Period/Frequency of a Rotational Object

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Homework Help Overview

The discussion revolves around a problem involving an object moving uniformly in a circular path, specifically focusing on its period, frequency, and angular speed. The subject area is rotational motion in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to derive frequency from the period and discuss the relationships between translational speed, frequency, and angular speed. Some question the definitions and units involved, particularly regarding angular speed.

Discussion Status

Participants are actively engaging with the problem, with some offering potential approaches to calculate the required values. There is a mix of attempts to clarify concepts and check assumptions, but no explicit consensus has been reached on the final answers.

Contextual Notes

Some participants express uncertainty about the relationships between the variables and the units involved, particularly in converting to radians per second. The original poster indicates feeling lost in the approach, which suggests a need for further clarification on the concepts involved.

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Homework Statement



An object moves uniformly around a circular path of radius .20 meters, making one complete revolution every 2.00 sec. What are (a) the translational speed of the object, (b) the frequency of the object, and (C) the angular speed of the object?



Homework Equations



x = A Cos (angular speed * t)

Angular speed = 2 pi frequency

T = 1/f


The Attempt at a Solution



I am lost as to how to approach this one... I think I might need to solve for the angular speed first?
 
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Looks to me like you could easily find the frequency from the given info.

making one complete revolution every 2.00 sec...

Casey
 
Frequency = 1 / period correct? So, 1 / 2 sec = .5 rev/sec?
 
Anyone?
 
Would this be suffice to solve it?

r = .2m and T = 2s
v = 2pi(r) / T
a) so just plug in.

b) f = 1/T so just plug in again

c) omega = v/r so just plug in.

On part c) how can I get the final answer to be in rad/sec?
 
BBallman_08 said:
c) omega = v/r so just plug in.
On part c) how can I get the final answer to be in rad/sec?
[itex]\omega = v/r[/itex] is in radians per second.
 
Oh okay, so in part a) the r is referring to 2 Pi Radians or not 2 pi radius?
 
You have the right answer for part (a), [itex]v = 2\pi\, r/T[/itex]. Here is how this expression arises. The circumference of the circle is [itex]2\pi r[/itex]. This is the arc length the object travels in In one revolution. Dividing by the time gives the average speed. Since the speed is constant in uniform circular motion, the average speed is the speed, period. Given [itex]v = 2\pi\, r/T[/itex], what is [itex]\omega=v/r[/itex]?
 

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