What Is the Average Vertical Velocity Halfway Through a Circular Path?

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Homework Help Overview

The problem involves a particle moving along a circular path, specifically half a circle, and seeks to determine the average vertical velocity at the halfway point. The particle travels at a constant speed, and the radius of the circular path is provided.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find the average vertical velocity, questioning the validity of their assumption that initial velocity is zero.
  • Some participants question the definition of average vertical velocity and its calculation, suggesting a focus on vertical displacement over time.
  • Others explore the implications of constant acceleration in circular motion and the appropriateness of different definitions of average velocity.

Discussion Status

Participants are actively engaging with the definitions and calculations related to average vertical velocity. Some have provided guidance on the correct approach to calculating average vertical velocity based on displacement and time, while others are exploring the nuances of using different definitions in the context of circular motion.

Contextual Notes

There is a noted confusion regarding the assumptions made about initial and final velocities, as well as the implications of constant speed in a circular path. The discussion highlights the need for clarity on vertical displacement and elapsed time in the context of the problem.

ortegavs
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Homework Statement


A particle moving from position 1 to position 2 moves along path C ( which is half a circle ). It travels at a constant speed of 5 pi m/s. At exactly half way ( 1/4 of a circle ) through the trip its average vertical velocity is? Radius is 5 meters.


Homework Equations


speed=distance/time
circumference=2 pi r
V avg= Vi + Vf/2
Vf=x/t

Unknowns t, Vavg, Vf



The Attempt at a Solution

I assumed that Vi is zero which simplifies my kinematic equation to V avg=Vf/2. I then substituted x/t for Vf to get Vavg=x/2t. I solved for t using the speed equation. t= distance/speed and then distance=πr/2=π5/2 and speed is 5π. The 5π cancels and the equation simplified to t=1/2s I then plugged in my value for t into the kinematic equation which gives x/2t=x/2/2 since x=5 then 5/2/2 and this gives you 5 m/s. The answer given is 10 m/s but this answer is the final velocity not the average. Am I right about this or does the book have right? The only thing that bothers me is that I set Vi to zero but the question says constant speed meaning that Vi and Vf should be the same that is 5π m/s at the beginning and at the end. Still the answer would not be 10 m/s.
 
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The book answer is correct, assuming points A and B lie on a horizontal axis (you didn't show a picture). You must first look at the definition of average vertical velocity. Average vertical velocity between 2 points is defined as total vertical displacement between those points divided by total elapsed time during that displacement. What is the total vertical displacement after it has traveled the 1/4 circle arc? What time has elapsed?
 
Jay

Points A and B do lie on a horiz. axis. The definition of average V is vi +vf/2. I don't understand why I can't use this definition? And if you use to points A and B you have to divide by two I don't understand. So with the your definition V avg= xf-xi/t which is 5/1/2 and this equal to 10 m/s which is the book answer. How do you know which definition to use?
 
Ok I think I got it. Since the object is moving in a circle then it is constantly changing its direction thus its acceleration is not constant as I had assumed. Given this I cannot use kinematic equation V avg = Vf+Vi/2 since this equation requires constant a. Thus I have to us the other definition provide by Jay V avg= displacement/time

Finally it makes sense
 
ortegavs said:
Ok I think I got it. Since the object is moving in a circle then it is constantly changing its direction thus its acceleration is not constant as I had assumed. Given this I cannot use kinematic equation V avg = Vf+Vi/2 since this equation requires constant a. Thus I have to us the other definition provide by Jay V avg= displacement/time

Finally it makes sense
Yes, that is correct, nice work. Both definitions of average velocity are the same only when acceleration is constant. Note, however , that the result of 10 m/s is the average vertical velocity (in the y direction), which is vertical displacement/time, which is what the problem asked, and which is therefore correct. I want to point out, however, that the average velocity would be different. The total displacement, d, would be (R)(sq rt 2) = 5(1.414) = 7.07 m, and the average velocity would be d/t = (7.07)/(1/2) = 14.14 m/s at an angle of 45 degrees with the horizontal. The reader should fully understand this before proceeding any further:wink:
 

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