# Calculating g using horizontal uniform circular motion

1. Jul 3, 2012

### Ebo

Hi!
I’ve recently done an experiment that I didn’t completely understand and now it seems I’ve lost all my notes as well (the notes written down from the teacher’s board, with a formula I really need, but didn’t understand). This work is supposed to be completed during the summer holiday, so I can’t really ask my teacher for help, either.

During the experiment, I was supposed to find gravity using “horizontal uniform circular motion.” I had two different masses attached to a piece of string and rotated one of the masses above my head while trying to make the other mass stay still.

This is what I’ve managed to figure out this far:
I believe T1 (in the part of the string not rotating) should be equal to T2 (the tension in the rotating string).
X (T1) = 0
Y (T2) = m(1)*g

X (T2) = T * sin(θ) = ((4*π^2*R)/T^2)*m(2)
Y (T2) = T * cos(θ) = g * m (2)
I did not calculate θ as my teacher said it would not be necessary.
This is the point where I think I’m starting to get lost. I’ve got the different components, but I’m not sure how to combine them. I’ve no idea if this works at all:
Y (T total) = g(M(1) – M(2))
X (T total) = ((4*π^2*R)/T^2)*m(2)

And now for my question. I’m sorry, it might sound really stupid, but like I said I’m totally confused and this is my first year with physics, so please bear with me:

Can I now assume that the Y-component of T total equals the X-component of T total?
If not, how can I calculate T2? I thought of using the Pythagoras theorem, but it seemed like I would end up with g on both sides of my formula, and g is what I am supposed to calculate.

2. Jul 3, 2012

### Simon Bridge

You need to draw the free-body diagrams for the two masses ... the centripedal force for the rotating mass is supplied by the weight of the hanging mass. The tension in each section of the string is equal - yes.

3. Jul 3, 2012

### Ebo

I have drawn (several) free body diagrams, but they don't seem to help me much. I really don't see how I can calculate T2 without knowing θ.

I tried to use this formula:
g = (4*π^2*R*M2)/(T^2*(M2-M1)

However, this gave me a value of nearly 20 m s^-2, so I guess it's wrong(?) I’m not completely sure, though, because there’s probably loads of error in the experiment.

My thinking was that this way I will only get g on one side (Ty = g(M2-M1); because Ty for T2 > Ty for T1 because M2 > M1). On the other side of the equation, I would have (4*π^2*R*M2)/(T^2), the Tx components.

Can I assume that the Tx components equals the Ty components?

4. Jul 3, 2012

### truesearch

If you consider the string connected to the rotating mass the tension has a horizontal component and a vertical component.
Can you see what these components must be equal to?....what do they do?

5. Jul 3, 2012

### Ebo

The vertical component equals M2*g, and the horizontal component equals the centripedal force. If I add them together (as vectors), I would get the tension in the string connected to the rotating mass and the tension in this string equals the tension in the hanging mass (=M1*g)

However, if I use the Pythagoras theorem to get the tension in the string connected to the rotating mass, I think I would end up with g as a component on both sides of my equation (T1=T2) ...

6. Jul 3, 2012

### Simon Bridge

Yes - yes you would. This is why we have algebra :)

Start out like this: $T_1=T_2=T$, then write: $T^2=$ ... complete the relation and solve for g.

7. Jul 3, 2012

### Ebo

Alright ...
T(2)^2 = M(2)^2 * ((4*π^2*R)/T^2)^2 + M(2)^2 * g^2
T(1)^2 = M(1)^2 * g^2

Solving for g ...

g^2*(M(2) - M(1))^2 = M(2)^2 * ((4*π^2*R)/T^2)

g = (4*π^2*R*M(1))/((T^2)*(M(1)-M(2))

The trouble is, this gives me a negative value for g (because M(1)<M(2)).

8. Jul 3, 2012

### Simon Bridge

Check the direction of the two tensions the way you calculated them.

9. Jul 3, 2012

### Ebo

After another look at my data I realise I was wrong. M(2) was in fact smaller than M(1), not bigger. I somehow forgot that. Which means that the formula works. I don't get the correct value, but my teacher told me I probably wouldn't, so that's okay.

Thank you so much! This has really helped me a lot :)

10. Jul 4, 2012

### Simon Bridge

Well fair enough - and so long as you got something that was the right order of magnitude. It's not a very sensitive experiment ... mostly to develop your ability to do the sort of reasoning above and train your observation skills.