Solving for Displacement in Circular Motion

[shawonna23]

One afternoon, a couple walks two-thirds of the way around a circular lake, the radius of which is 1.60 km. They start at the west side of the lake and head due south to begin with.

What are the magnitude and direction (relative to due east) of the couple's displacement?

Can someone tell me if my answers are right?

The magnitude is 5.347

The direction is 60 degrees

If these answers are wrong, what do I need to do to get the correct answers?

 

[christinono]

The angle is right (I told you that before). Let me just verify the displacement (1 sec)...

 

[christinono]

Actually, I got another answer for the displacement. What was your method?

 

[MathStudent]

hmm... I think that is the angle wrt to the center of the lake... not from where the couple started

 

[christinono]

Actually, I'm pretty sure the angle is good.
Like I said earlier, draw a circle at the center of a x and y graph. The radius of the circle is 1.6 km. Make a dot at the left of the circle (coordinates are:
(-1.6 km, 0)). Now make another dot at where the couple ends up (in the 1st quadrant). Now make a triangle with both dots and the origin. Now, all U got to do is use simple trig...

 

[MathStudent]

Actually, I'm pretty sure the angle is good.
Like I said earlier, draw a circle at the center of a x and y graph. The radius of the circle is 1.6 km. Make a dot at the left of the circle (coordinates are:
(-1.6 km, 0)). Now make another dot at where the couple ends up (in the 1st quadrant). Now make a triangle with both dots and the origin. Now, all U got to do is use simple trig...

right, that's the diagram that I had set up in the orginal thread. The question asks for the angle of the displacement, that is the angle the displacement vector makes with the x-axis . Which angle are you referring to? The triangle with the vertices you gave is not a right triangle, so what trig methods did you use? If you want, you can email me with your methods so as not to give away any answers to the OP. Or just set up another thread.

 

[MathStudent]

One afternoon, a couple walks two-thirds of the way around a circular lake, the radius of which is 1.60 km. They start at the west side of the lake and head due south to begin with.

What are the magnitude and direction (relative to due east) of the couple's displacement?

Can someone tell me if my answers are right?

The magnitude is 5.347

The direction is 60 degrees

If these answers are wrong, what do I need to do to get the correct answers?

Those answers are incorrect... please post your work so we can try to figure out what went wrong

 

[shawonna23]

Physics: Displacement

I am really lost on that part of the problem. I know that the distance they travel is 6.70206 km. I drew the diagram like you guys said, but I still don't know what to do from there. I don't want you to give me the answer, but could you walk me through the steps of deriving the answers because I am totally loss and the answers are due by midnight.

 

[MathStudent]

You need to find the coresponding x, y coordinates of their final position.

We know they walked a distance of 6.7 km around the circumference of the circle.

Since they headed south, they walked towards -y axis.

1/2 < 2/3 < 3/2 so they must be somewhere in the first quadrant (positive x, positive y)


To Do:
plot an arbitrary point on the circle that is in the first quadrant and let this point denote their final position. Draw a line from the origin to this point, and let $\phi$ denote the angle that this line makes with the positive x axis. If we knew $\phi$, then we could find the x, and y coordinates using the trig functions sin and cos.

a formula that might help here is,

$$s = r \theta$$

where s is the arc length around a circle, r is the radius, and $\theta$ is the angle subtended by the arc in radians not degrees.

see if you can find the angle $\phi$.

hint: first find the angle $\theta$, the angle that is subtended by the arc corresponding to the path they travelled.

 

[shawonna23]

theta= 4.189 radians or 240 degrees

 

[MathStudent]

yep... now to find phi..

 

[shawonna23]

is this correct, and if so what do I do now?

 

[MathStudent]

that is correct. They started at one end of the x axis, and passed the other end. Phi is the angle subtended by their path from the +x axis to their final position. So phi is then the angle subtended by their total path minus the angle subtended by the arc from the -x axis to the +x axis. ( hint, what are the angles in between the coordinate axis? )

 

[shawonna23]

Since I know theta, do I find the x and y coordinates using sin and cos?
Is it 1.6 sin 240 and 1.6 cos 240? I really have trouble finding x and y coordinates.

 

[shawonna23]

is phi 60 degrees? are the angles 180 and 240? I really don't know how to find those angles.

 

[MathStudent]

yes... phi is 60 degrees.

You can think of it this way. as the couple heads south from (-1.6, 0) towards the negative y axis, their angle displacement passes the -y axis, and then passes the +x axis, which make up 180 degrees. Thus the rest of their angle displacement is theta - 180 degrees.

 

[shawonna23]

are you still there? I'm stuck!

 

[christinono]

Sorry, had to leave for a while, but I'm back now. Look, this is a really simple problem. I kinda got the angles mixted up. Yes "math student", the angle is not 60 degrees, since it's from the original starting point.

Try this:
Try what I said to do at the beginning (the whole circle thing). You should come up with an isoceles triangle with an angle of 120 degrees and 2 of 30 degrees. The 2 shortest sides are 1.6 km, and the longest is what you are looking for. Use the sine law to find it. Does this make sense?

 

[shawonna23]

Physics: Displacement

what do I do to find the x and y coordinates now that I know phi is 60

 

[christinono]

You don't need to. You are making this a lot harder than it really is. Ok, I'll walk you through it. First, draw the circle and place the initial and final positions of the couple.

 

[christinono]

Are you still with me Shawonna??

 

[shawonna23]

To find the longest angle, could I do square root of 1.6^2 + 1.6^2 = 2.263km? or is this wrong?

 

[christinono]

To find the longest angle, could I do square root of 1.6^2 + 1.6^2 = 2.263km? or is this wrong?

Look, I don't have the slightest idea what you are talking about. What exactly are you doing, or what have you done so far?

 

[shawonna23]

ok, I drew the diagram

 

[shawonna23]

is the longest angle found by doing 1.6 divided by sin 30 or 1.6 divided by sin 120? I don't know which one to use?

 

[christinono]

is the longest angle found by doing 1.6 divided by sin 30 or 1.6 divided by sin 120? I don't know which one to use?

Ok, you're on the right track. Now, just tell me what your triangle looks like.

 

[shawonna23]

I have the angles 30, 30, and 120. Now I am trying to find the longest side by using the sine law like you said. Do I do this 1.6 divided by sin 30 or do I use the angle 120?

 

[christinono]

Good, I've got the exact same triangle. Just use the sine law:

$$\frac{Sin A}{a} = \frac{Sin B}{b}$$

 

[shawonna23]

its an isoceles triangle with angles 30, 30, and 120

 

[christinono]

yeah, that's right. Now, all you've got to do is find the length of the longest side.

 

[shawonna23]

when using the sine law am I trying to find c?

 

[shawonna23]

Is it sin30/1.6=sin120/c

 

[christinono]

Here is the formula:
$$\frac{Sin A}{a} = \frac{Sin B}{b}$$

Now, place values in:

$$\frac{sin 30}{1.6km} = \frac{sin 120}{x}$$

isolate x:

$$x = \frac{sin 120}{sin 30} \times 1.6 km$$

Now, solve for x.

 

[shawonna23]

x=2.77 Now what do I do now?

 

[christinono]

What do you mean, that's the answer (I hope). Does you texbook or you worksheet give the answer?