How do yo-yos work? Why do they come back up after being thrown down?

Asked by: Anita Kolat


When you release a yo-yo, gravity acts on its center of mass to pull the yo-yo downward. Because the string of the yo-yo is wrapped around the yo-yo's axle, and because one end of the string is attached 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 of drop increases with time and so, necessarily, does the rotation rate of the yo-yo. The rate of drop and the rotation rate are greatest when the bottom is reached and the string is completely unwound. The spinning yo-yo contains angular momentum (or rotational kinetic energy) derived from the gravitataion potential energy through which the yo-yo has dropped.

Usually, the string is tied loosely around the axle so that the yo-yo can continue to spin at the bottom. Because the full length of the string has been paid out, the yo-yo can drop no further and, consequently, the rotation rate cannot increase further. If left in this condition, the friction between the axle and the string will eventually dissipate the energy of rotation or, equivalently, the angular momentum of the yo-yo and the yo-yo will come to rest.

However, a momentary tug on the string causes the friction between the string and the axle briefly to increase so that the axle no longer slips within the string. When the axle thus stops slipping, the angular momentum of the spinning yo-yo is sufficient to cause the string to wind around the axle. This, of necessity, causes the yo-yo to begin to 'climb' back up the string. After the first one or two rotations, the string can no longer slip, so the process of climbing up the string continues beyond the momentary application of the tug.

As the yo-yo continues to climb back up the string, the angular momentum (or kinetic energy of rotation) of the yo-yo is converted back into gravitational potential corresponding to the increasing height of the center of mass of the yo-yo. For this reason, the yo-yo's angular momentum and, hence, its rotation rate, steadily decreases as the yo-yo rises. This is, of course, the reverse of the process when the yo-yo was dropped.

If not for frictional losses, the yo-yo would climb all the way back up the string to your hand just as its rotational rate decreases to zero. But, due to friction, the yo-yo does not in fact quite get back up to your hand before it stops rotating.

Thereafter, the process repeats, with the yo-yo returning short of its previous height on each cycle. Eventually, the yo-yo comes to rest at the bottom.

Of course, as everyone knows, it is possible to keep the yo-yo going indefinitely by giving it a slight upward pull on each cycle. This pull can be combined with the tug required to initiate the climb back up the string. The pull serves to give the center of mass of the yo-yo a little extra kinetic energy to compensate for frictional losses, so that the yo-yo can be kept going indefinitely.

Yo-yos can also be thrown horizontally, or launched in other directions. The principle of operation is then just the same except that the kinetic energy of the center of mass, which is converted into spin as the string unwinds, results from being thrown, rather than from falling through a gravitational potential.

Answered by: Warren Davis, Ph.D., President, Davis Associates, Inc., Newton, MA USA



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