MadSci Network: Physics
Query:

Re: Physics of Yo Yos

Date: Wed Oct 7 12:15:19 1998
Posted By: Tom Cull, Research Associate, Washington U Med. School
Area of science: Physics
ID: 906651258.Ph
Message:

Yo-yo's simple are very cool, analyzing their behavior in anything but the simplest set-up is very difficult. Basically, a yo-yo transfers potential energy into kinetic energy as it falls (or really is thrown) down the string. The string provides a constraint that converts much of the translational kinetic energy into rotational kinetic energy.

Older yo-yo's or some war yo-yo's (yes, the legend is that yo-yo's started out as weapons) have the string fixed through the middle perpendicular to the axis of rotation. This causes the yo-yo to snap back when it reaches the end of the string because it starts to rewind.

Modern yo-yo's are connected to the string by a loop instead of by a fixed point. With a loop, the yo-yo can "sleep" at the end of string until acted upon by the user. If the friction is low and the moment of inertia of the yo-yo is high, the yo-yo can "sleep" for quite some time.

The yo-yo spins the same direction throughout the trip but the string alternates its winding orientation (clockwise then counter-clockwise then clockwise) when viewed from one side. The user can let the yo-yo fall or give it some initial speed by tossing it while holding the string. Care should be taken to make sure that all rotation is along the axis through the center of the yo-yo (i.e. perpendicular to the string). Any other rotation can cause the yo-yo to twirl of axis and sooner or later it will catch the string and get tangled.

Let's look at the simplest case of solving the yo-yo equation for the case of just vertical motion and no initial speed provided by the user. A diagram of a yo-yo and the selected axis and directions of positive translation and rotation are shown in Figure 1.

Newton's 2nd law in its force Eqn. 1 and torque Eqn. 2 form are all we need to solve for the equation of motion. The torq ue is calculated about the axis of rotation through the center of mass of the yo-yo.

Where Icm is the moment of inertia of the yo-yo about the axis of rotation through its center of mass, < i>g is the acceleration of gravity,T in the tension in the string, and r0 is the radiu s of the inner core of the yo-yo.

Eqn. 3 relates angular acceleration to translational (linear) acceleration and when substituted into Eqn. 2 results in a s ystem of two equations and two unknowns (a and T).

The solutions of (a and T are shown in Eqn. 4 and Eqn. 5, respectively.

The reader should note that the acceleration of the yo-yo is smaller than it would be if it were free-falling toward the ground as might be expected since the kinetic energy is being divided between rotational and translation modes. Also, of interest is that the tension required in the string is less than the weight of the yo-yo until the yo-yo gets to the end of the string.

The moment of inertia Icm can be measured in terms of mr0 -- typically Icm/mr 02 will much greater than 1. This means the acceleration will be very low and a lot of the kinetic energy will be in the form of rotational energy -- just what one wants for cool "sleeper" tricks.

Of course, the motion of a real yo-yo is more complicated that this simplified version -- but including anymore complication is will clo ud the fun. For example, returning the yo-yo to the hand from the hand of the string requires a quick jerk on the string which causes t he yo-yo to "leap" back onto the string. Solving for how this happens is not easy nor is it perfectly predictable -- if it were, we wou ld all be "Kings of Yo."

Sincerely, Tom "Rocking the Baby" Cull


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