Centripetal Force + Law of conservation of energy question.

[anonymous12]

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:

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

 

[Thundagere]

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.

 

[anonymous12]

Oops. But what do I do next? How do I solve the question?

 

[PhanthomJay]

[tex]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's why I put the 2 infront of m. Then I crossed out the m's and you get.

$$gh_1 + \frac{v^{2}}{r}_1 = 2gh_2 + \frac{mv^{2}}{r}_2$$

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'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 .

 

[asteriatic]

Your solution is on the right track, but there are a few errors. First, the equation for centripetal force should be F_c = \frac{mv^2}{r}, not F_c = mv^2. Also, you should use the height of the building (24m) as the radius in this equation, not the length of the rope. So the correct equation would be:

mgh_1 + \frac{mv^2}{r}_1 = 2mgh_2 + \frac{mv^2}{r}_2

Next, you need to substitute the values for the initial and final positions and velocities. Since the teacher starts at rest, the initial velocity is 0, and the final velocity is the velocity at the bottom of the swing, which can be found using conservation of energy:

mgh_1 = \frac{1}{2}mv^2

Solve for v and substitute it into the equation, along with the given values for g and r. You should end up with:

235.2 = 19.6h + \frac{1}{2}v^2

Now, you can solve for h by setting this equation equal to the maximum tension force (2mg) and solving for h. You should get h = 8.0m, the correct answer.