Please Summarize the follwing article (this is for apresentation)
\"When you release a yo-yo, gravity acts on its center of mass topull the yo-yo downward. Because the string of the yo-yo is wrappedaround the yo-yo's axle, and because one end of the string isattached to your finger, the yo-yo is forced to rotate as it drops.If the yo-yo could not rotate, it would not drop.
Just as any object falling in a gravitational field, the rate ofdrop increases with time and so, necessarily, does the rotationrate of the yo-yo. The rate of drop and the rotation rate aregreatest when the bottom is reached and the string is completelyunwound. The spinning yo-yo contains angular momentum (orrotational kinetic energy) derived from the gravitataion potentialenergy through which the yo-yo has dropped.
Usually, the string is tied loosely around the axle so that theyo-yo can continue to spin at the bottom. Because the full lengthof the string has been paid out, the yo-yo can drop no further and,consequently, the rotation rate cannot increase further. If left inthis condition, the friction between the axle and the string willeventually dissipate the energy of rotation or, equivalently, theangular momentum of the yo-yo and the yo-yo will come torest.
However, a momentary tug on the string causes the friction betweenthe string and the axle briefly to increase so that the axle nolonger slips within the string. When the axle thus stops slipping,the angular momentum of the spinning yo-yo is sufficient to causethe string to wind around the axle. This, of necessity, causes theyo-yo to begin to 'climb' back up the string. After the first oneor two rotations, the string can no longer slip, so the process ofclimbing up the string continues beyond the momentary applicationof the tug.
As the yo-yo continues to climb back up the string, the angularmomentum (or kinetic energy of rotation) of the yo-yo is convertedback into gravitational potential corresponding to the increasingheight of the center of mass of the yo-yo. For this reason, theyo-yo's angular momentum and, hence, its rotation rate, steadilydecreases as the yo-yo rises. This is, of course, the reverse ofthe process when the yo-yo was dropped.
If not for frictional losses, the yo-yo would climb all the wayback up the string to your hand just as its rotational ratedecreases to zero. But, due to friction, the yo-yo does not in factquite get back up to your hand before it stops rotating.
Thereafter, the process repeats, with the yo-yo returning short ofits previous height on each cycle. Eventually, the yo-yo comes torest at the bottom.
Of course, as everyone knows, it is possible to keep the yo-yogoing indefinitely by giving it a slight upward pull on each cycle.This pull can be combined with the tug required to initiate theclimb back up the string. The pull serves to give the center ofmass of the yo-yo a little extra kinetic energy to compensate forfrictional losses, so that the yo-yo can be kept goingindefinitely.
Yo-yos can also be thrown horizontally, or launched in otherdirections. The principle of operation is then just the same exceptthat the kinetic energy of the center of mass, which is convertedinto spin as the string unwinds, results from being thrown, ratherthan from falling through a gravitational potential.\"
