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Centre of Gravity

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Centre of Gravity

- Definition
- Finding C.G. of an irregularly shaped thin piece of card

- The Centre of Gravity (C.G.) of a body is the point through which the whole weight of the body seems to act
- For a regular object, the C.G is always at the centre

We can think of the uniform bar as being made up of a lot of tiny particles. Each particle will have a force of gravity pulling it downwards. (weight) The bar will balance at one particular point, P, where the sum of all the clockwise moments of the individual forces is equal to the sum of all the anticlockwise moments of the individual forces.

The effect will be the same as if we had a single force acting downwards at P. So we can think of the gravitational force on the bar as a single force acting downwards at P.

The point, P, is called the centre of gravity of the bar

- If a body is hanging freely at rest, its centre of gravity is always vertically below the pivot.

When the card is released, the weight of the card acts like a single force through the C.G. This is balanced by an equal and opposite force – the reaction of the pin. If the C.G. is not directly beneath the pin, the two forces will form a rotating motion which will make the card swing downwards until the C.G. is directly below the pin. We then hang a plumb line in front of the card and draw a vertical line downwards through the pinhole. The C.G. lies somewhere on this line, AB. We repeat this with the pin at a different position and obtain a second line, CD. Since the C.G. must lie on both lines, it is at the point where the lines intersect.

The Centre of Gravity (C.G.) of a body is the ____________________________________________________________________________________________________________

For a regular object, the C.G is always at the _________

We can think of the uniform bar as being made up of a lot of ____________________. Each particle will have a force of gravity pulling it downwards. (weight) The bar will ________ at one particular point, P, where the ____________________ ____________________ of the individual forces is equal to the __________________________________________ of the individual forces.

The effect will be the same as if we had a ________________ acting downwards at P. So we can think of the gravitational force on the bar as a single force acting downwards at P.

The point, P, is called the _____________________ of the bar

When the card is released, the weight of the card acts like a single force through the C.G. This is balanced by an equal and opposite force – the reaction of the pin. ___________ ____________________________________________________________________which will make the card swing downwards until the C.G. is directly below the pin. We then hang a plumb line in front of the card and draw a vertical line downwards through the pinhole. The C.G. lies somewhere on this line, AB. We repeat this with the pin at a different position and obtain a second line, CD. Since the C.G. must lie on both lines, it is at the point where the lines intersect.