Im stuck: Energy in rotational motion

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Homework Help Overview

The problem involves a uniform rope that is initially hanging and then falls to the floor, prompting a discussion on the change in gravitational potential energy associated with this motion.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for gravitational potential energy and the height from which the rope falls. Some question whether to consider the entire length of the rope or just the center of mass in their calculations.

Discussion Status

There is a mix of interpretations regarding the height to be used in the potential energy calculation. Some participants suggest using the full length of the rope, while others propose using the height of the center of mass, indicating an ongoing exploration of the problem.

Contextual Notes

Participants are navigating assumptions about the rope's configuration and the relevant height for calculating potential energy, which may affect their conclusions.

Edwardo_Elric
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Homework Statement


A uniform rope 12.0m long and with a mass 3.00kg is hanging with one end attached to a gymnasium ceiling and the other end just touching the floor. The upper end of the rope is released, and the rope falls to the floor. What is the change in the gravitational potential energy if the rope ends up flat on the floor(not coiled up) ?


Homework Equations


Fx = 0
Fy = 0
U = mgh1 - mgh2



The Attempt at a Solution


Fy = 0 = Tsin(theta) - W
Tsin(theta) = W
T = ((3(9.8))/(sin(theta))

Fx = 0 = Tcos(theta)
No other force acts on the x-axis
 
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change in gravitational potential energy = mgh2 - mgh1

that's all you need.
 
I thought it was inclined


U = mgh2 - 0
= (3.00kg)(9.8m/s^2)(12.0m)
= 352.8 J

thankss
 
Edwardo_Elric said:
I thought it was inclined


U = mgh2 - 0
= (3.00kg)(9.8m/s^2)(12.0m)
= 352.8 J

thankss

wait. use the center of mass... what's the height of the center of mass? you need to use 6.0m not 12.0m.
 
Last edited:

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