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It seems as though the outside edge of a spinning CD is moving faster than the middle of a CD. This can't be true, right? Why is it not true mathematically speaking?

Thank you.

Donna

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- #1

- 2

- 0

It seems as though the outside edge of a spinning CD is moving faster than the middle of a CD. This can't be true, right? Why is it not true mathematically speaking?

Thank you.

Donna

- #2

Gold Member

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- #3

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Consider a point on the outer circumfrance of the disk.

Draw a tangent to that point on the curve and draw a line @ 90 degrees to the tangent to the inner circumfrance of the disc.

This gives a line which is always the same, the 2 points are always in the same position relative to one another - this is true even as the disc is turning.

So as the disc rotates 90 degrees round the two points do too. However relative to each other they have not moved.

So the outer circum point has moved pi * d.1 * 90/360

d.1 = diameter to of entire disk

And the inner circum point has moved pi * d.2 * 90/360

d.2 = diameter of 'hole at centre of disk'

It is clear that d.1 > d.2 and so the point on the outer circumfrance will have traveled further in the same time.

As avg. speed = distance / time

And d.1 distance > d.2 distance, speed.1 will be greater than speed.2.

Thus the outside of the disc has a greater average speed.

Draw a tangent to that point on the curve and draw a line @ 90 degrees to the tangent to the inner circumfrance of the disc.

This gives a line which is always the same, the 2 points are always in the same position relative to one another - this is true even as the disc is turning.

So as the disc rotates 90 degrees round the two points do too. However relative to each other they have not moved.

So the outer circum point has moved pi * d.1 * 90/360

d.1 = diameter to of entire disk

And the inner circum point has moved pi * d.2 * 90/360

d.2 = diameter of 'hole at centre of disk'

It is clear that d.1 > d.2 and so the point on the outer circumfrance will have traveled further in the same time.

As avg. speed = distance / time

And d.1 distance > d.2 distance, speed.1 will be greater than speed.2.

Thus the outside of the disc has a greater average speed.

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- #4

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what did you mean by it has moved pi r1^2 *90/ 360 ?

- #5

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i meant pi*d! my mind went absent...will amend post!

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BJ

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