# Homework Help: Finding an angle of a rope swing involving Energy

1. Oct 17, 2011

### Smartguy94

1. The problem statement, all variables and given/known data
The rope of a swing is 3.00 m long. Calculate the angle from the vertical at which a 80.0 kg man must begin to swing in order to have the same KE at the bottom as a 1420 kg car moving at 1.51 m/s (3.38 mph).

2. Relevant equations
K= (1/2)mv^2
U= mgh
ma=v^2/r=F

3. The attempt at a solution

K + U = Total Energy
1618.871 + 2352 = 3970.871 J

i got really lost on this one

2. Oct 17, 2011

### cepheid

Staff Emeritus
Welcome to PF Smartguy94!

You're correct that a car of that mass moving at that speed will have 1618.871 J of kinetic energy. However, you're not quite applying the conservation of energy to the swing correctly. It's certaintly true that K + U = total energy. And since total energy is conserved, it must be true that:

(K + U)before = (K + U)after

where "before" refers to the state of things before the swing starts dropping, and "after" refers to the state when the swing is at the bottom (i.e. entirely vertical). Now, in the "before" case, the swing is not moving, so all of its energy (which is the total energy of the system) is potential energy. Furthermore, at the bottom of the motion (when the swing is entirely vertical), we can define the height to be zero, so that none of the energy is potential energy, and therefore all of it must be kinetic. This gives us:

Ubefore = Kafter

In other words, initially, the system has a certain amount of potential energy. By the time we reach the bottom of the swinging motion, all of this potential energy has been converted into kinetic energy. However, the total amount of energy in the system remained constant at all times.

You just have to figure out what value of U corresponds to the required value for K at the bottom. That will give you an initial height h above the lowest point. Once you have that, you'll probably have to do a small amount of trigonometry in order to determine the initial angle of the swing that would have it be raised above the lowest point by this height.

3. Oct 17, 2011

### Smartguy94

right, i tried to find the U in terms of that, but I have a hard time figuring out what the height is at that point because i know that H=3 so what I am trying to do is

H=3m
the heigh that we have to find would be

H - H*cos(theta)

now i stuck with a big dilema here because theta is the thing that we are trying to find out, and in order to find the height to find the theta, I need the theta...

4. Oct 18, 2011

### cepheid

Staff Emeritus
Did you try drawing a diagram?

Code (Text):

|\
| \
|  \
|   \   L
|    \
|     \
|------ h

If you look at the right triangle in the figure, you can see that the hyptoenuse of course has length L = 3.00 m. Now, the height above the ground h that the swing must be at corresponds to the difference between the full length L and the length of the vertical arm of the right triangle, which is given by Lcos(θ)

So, you have h = L - Lcos(θ). But you know h from your energy computations. You know L. So you can solve for θ.

5. Oct 18, 2011

### Smartguy94

I did draw a graph like this

Code (Text):

|\
| \
Lcosθ  |  \
|   \   L
|    \
|     \
|------ h
|        \
|         \

and when you said that i have h from my energy computations, I am quite confused to be honest, because I was trying to find what h is, and I can't figure out how.

6. Oct 18, 2011

### cepheid

Staff Emeritus
That was the whole point of the explanation in my first post, and the whole point of the problem: apply the conservation of energy.

Ubefore = Kafter

You know Kafter, because it is specified that it has to be the same as a car of 1420 kg moving at 1.51 m/s, which works out to 1618.871 J.

Ubefore = 1618.871 J

At what height above the ground does the swing have to be initially in order to have this amount of potential energy?

7. Oct 18, 2011

### Smartguy94

OOHH!!!! my god!!! i feel super retarded right now, I'm sorry man my brain is literally dead right now thank you so much though, i got the answer

so it's

mgh = U
h= U/mg
h = 2.065

h=L-Lcosθ
θ=71.8