Does a CD's Outer Edge Spin Faster? Math Explained!

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In summary, the outer edge of a CD travels at a faster linear velocity than the inner edge due to the varying distance from the center. However, all points on the CD have the same angular velocity. The math involved is calculating the peripheral velocity using the equation v = wr, where v is the peripheral velocity, w is the angular velocity, and r is the radius. More information can be found at https://www.physicsforums.com/showthread.php?t=86445.
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My son has asked me a question I cannot answer. Can someone help?

Does any given point on the outer edge of a CD (or any spinning object, like a record album), turn at a faster speed than a given point on the inside edge, because it is going a farther distance in the same amount of time?

What is the math involved?

Thanks.

Donna
 
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All points on the CD make the same number of revolutions per unit time, but the linear distance traveled varies with the distance from the center. The outermost part of the CD travels faster than the innermost part.

Linear velocity (speed) = Radius (Distance from center) * Angular velocity (revolutions per unit time)
 
  • #3
Donna said:
My son has asked me a question I cannot answer. Can someone help?

Does any given point on the outer edge of a CD (or any spinning object, like a record album), turn at a faster speed than a given point on the inside edge, because it is going a farther distance in the same amount of time?

What is the math involved?

Thanks.

Donna
Yes, the points further from the centre (of rotation) will move faster.

All the points have the same speed of rotation, though. e.g. 33 rpm.

But, the speed of rotation (called the angular velocity) is related to the circumferential speed of the point like this,

v = wr

where v = peripheral velociity
w = angular velocity
r = radius, i.e distance from the centre of rotation.
 

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