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

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Points on the outer edge of a CD do indeed spin faster than those on the inner edge due to the greater distance they cover in the same time frame. While all points on the CD complete the same number of revolutions per unit time, the linear velocity increases with distance from the center. The relationship is expressed mathematically as v = wr, where v is peripheral velocity, w is angular velocity, and r is the radius. This means that the outermost part of the CD has a higher circumferential speed compared to the innermost part. Understanding this concept clarifies the difference in speed between various points on a spinning object.
Donna
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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)
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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