How Do You Calculate the Velocity of a Point on a Rotating Disc?

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To calculate the velocity of a point on a rotating disc at 800 rpm, first convert the rotation speed to revolutions per second (40/3 rps). The period T is then determined as 3/40 seconds. The velocity v can be calculated using the formula v = (2πr) / T, resulting in v = (80πr) / 3 m/sec. It's crucial to maintain the "seconds" unit throughout the calculation to ensure the final result is in distance/time. This method confirms the correct approach to finding the velocity of a point on the rotating disc.
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Homework Statement



Find the velocity v of a general point on a plate rotating at 800 rpm which is r meters from the center.

This is literally a copy and paste job.

Homework Equations



period = T = 1 / f ; f = frequency

The Attempt at a Solution



This is what I did please tell me if it is correct

800 rpm = 40 / 3 rps = f

T = 1 / f = 3 / 40

v = distance / time = (2 pi r) / (3 / 40) = (80 pi r) / 3
 
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That is correct.
You could also have seen that 40/3 rps can be directly translated into a speed by replacing the r in rps with the distance covered in 1 revolution (2 pi r).
That would give you the same result ##\frac{80 \pi r}{ 3}##m/sec.
 
Yes, that works. You need to carry the "seconds" unit through to the end though. The period T has seconds as its unit.

Otherwise you'll end up with a "velocity" with units of distance rather than distance/time.
 
RUber said:
That is correct.
You could also have seen that 40/3 rps can be directly translated into a speed by replacing the r in rps with the distance covered in 1 revolution (2 pi r).
That would give you the same result ##\frac{80 \pi r}{ 3}##m/sec.
Awesome! Thank you.
 
gneill said:
Yes, that works. You need to carry the "seconds" unit through to the end though. The period T has seconds as its unit.

Otherwise you'll end up with a "velocity" with units of distance rather than distance/time.
Ohh... I see. Thank you
 
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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