't know how you obtained the angle for the vector $\left \langle -4,4\sqrt{3} \right \rangle$. I guess you can use $\tan^{-1}\left(\frac{4\sqrt{3}}{4}\right)$ but without calculator? $\endgroup$ – Chris Steinbeck Bell Oct 27 '19 at 6:51
  • $\begingroup$ I overlooked the fact there's the $30-60$ triangle so it comes from there. But again doing the above operation it would result into $\omega=\frac{\pi}{3}$?. $\endgroup$ – Chris Steinbeck Bell Oct 27 '19 at 6:57
  • $\begingroup$ @ChrisSteinbeckBell : At position angles $0$ and $\pi$, the velocity is in the $j$ direction. Whether it's positive or negative depends on whether the motion is clockwise or anticlockwise. $\endgroup$ – Eric Towers Oct 27 '19 at 7:03
  • $\begingroup$ @ChrisSteinbeckBell : $\langle -4,4\sqrt{3} \rangle = 8 \langle -1/2, \sqrt{3}/2 \rangle$. That is, factor out the speed to get a recognizable point on the unit circle. This point is in quadrant $2$, at angle $2 \pi/3$. $\endgroup$ – Eric Towers Oct 27 '19 at 7:05
  • $\begingroup$ It seems I overlooked that, I have noticed that as you mentioned it must go in the the top or it could be in the bottom and clock or counterclockwise. But what about the angle of the other vector. Wouldn't it be $\frac{\pi}{3}$? $\endgroup$ – Chris Steinbeck Bell Oct 27 '19 at 7:06

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