Max Height of Swing in Circular Motion

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Homework Help Overview

The problem involves a child's playground swing supported by chains, focusing on determining the maximum height of the swing based on its initial conditions, including its height above the ground and speed when the chains are vertical. The subject area relates to concepts in circular motion and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Energy conservation

Approaches and Questions Raised

  • Participants discuss the relevance of circular motion versus energy conservation in solving the problem. Some express confusion about the necessity of mass, time, or angle in their calculations. Others explore the relationship between kinetic and potential energy, questioning how to incorporate the length of the swing chain into their reasoning.

Discussion Status

Some participants have suggested using conservation of energy to approach the problem, while others are still grappling with the implications of the swing's chain length and its role in the solution. There is a mix of interpretations regarding the necessary components for solving the problem, indicating an ongoing exploration of ideas.

Contextual Notes

Participants note the absence of certain variables, such as mass and time, which they feel may be critical to fully understanding the problem. The inclusion of the swing chain's length raises questions about its significance in the context of the problem.

ally1h
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Homework Statement


A child's playground swing is supported by chains that are 4.0m long. If the swing is 0.50m above the ground and moving at 6.0 m/s when the chains are vertical, what is the maximum height of the swing?



Homework Equations





The Attempt at a Solution


Back again.. this time I am LOST. I feel like this should be a circular motion type of problem, but that doesn't entirely make sense since I'm working on energy conservation, linear momentum, torque, and angular momentum. This is what I DO understand:

radius = 4.0m
vi = 6.0 m/s
vf = 0 m/s
The total height is whatever it is plus 0.50m.

I'm lost because I have no clue how to do the problem without a time component or a mass component or an angle. I understand that gravity plays a vertical role, slowing down the speed of the swing.
 
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ally1h said:

Homework Statement


A child's playground swing is supported by chains that are 4.0m long. If the swing is 0.50m above the ground and moving at 6.0 m/s when the chains are vertical, what is the maximum height of the swing?

Homework Equations


The Attempt at a Solution


Back again.. this time I am LOST. I feel like this should be a circular motion type of problem, but that doesn't entirely make sense since I'm working on energy conservation, linear momentum, torque, and angular momentum. This is what I DO understand:

radius = 4.0m
vi = 6.0 m/s
vf = 0 m/s
The total height is whatever it is plus 0.50m.

I'm lost because I have no clue how to do the problem without a time component or a mass component or an angle. I understand that gravity plays a vertical role, slowing down the speed of the swing.
It may be easier in this case to consider conservation of energy rather than focusing on circular motion.
 
The conservation of energy should be sufficient to answer this problem.

You don't need mass or time. I am pretty sure it is assumed that the ground is flat, otherwise the problem wouldn't be solvable. In the case the ground is flat then you are given an angle which is 90 degrees when the swing as at the bottom of its swing.

The swing is at is maximum velocity when it is at the bottom of its swing (when its not moving upwards). Like you said the swing has no velocity at vf of this problem. This means that at this time it has transferred all of its kinetic energy into ...
 
Sorry, I had a dr. appointment...The kinetic energy was transferred into potential energy. This helps a bit, but I feel like I'm still missing something.

I know the change in potential energy is U = mgΔy; I don't know m, but I can figure Δy from the equation: Δy = vi^2 / 2g = (6.0 m/s)^2 / (2)(9.8 m/s) = 1.84 m. So am I just adding 1.84m + 0.50m = 2.34m ??

I feel like I'm missing something, otherwise why would the length of the swing chain (4.0m) be included?
 
ally1h said:
I know the change in potential energy is U = mgΔy; I don't know m, but I can figure Δy from the equation: Δy = vi^2 / 2g = (6.0 m/s)^2 / (2)(9.8 m/s) = 1.84 m. So am I just adding 1.84m + 0.50m = 2.34m ??
That's all there is to it. Realize that the equation you used is just an application of energy conservation.
I feel like I'm missing something, otherwise why would the length of the swing chain (4.0m) be included?
I can think of several reasons: (1) Just to see if you know what matters and what doesn't, or (2) The problem might have another part that will require the length of the swing. :smile:
 

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