Centripetal Force + Law of conservation of energy question.

AI Thread Summary
The discussion revolves around a physics problem involving centripetal force and the law of conservation of energy, where a teacher swings from a height using a rope that can break under excessive tension. The key equation for energy conservation is misapplied, as potential energy and force units are incorrectly combined. To solve the problem, one must correctly identify the centripetal force at the moment the rope breaks, factoring in the teacher's weight and the angle of swing. A proper application of energy conservation is necessary to derive the height at which the rope will break. The correct answer, as noted, is 8.0 meters above the ground.
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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
 
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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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