00:01
We know that the art spins about its own axis.
00:03
Let omega be the angular velocity of this spinning motion.
00:09
Let o be the center of the art and e be a point on the surface of the art at the equator.
00:16
Now let us consider an object of mass m placed at a position p on the surface of the art.
00:24
Now to measure the angular motion of this mass m, we have a position.
00:31
To draw a line from the position p to the axis of rotation which is o prime and this line should cut the axis of rotation perpendicularly then if capital r is the radius of the art then o prime p is given by small r and smaller is related to capital r the radius of that by this relation that smaller equal to capital r cost theta.
01:03
Here theta is the angle which opi makes with oe.
01:09
Now if we go to the frame which is rotating along with the art then we have to consider the centrifugal force at the point p which is given by e.
01:20
So fc is in this direction shown here which is along the line o prime p and that is given by am omega square small r.
01:29
Now there is also the weight of the particle which is this m times g acceleration due to gravity now because of this mc the effective weight of the particle will be given by w as a function of theta that is equal to m g minus the component of f c which is along this radial direction op so it will be f c c cause that gives us m g minus m omega square r cos square theta and if we just take m common that gives us m times g minus omega square r cause square theta now let us see at the equator at the equator theta are equal to 0 and therefore we have to consider w as a function of 0 and that is given by m g minus omega square r because cos 0 is just equal to 1 now what happens at the poles at the poles theta is equal to 90 degree so we have to compute w at apparent weight at an angle of 90 degrees and that gives us mg minus 0 because cos of 90 degree is equal to 0.
03:06
So it just gives us mg...