Centripetal Force + Law of conservation of energy question.

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Homework Help Overview

The problem involves a physics scenario where a teacher attempts to swing from the roof of a 24-meter high building using a rope. The question focuses on determining the height at which the rope breaks due to tension exceeding twice the teacher's weight, applying the law of conservation of energy in the analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy and the relationship between centripetal force and tension. There are attempts to set up equations based on energy conservation, but some participants question the validity of combining potential energy and force units.

Discussion Status

The discussion is ongoing, with participants providing insights into the correct application of physics principles. Some guidance has been offered regarding the need to correctly identify centripetal force and the components involved, but no consensus has been reached on the next steps to solve the problem.

Contextual Notes

There is a hint provided in the original problem statement to apply the law of conservation of energy, but participants express uncertainty about the setup and calculations involved. Additionally, there are concerns about the interpretation of the equations being used.

anonymous12
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Homework Statement


Your favourite physics teacher who is late for class attempts to swing from the roof of a 24-m high building to the bottom of an identical building using a 24m rope as shown in Figure 5. She starts from rest with the rope horizontal, but the rope will break if the tension force in it is twice the weight of the teacher. How high is the swinging physicist above level when the rope breaks? (Hint: Apply the law of conservation of energy.)

Figure 5:
Screen shot 2011-12-14 at 9.58.37 PM.png

Homework Equations


F_c = \frac{mv^{2}}{r}

The Attempt at a Solution


mgh_1 + \frac{mv^{2}}{r}_1 = 2mgh_2 + \frac{mv^{2}}{r}_2 Since the questions states that when tension force is twice the weight of the mass, then the rope will break. That's why I put the 2 infront of m. Then I crossed out the m's and you get.

gh_1 + \frac{v^{2}}{r}_1 = 2gh_2 + \frac{mv^{2}}{r}_2
(9.8)(24) + 0 = 2(9.8)h + \frac{v^{2}}{r}
235.2 = 19.6h + \frac{v^{2}}{r}
I don't really know what to do next.

The answer in the back of the book is 8.0m
 
Last edited:
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Before going anywhere, keep in mind that
ac=mv2r

Is not true. v2/r is the acceleration, meaning that mv2/r is the centripetal force.
 
Oops. But what do I do next? How do I solve the question?
 
anonymous12 said:
F_mgh_1 + \frac{mv^{2}}{r}_1 = 2mgh_2 + \frac{mv^{2}}{r}_2 Since the questions states that when tension force is twice the weight of the mass, then the rope will break. That&#039;s why I put the 2 infront of m. Then I crossed out the m&#039;s and you get.<br /> <br /> gh_1 + \frac{v^{2}}{r}_1 = 2gh_2 + \frac{mv^{2}}{r}_2
what equation is this? You are adding PE units with force units. This is not the conservation of energy equation. You first have to identify the centripetal force when the rope breaks at a tension of 2mg. The centripetal force is the net force in the centripetal direction which is the tension force less the component of the person&#039;s weight in the centripetal direction at an unknown height and angle (draw a sketch). Then apply conservation of energy correctly to get a second equation. Note also that hsintheta is h/24, where theta is the angle with the x axis, and h is measured from the top of the roof .
 

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