Distance

[Guest]

A phonograph record is 12" across.
An outer non-playing area 1" in width.
A non-playing central area 4" in diameter.
An average of 90 grooves per in.

Determine distance the needle travels when the record plays once.

12 - 1 - 4 = 7"

I don't think that is correct. Where did I make the error(s) ?

[Guest]

Consider the groove as a long strip (1/90) inches wide.
The disk is 5" outer and 2" inner radius

Area of disk is pi x (5^2 - 2^2)
= pi x (25 - 4)
= pi x 21

length of groove = pi x 21 / (1/90) / 12 ..... expressed in feet
= pi x 21 x 90 / 12
= 22/7 x 21 x 90 / 12
= 66 x 90 / 12
= 11 x 90 / 2
= 11 x 45 = 495 feet

[Guest]

"Area of disk is pi x (5^2 - 2^2)" Would the disk also be considered a ring?

I have a different answer provided:

The grooves are irrelevant. The needle only travels 3 inches. Please explain that. Thanks.

[Guest]

If 3 is the answer you want, then that is a trivial solution... ie 5 - 2 = 3. and you don't need to know the grooves are 90 per inch
I thought the question was asking how far does the needle around the surface of the disk as it rotated to play the recording... ie the length of the recording, which is 495 feet.

[Guest]

"I thought the question was asking how far does the needle around the surface of the disk as it rotated to play the recording... ie the length of the recording." It is asking that.

Would the disk be considered a ring ?

[Guest]

Well then, what is wrong with my original answer of 495 feet?.

is it near the answer you have?. Post your answer and I can see if I can agree.

[Guest]

"Well then, what is wrong with my original answer of 495 feet?." Nothing is wrong.

Only answer I have is the 3 inches.

[Guest]

I don't understand how you consider that "Nothing is wrong" with the 495 feet answer if you consider the 3 inch answer to be correct.
3 inches is the radial distance from the outer groove (at radius of 5 inches) to the inner groove (at a radius of 2 inches)

[Guest]

So, would the 3 inches also be the width traveled and the 495 feet total distance traveled ?

Also," the outer groove (at radius of 5 inches) to the inner groove (at a radius of 2 inches)" How was each radius calculated ?

[Guest]

A phonograph record is 12" across. That is 12 inches diameter or 6 inches radius to outer edge
An outer non-playing area 1" in width. width at outer edge so take 1 inch from radius 6 -1 = 5
A non-playing central area 4" in diameter. inner circle is 4 inches diameter or 2 inches radius
Distance across playing area radially is 5 - 2 = 3
An average of 90 grooves per in. We don't need this information for your answer

[Guest]

Thanks.

So, with the omission of the grooves, no matter how the distance is calculated, the answer will be 3 in. ?

[Guest]

I don't know the question. I calc the length of the record groove. Your answer seems to be the dist moved radially but you won't show your working. So what is the question

[Guest]

Using the circumference. could the distance be determined ?

[Guest]

There are several ways to do it what is the question

[Guest]

Circumference - Pi x diameter
3.1416 x 12 = 37.6992 x 3 (distance radially) = 113.0976 = 113 in. / 12 = 9.4166 = 9 ft.

Would that be correct ?

[Guest]

What is the question you trying to solve my earlier solution works out the length of the groove based on the area of the recording and the width of the groove i dont know what you are trying to work out

[Guest]

I was trying to determine length of travel from circumference and width similar to your earlier post. I guess that doesn't work. You said there are several ways to do it. Please post another way. Thanks.

[Guest]

Groove is a spiral from 5 inch rad to 2 inch rad each turn getting shorter find the total of these using series and see if it comes to 495 feet

[Guest]

"find the total of these using series " I have no idea.

[Guest]

Please respond - thanks.