Circle Promblem including degrees, radians, and revolutions.

[Help me in math now!]

Hi! I do not understand how to complete this problem. Please help me complete this!

For this question would I have to find where the 2.8 revolutions have ended and find the coordinates of that point. If so, would I divide 2.8 revolutions by 360 degrees and find the total degree number, then find the coordinate? Or, I could go for radians and dive it by 2pi and find the total radians? If so, are there any more sites that could help me with the understanding of these kinds of problems, like give me more of these types of problems to do? Thank you very much, and for your time in helping me with this problem!

Imagine a coordinate system superimposed on the wheel of fortune
with its origin at the center of the wheel. Suppose a contestant spins
the wheel releasing it at the “3 o’clock” position on the coordinate
system, and the wheel completes 2.8 revolutions in 20 seconds.

A. What are the coordinates of the release point when the wheel
stops at the end of 20 seconds?

 

[srmichael]

Hi! I do not understand how to complete this problem. Please help me complete this! I do not understand how to obtain the information to solve for the angular velocity, would you only multiply 20 sec by 2.8? I also do not understand the rest. Thank you very much, and for your time in helping me with this problem!

Imagine a coordinate system superimposed on the wheel of fortune
with its origin at the center of the wheel. Suppose a contestant spins
the wheel releasing it at the “3 o’clock” position on the coordinate
system, and the wheel completes 2.8 revolutions in 20 seconds.

A. What is the angular velocity in revolutions per second? In radians
per second?

B. If the wheel is 10 feet in diameter, what is the average linear
velocity in feet per second of a point on the edge of the wheel?

C. What are the coordinates of the release point when the wheel
stops at the end of 20 seconds?

In A it is asking for revolutions per second? That implies division with revolutions in the numerator and time in seconds in the denominator. Thus,

\(\displaystyle \omega=\frac{v}{r}=\frac{2.8}{20}=0.14\) revs/sec

There's a start. Now, how many radians are in a revolution?

 

[mmm4444bot]

revolutions per second

multiply 20 sec by 2.8?

No.

The word "per" denotes a ratio. In other words, the units revolutions per second look like this:

\(\displaystyle \frac{revolutions}{second}\)

Hence, you do not multiply.

Divide the number of revolutions by the number of seconds.

do not understand the rest

Why is that? How many examples have you already studied? What specifically are your questions about the rest? Vague statements like the one above provide no information for tutors to assess what you need.

Please ask specific questions or explain what you're thinking about this exercise.

If you need to study some examples, do that first. (Let us know, if you need links to lessons or examples about angular and linear velocity.)

Cheers :cool:

 

[mmm4444bot]

I deleted your duplicate thread, posted five minutes ago.

Please do not repeat previous threads; if you would like to continue discussing the exercise above, do that by adding your posts to this thread.

If you would like to discuss a different exercise, then start a new thread for that.

Cheers :cool:

 

[Help me in math now!]

Reply to previous comments

Okay thank you, and I have added on what I need help with. Is there anything else that I have to do?

 

[HallsofIvy]

​

What you had as "C" before is not "A". Does that mean you have alread done the previous "A" and "B"? If the wheel has turned 2.8 revolutions, the first 2 revolutions just put it back to the starting point. The final .8 revolution will move it to \(\displaystyle .8(2\pi)\) radians or \(\displaystyle .8(360)= 288\) degrees. Now, do you have a formula for changing from angle and radius to x and y coordinates?

 

[mmm4444bot]

Another way to visualize:

The end location of the release point and the origin are endpoints of the hypotenuse of a right triangle, formed with a segment of the positive x-axis and the "height" of the triangle.

This is what I'm seeing, in my mind. Did you draw a picture?

 

[Help me in math now!]

Reply to previous

What you had as "C" before is not "A". Does that mean you have alread done the previous "A" and "B"? If the wheel has turned 2.8 revolutions, the first 2 revolutions just put it back to the starting point. The final .8 revolution will move it to \(\displaystyle .8(2\pi)\) radians or \(\displaystyle .8(360)= 288\) degrees. Now, do you have a formula for changing from angle and radius to x and y coordinates?

No I do not have the formula for chanfing the angle and radius x, y coordinates.

 

[Help me in math now!]

Another way to visualize:

The end location of the release point and the origin are endpoints of the hypotenuse of a right triangle, formed with segments of the vertical and horizontal axes.

Did you draw a picture?

Well, first of all would I take the degree measure of this which is 288 degrees and measure it out on the circle? Would I then take this point and create a triangle from it? I did draw a picture, though it doesn't give much help towards the solution of the problem. And thank you for your time in helping me in this problem!

 

[mmm4444bot]

OIC, you added information to this discussion by editing it into (or out of) your original post. I somehow missed those edits, and I'm not into comparing the different versions, but I'm willing to continue helping you to locate the coordinates of the release point.

There are many different approaches; I regret that I do not know your instructor's preferences. Halls suggested a formula, and I suggested Right-Triangle Trigonometry. Either way is fine, but do you have a sense of what your class has been working on lately?

I'll stick with the triangle, in this thread. If you have a preference, please say so. I'm also puzzled by your statement that a diagram is not much help. I'm wondering whether you have yet learned about Right-Triangle Trigonometry (i.e., sine is opposite/hypotenuse, cosine is adjacent/hypotenuse)?

The radius of the circle and the hypotenuse of the right triangle are the same. Hypotenuse*sin(angle) gives you the length of the side of the triangle opposite the angle and hypotenuse*cos(angle) gives you the length of the side of the triangle adjacent to the angle, and those two lengths are the (x, y) coordinates you seek.

Above, angle is the reference angle for 288° because The Wheel of Fortune always spins in a clockwise direction.

Have you used reference angles before? Again, if you need links to lessons, let us know.

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NOTE: I edited my misstatement, in the previous post where I stated that segments of both axes form the legs of the triangle.