# Potential energy and a pendulum

1. Nov 29, 2008

### Mikejax

1. The problem statement, all variables and given/known data
A 2.kg ball is attached to the bottom end of a length of fishline with a breaking strength of 44.5N. The top end of the fishline is held stationary. The ball is released from rest with the line taut and horizontal (θ = 90 degrees). At what angle θ (measured from the vertical) will the fishline break?

2. Relevant equations
mac = m * v^2/r = sum of radial forces

3. The attempt at a solution

I have the solution to the problem. There is one line I cannot understand:
It says "eliminate v^2/r = 2gcosθ"
You don't have to solve the problem for me, I was just hoping that someone could explain that step to me.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 29, 2008

### alphysicist

Hi Mikejax,

If you use conservation of energy to equate the energy at the starting point and when the pendulum is at angle theta you will get that result. You can then plug that into your force equation to solve the problem.

3. Nov 29, 2008

### Mikejax

Could one treat that as a general formula for pendulum motion?

that (v^2)/r = 2gcosθ?

Can I apply this to any problem of this nature?

4. Nov 29, 2008

### alphysicist

This formula was derived for this particular problem (starting with the string horizontal with theta=90 degrees and released with zero velocity). If that is changed the specific formula will change also.

5. Nov 29, 2008

### Mikejax

Ok so, how does one link the two energies together?

I assume forces here are conservative so

Ui + Ki = Uf + Kf

mgh + 0 = mgh + 1/2mv^2....

how does cosθ incorporate into this equation, or what is a better way to equate the problem?

6. Nov 29, 2008

### alphysicist

Draw a diagram of the pendulum at an angle theta relative to the vertical. Then draw a right triangle with the length of the pendulum string being the hypotenuse.

The length of the vertical side of this right triangle is the vertical height h in the energy equation; since the vertical side is adjacent to the angle, it is related to the hypotenuse through the cosine. Do you see that this gives $h=r\cos\theta$.

(Of course it might need a minus sign, depending on where you set the h=0 for the potential energy in your energy equation.)

7. Nov 29, 2008

### Mikejax

Ok, so my text book uses the following equation for energy of the ball as it swings

mgrcosθ = 1/2mv^2.

I understand that formula means that the potential energy (left side) = kinetic energy (right side) and that the energy has been conserved. And I understand using trig to reflect the height of the ball. However...

What I don't get is that the right side has no potential energy listed...even though the ball is clearly falling, has not hit the ground, and therefore has potential energy due to gravity....

How do they arrive at the equation above without writing a value for potential energy on the right side of the equation?

8. Nov 29, 2008

### Mikejax

any ideas?

9. Nov 29, 2008

### thejinx0r

It's very simple:

The change in potential energy is equal to the kinetic energy.
eg: $$\delta U = K_E$$ in this case since you're initial KE = 0.

But more generally, (if there are no friction forces)
$$\DELTA U = \DELTA K_E$$

So, in a sens, you are right. You would have to write the PE (Potential Energy) on the RHS. But, if you do that, you must also add it to left hand side since the LHS must represent the total PE. But as you can see, they cancel out so you are left with what you had before.

10. Nov 29, 2008

### alphysicist

Hi thejinx0r,

This is not quite right; you need a minus sign in this equation.

11. Nov 29, 2008

### alphysicist

It's there. The energy equation is:

$$m g h_i + \frac{1}{2} m v_i^2 = m g h_f + \frac{1}{2} m v_f^2$$

The question is, where do you set the h=0 level at (that is, where is the origin of your y-coordinate axis)?

If you set $h_i=0$, then $h_f = - r \cos\theta$ (it's negative because $h_f$ is below $h_i$), giving:

$$0 + 0 = m g (-r \cos\theta)+ \frac{1}{2} m v_f^2$$
which is the equation in your post.

If on the other hand you set $h_f=0$, then $h_i=+r \cos\theta$ (positive since $h_i$ is above $h_f$), giving

$$m g (r \cos\theta)+ 0 = 0+\frac{1}{2} m v_f^2$$
which again is the same as your equation.

So the potential energy terms are there on both sides; it's just that with $PE=mgh$ you have the freedom to set one height equal to zero.