Centre of Gravity
Centre of Gravity (C.G.)

Definition

Finding C.G. of an irregularly shaped thin
piece of card
Definition

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
Finding C.G.

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.