Circular Motion (standing on a carousel)

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The discussion focuses on solving problems related to circular motion on a carousel, specifically finding velocity and time given acceleration and distance from the axis. The equation for acceleration, a = V^2/r, is emphasized, alongside the relationship between linear speed and angular speed (v = ωr). Participants clarify the calculation of the period using the formula T = 2π/ω, confirming that the period calculated as approximately 14.65 seconds is correct. The conversation highlights the importance of understanding angular speed in relation to circular motion. Overall, the thread provides guidance on applying formulas to solve circular motion problems effectively.
dragonladies1
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Hi again,

I'm sorry for all of the questions in this section. I am having the worst time trying to understand how to set up the problems. I know the equation for acceleration is a= V^2/r; however for this problem I am looking to find the velocity and time. Again, I would appreciate any kind of help possible.

What are the period and speed of a person on a carousel if the person has an acceleration with a magnitude of 0.7 m/s2 when she is standing 3.8 m from the axis?

What are her acceleration magnitude and speed if she then moves to a distance of 2.1 m from the carousel center and the carousel keeps rotating with the same period?

Thank you!
 
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Hi dragonladies1! :smile:

(try using the X2 tag just above the Reply box :wink:)

Just apply your v2/r formula (and use v = ωr). :smile:
 
Thank you. Using my v2/r formula I came up with 55.63 and I"m not sure what the other formula is. My teacher hasn't mentioned anything using omega.
 
oh, it's just the fairly obvious formula that if your radius is r cm, and your angular speed is ω radians per second, then your actual speed (tangential) is ωr cm per second. :smile:

(and the period is 2π/ω seconds)
 
Ok, so I determined that the speed is 1.63, but I don't know why I can't figure out how to find the period for my problem. I plugged in

Period= 2π/.43=14.65.

Is this correct?
 
dragonladies1 said:
Ok, so I determined that the speed is 1.63, but I don't know why I can't figure out how to find the period for my problem. I plugged in

Period= 2π/.43=14.65.

Is this correct?

Yup! :biggrin:

(or 14.64)

(The reason why it's 2π/(angular speed) is because the period T is the time it takes to go 2π: so (angular) speed = (angular) distance over time, so ω = 2π/T. wink:)
 
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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