Two pulleys and same masses - velocity?

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Homework Help Overview

The problem involves two equal masses suspended over pulleys, initially at rest, and asks for the velocity of the leftmost mass when the separation between the two masses is a specified height, h. The scenario assumes frictionless pulleys and extensionless strings, placing it within the context of classical mechanics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles and the relationship between the velocities of the two masses. There is an exploration of how the distances traveled by each mass relate to their speeds, with some participants questioning the assumptions made about their motion.

Discussion Status

The discussion is ongoing, with some participants providing insights into the relationship between the speeds of the two masses based on their respective distances traveled. There is acknowledgment of a potential solution, but the details of the relationship and the reasoning behind it are still being explored.

Contextual Notes

Participants note the lack of explicit values for certain variables, such as the initial height, and the implications of the problem's constraints on the analysis being conducted.

riverguardian
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Homework Statement



If two equal masses are suspended as in the following, and they are both initially at rest, after they are released, what is the velocity of the leftmost mass at the instant the separation between the two masses is h ? Assume frictionless pulleys and extensionless strings.

yPOxJ.png


Homework Equations



KE(i) + PE(i) = KE(f) + PE(f)

The Attempt at a Solution



Set x as the height initially. For every extension of the string immediately left, the pulley system pulls the second mass up half the distance, so the left cord will have traveled 2/3h and the right cords will have moved 1/3h, for a total separation of h.

mgx + mgx = mg(x+1/3h) + mg(x-2/3h) + 1/2*m*v1^2 + 1/2*m*v2^2

We don't know what the velocity of each mass is, correct? We also know that each of the tensions a/b/c or T, and there's a gravitational force downwards with magnitude mg.
 
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The answer, apparently is v = sqrt(8/15gh)
 
help?
 
You know what the speed of each is relative to the other. Therefore you only have one unknown in your energy balance. Answer is correct.
 
LawrenceC said:
You know what the speed of each is relative to the other. Therefore you only have one unknown in your energy balance. Answer is correct.

How are the speeds related? What do we use to determine their relation?
 
You know how far each has traveled from its starting point...if one mass went twice as far (which you have already determined) in the same amount of time, how must their speeds relate given that the cord does not streach?
 

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