How to find the center of gravity of an object

[Zarich12]

Is there a relatively simple way to find the center of gravity of an object? I need to know to see if my model rocket has a high center of gravity. The bulk of he weight is in the bottom.

 

[jbriggs444]

Balance it on something narrow. See where the balance point is.

 

[Zarich12]

OK. Why didn't I think of that. I guess I'm thinking too hard :)

 

[Zarich12]

Ok. One mor question. If that same rocket is too tippy in the air, does that mean it has a high or low center of gravity. I think it's high, but just wanted to check.

 

[Tom MS]

Ok. One mor question. If that same rocket is too tippy in the air, does that mean it has a high or low center of gravity. I think it's high, but just wanted to check.

I wouldn't think it would matter because the torque is proportional to the radius from the pivot point AND the sine of the angle. If the center of mass was higher, the radius would increase but the angle would decrease.
"Tippy", of course, is a relative term that I suspect in this case describes a mass with a high surface density along the cross-section of the object whose edge runs from the pivot point to the top of the object. Such an object, when any horizontal adjustment to the pivot point was made, would have the torque would change greatly.

 

[Zarich12]

Look, I understand most of what you just said but some of it went right past me. I'm 13. To answer your last question, yes, it does change greatly. If the ballance of this model rocket is off by just a few tenths of a ounce, the rocket will tip over. It will also do so if the center of gravity is too high. Then the rocket loses vertical distance because it uses it's burn time to fire horizontally. I just want to make the most of the 0.7 seconds of firing time the motor has.

 

[sophiecentaur]

Look, I understand most of what you just said but some of it went right past me. I'm 13. To answer your last question, yes, it does change greatly. If the ballance of this model rocket is off by just a few tenths of a ounce, the rocket will tip over. It will also do so if the center of gravity is too high. Then the rocket loses vertical distance because it uses it's burn time to fire horizontally. I just want to make the most of the 0.7 seconds of firing time the motor has.

The motor thrust is acting against the bottom of the rocket (obviously but needs to be said). This is like balancing it on your finger but with no 'skill' to help it stay there. Having the CM as low as possible will make it less unstable. Until is actually gets moving, it will be tending to fall over. Space rockets have complicated servos to keep them upright for the first bit of the ascent. Once the rocket builds up some speed, the aerodynamics (fins and stuff) will keep it 'flying' straight so the better the initial acceleration, the better. It's a race to get it fast enough before the tipping becomes significant.
'Firework' rockets use a stick which puts the CM below the propulsion and that solves the problem (and also the aerodynamics help it to fly straight) - but doesn't make it look like a proper rocket and adds drag, I guess.
A short length of takeoff track could help to aim the rocket till it has built up speed and the fins start to work.
Best to look on some amateur rocket enthusiast sites. They have all been there before and will be full of appropriate advice (from a different angle from the PF comments).

 

[Tom MS]

Ok. One mor question. If that same rocket is too tippy in the air, does that mean it has a high or low center of gravity. I think it's high, but just wanted to check.

My bad. A better explanation would be that a measure of tipiness is the angular acceleration, which is basically how fast the speed of angular movement changes. This value can be related by: \alpha = \frac {\tau} {I}
In English, the above equation means the angular acceleration equals the torque divided by the moment of inertia. Torque (the numerator), can be related by the distance from the pivot point to the center of gravity. The moment of inertia (the denominator) can be related to the square of the distance from pivot point to the center of gravity. Because the moment of inertia is related by one higher a power, that means that it "wins out" in the battle for who decides the angular acceleration. This means that the farther away the center of gravity the pivot point is, the less angular acceleration there will be.
In conclusion, more distance from the pivot point means less tipiness; however, rocket science may be different and is certainly beyond the scope of what I know.

 

[SF cookie]

Keep the centre of gravity as low as you can get it. Fins will help too, provided they are accurately aligned with the axis of the body of the rocket. More specifically, fins near the base of the rocket help to move the centre of gravity lower and whilst still within the atmosphere will help to keep the back end behind the front - thanks to atmospheric drag.

Rockets are fun, and whatever they say, it's not all rocket science. Think about the forces at play and what you can do to keep them working in your favour. Common sense should be your touch-stone, the formulae can come later!

 

[Zarich12]

Right, thanks!