Wind the strings around the axle ends (in one layer) until the wheel reaches the support rod, then allow the wheel to fall. Because the strings force the wheel to rotate as it falls, it accelerates downwards at a lower rate than it would if it were in free fall. Because the radius of the axle is much less than that of the wheel, the moment the string has to rotate the wheel is small compared to that which the wheel possesses by virtue of its moment of inertia. Its rotational acceleration is thus quite slow, and it takes the wheel several seconds to fall (just over 5 seconds).

As is true for 28.27 – Roll cylinders down incline, we can analyze this system in two ways. We can examine its energetics or its dynamics. When the yo-yo is at the top of its travel, it has potential energy of MgL, where L is the length of the string that has unwound from the spindle when the yo-yo reaches the bottom. When we let it fall from the top, as noted above, the wheel falls downwards in a straight line, but it also rotates. Thus the potential energy is converted into both linear and rotational kinetic energy thus: MgL = (1/2)Mv² + (1/2)Iω², where v and ω are the wheel’s final linear velocity and angular velocity, respectively. If we call the radius of the axle r, then ω = v/r, and this becomes MgL = (1/2)Mv² + (1/2)I(v²/r²), which we can rearrange to gl = (v²/2)(M + I/r²), or 2gL = v²(1 + I/Mr²), which gives v² = 2gL/(1 + I/Mr²), and v = √[(2gL)/(1 + I/Mr²)], which we can also write as v = √[2gLr²/(r² + I/M)]. Since L = (1/2)(v_i + v_f)t, and the wheel started at rest, we have L = (1/2)vt, and the time it takes the wheel to fall is t = 2L/v, or t = 2L√[(1 + I/Mr²)/2gL], which gives t = √[2L(1 + I/Mr²)/g], which we can also write as t = √[2L(r² + I/M)/gr²]. We see that the smaller the yo-yo’s moment of inertia is or the greater r is , the greater the yo-yo’s final velocity (both linear and angular), and the shorter the time it takes the yo-yo to fall. If its moment of inertia were zero or r were very large, the above would give v = √(2gL) and t = √(2L/g), which is what v and t would equal if the yo-yo were in free fall. We also see that the greater the moment of inertia or the smaller r, the smaller the velocity (both linear and angular), and the greater the time for the wheel to fall.

To analyze the dynamics of the system, we note that gravity pulls downwards on the wheel with Mg, the tension in the strings (which we will handle as if they are one; the tension is equally split between them), T, pulls upwards on the wheel, and since the wheel accelerates downwards, T is less than Mg, or Mg - T = Ma, or T = Mg - Ma. The tension, applied at the axle radius, r, exerts a torque on the wheel, thus: Tr = Iα. So T = Iα/r. Since a = rα, T = Ia/r². Combining the two equations for T and solving gives a(M + I/r²) = Mg, or a = g/(1 + I/Mr²), or a = gr²/(r² + I/M). From the equations of linear motion, we know that after the yo-yo has fallen through the distance L, the square of its (linear) speed, v², equals 2aL, or v = √(2aL). This gives v = √[2gLr²/(r² + I/M)], which agrees with what we arrived at above, as it should. We can use either L = (1/2)at² or t = 2L/v to find, again, t = √[2L(r² + I/M)/gr²].

We note, as for the cylinders rolling down the incline, that the mass term in the moment of inertia cancels with the mass term in the expressions above, and the acceleration, velocity and time to fall a particular distance are thus independent of the mass of the yo-yo, but depend only on how that mass is distributed. We also see that in order to find the tension in the strings, we must use the dynamical analysis, which gives us, if we substitute what we found for a into either equation above for T, T = Ig/(r² + I/M). From this, we can see that the smaller the moment of inertia or the greater r, the lower the tension in the strings, and the greater the moment of inertia or the smaller r, the greater the tension. We can also write this as T = IMg/(Mr² + I), which may make it easier to see that when I >> M or r² is very small, or both, T ≈ Mg. As noted above, whatever the tension is, each string provides half of it.

In addition, we can determine the work done by the tension in causing the wheel to rotate, and verify that it equals the rotational energy that the wheel accumulates during its fall. This energy, as noted above, is (1/2)I(v²/r²), or gLI/(r² + I/M), but T = Ig/(r² + I/M), so this equals TL, the work done by the tension during the wheel’s fall through the distance L. The linear kinetic energy, (1/2)Mv², equals MgL/(1 + I/Mr²), or MgLr²/(r² + I/M). We see from these that for very large I, very small r² or both, TL approaches MgL, and the linear kinetic energy becomes very small; most of the potential energy goes into turning the wheel, and very little into accelerating it downwards. On the other hand, for very small I (and/or large r²), the rotational kinetic energy becomes very small (it goes to gLI/r²), and the linear kinetic energy approaches MgL; most of the energy goes into accelerating the wheel downwards, and not so much into turning it.

Though our yo-yo has a slightly thickened rim area, the rim thickness is not greatly different from that of the inner part of the wheel, and the hub is quite thick. Given this wheel shape, and that the wheel is mounted on a long axle, it is not unreasonable to treat the yo-yo as if it were a uniform disc. In this case, its moment of inertia is (1/2)MR², and the expressions above for v and t become v = √[2gLr²/(r² + (1/2)MR²/M)], or v = √[2gLr²/(r² + R²/2)], which we can also write as v = 2r√[gL/(2r² + R²)], and t = √[2L(r² + (1/2)MR²/M)/gr²], or t = √[2L(r² + R²/2)/gr²], which we can also write as t = (1/r)√[L(2r² + R²)/g]. For our yo-yo, the outer radius is 60.5 mm, and the axle radius is 2.4 mm. (Here we neglect the thickness of the string, which can be wound up in one layer along the axle, so as not to change r too much.) So we have v = 2(2.4 × 10^-3 m)√[(9.8 m/s²)(0.51 m)/(2(2.4 × 10^-3 m)² + (6.05 × 10^-2 m)²)], which gives v = 0.18 m/s, and t = (1/2.4 × 10^-3 m)√[0.51 m(2(2.4 × 10^-3 m)² + (6.05 × 10^-2 m)²)/9.8 m/s²], which gives t = 5.8 s. A rough measurement of how long it takes the yo-yo to fall gives about 5.2 s. These are fairly close, considering that the thickness of the string may add noticeably to the moment it has in rotating the wheel, and that we don’t really know the mass distribution in the wheel. As noted above, in free fall, the yo-yo should take t = √(2(0.51 m)/(9.8 m/s²), or 0.32 s to fall the same distance (and its final velocity would be v = √(2(9.8 m/s²)(0.51 m), or 3.2 m/s).

References:

1) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 237, 241, 243-250.