Velocity of an Object in Uniform Circular Motion

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An object with a mass of 0.5 kg is in uniform circular motion with a radius of 2 meters and a force of 4 N acting on it. The attempt to calculate the velocity using the centripetal force equation revealed an algebra mistake, specifically in treating a product as a sum. The discussion emphasizes the importance of maintaining significant figures only in the final answer to avoid compounding errors. The correct approach to solving the problem involves careful algebraic manipulation and understanding of the equations involved. Overall, the thread highlights common pitfalls in physics calculations and the significance of accuracy in mathematical operations.
Argella
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Homework Statement


An object of mass 0.5 kg is swung in uniform circular motion. The radius is 2 meters, and the force exerted is 4 N. Calculate the magnitude of the velocity.

Answer Choices:
a) 0.25 m/s
b) 1 m/s
c) 4 m/s
d) 16 m/s

Homework Equations


v = 2piR/T
F = mv^2/R

The Attempt at a Solution


Since I didn't know the variable time, I couldn't use the velocity equation. Next, I tried plugging in the variables for the centripetal force equation and solving for velocity:

4 = (0.5)v^2/2
4 = 0.5/2 + v^2/2
4 = 0.3 + v^2/2 <- 0.3 because of significant figures
(subtracted 0.3 from both sides)
4 = v^2/2 <- 4 because of significant figures
(multiplied both sides by 2)
8 = v ^2
(took the square root of both sides)
v = 3 which isn't one of my answer choices.

Maybe I'm not doing my algebra correctly or maybe I'm not using the right equation, but I couldn't find another equation using the variables I have. If I knew the time it took for one revolution I could use the velocity equation, but I don't know how to find time and I couldn't find an equation for finding it.
 
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Argella said:
4 = (0.5)v^2/2
4 = 0.5/2 + v^2/2
There is an algebra mistake in going from the first to the second step. You have essentially turned a product into a sum instead.

On a side note, when taking significant figures into account, do so only on the final answer. If you keep rounding off at each step, you will find that the rounding-off error grows substantially larger after each step, and the final answer you obtain might be significantly off the correct one.
 
Ohh, okay, that makes a lot more sense. I'll be sure to keep that in mind about significant figures in the future, too.

Thanks a lot!
 
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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