# Why is the constant 5?

1. Mar 27, 2005

### Logik

Alrigth we had a lab were we had to find out what is the initial speed to give a particle to go through a vertical loop with no friction (like a rollercoaster) the equation we found was this :
$$V_o=\sqrt{(c*G*r)}$$
$$V_o = initial speed$$
$$c = constant$$
$$G = 9.8 m/s^2$$
$$r = radius of circle$$
If you fill it in for the condition we had during lab we get :
$$V_o=\sqrt{(5*9.8*r)}$$
The question is were does that 5 comes from... what does it represent...

2. Mar 27, 2005

### cepheid

Staff Emeritus
Think about the theory:

There are no frictional forces, so only conservative forces (gravity) act on the particle. Therefore, its mechanical energy will be conserved. The sum of its potential and kinetic energies will be constant, and must add up to the total energy it had at the start (which was entirely kinetic). We want it to have just enough initial kinetic energy to make it to the top of the loop, i.e. it must have enough initial kinetic energy that when gravity (pointing downward) starts to do negative work on it, slowing it down (changing its kinetic energy to potential energy), it has enough that it reaches the height r above the centre of the circle before coming to a stop. In other words, it must have initial kinetic energy equal to the potential energy it would have once it made it to the top of the loop. Mathematically, this statement is:

$$\frac{1}{2}mv_0^2 = mgh$$

Note: You should be using lowercase g here.

$$\frac{1}{2}v_0^2 = gh$$

$$v_0^2 = 2gh$$

$$v_0 = \sqrt{2gh} = \sqrt{2(9.81)r}$$

In short, I have no idea where the five comes from.

This extra factor of 5/2 in your square root...doesn't jive with the theory.

Edit: is five just some value you found you needed experimentally in order to get an inital velocity high enough that your roller coster made the loop de loop? If so, there will be frictional losses in real life, and so the particle will need to be going faster initially. That's all I can think of.

Last edited: Mar 27, 2005
3. Mar 27, 2005

### Staff: Mentor

To keep the car in contact with the track at the top of the loop, the car must have a minimum speed at the top. If the speed were zero it would fall off the track, never making it to the top. To find the minimum speed, use Newton's 2nd law for centripetal acceleration:
$$mg = mv_{top}^2/r$$​
Thus the KE at the top of the motion must equal:
$$1/2 mv_{top}^2 = 1/2 mgr$$​
Thus the total energy (KE + PE) equals:
$$1/2 mgr + 2mgr$$​
To find the initial speed needed at the bottom, just set that total energy equal to KE and solve for the speed. Then you'll see where the 5 comes from.

4. Mar 27, 2005

### Logik

$$1/2mv_o^2=5/2mgr$$
$$v_o^2=5gr$$
$$v_o=\sqrt{5gr}$$

w00t

5. Mar 28, 2005

### cepheid

Staff Emeritus
Yeah ok, I was quite a bit off on this one, but just to clarify:

Why wouldn't it make it to the top using my (stupid) calculation? It has just enough KE to do so. Wouldn't it make it to the top and then come to a dead stop, dropping straight down like a stone?

Although the formula almost makes sense to me, I don't remember exactly how it follows naturally from the statement above it. How does wanting to make the speed at the top a minimum lead us to this? Sorry, it's been a while since these sorts of problems for me. The only thing I can think of is that the faster you are going, the more centripetal acceleration you have (you sort of slam into the top of the track with more force, which redirects you to follow its path. So the normal force on you would be much greater than just your weight. But if you go up with just enough speed, the normal force on you is equal to your weight? I guess I'm asking, why is this the condition ( centripetal acceleration = g) required to just barely remain in contact with the rails?

2mgr? Oh ok, so we are letting the PE be zero at the bottom of the loop. For a while there, I thought maybe the height was being measured from the centre.

6. Mar 28, 2005

### Staff: Mentor

If the car were being shot straight up into the air, then it would be able to make it. But it's not being shot straight up, it's being forced to follow the curved track as long as possible. If its speed is insufficient, then the car leaves the track and follows a parabolic trajectory like any projectile. (Note that the track gives the car a horizontal component of velocity.)

To barely remain in contact with the track means that the normal force (N) between car and track just goes to zero: thus $N + mg = mv^2/r$ --> $mg = mv^2/r$ when N = 0.

7. Mar 28, 2005

### cepheid

Staff Emeritus
*smacks forehead* yeah ok, thanks. :rofl: