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When a yo-yo that has a string wrapped tightly around its axle reaches its full extension, it automatically bounces back upwards, the string re-winding in the process. What causes it to do that? I haven't been able to find a satisfactory explanation anywhere.

Jay Gee
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2 Answers2

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The string is terminated by a small loop around the axle. The loop is just a bit larger than the diameter of the axle so that, when the yoyo is fully unwound and spinning rapidly, the yoyo can continue spinning with its axle sliding on the lower part of the small loop. The tension due to the weight of the yoyo stops the string being swept around the axle. To make the yoyo come back you must slighly drop you hand so the tension in the string falls to zero and the spining yoyo drags the string round the axles, catching it, and starts winding it up --- so trading rotational KE for gravitational PE. There is an art to this manoeuvre! It takes some practice.

mike stone
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Consider the system as the yo-yo and a short portion of the string attached to the yo-yo. For this system, the external forces are: gravity and the tension on the string. Friction between the yo-yo and the string is an internal force for this system. The motion of the Center of Mass (CM) of the yo-yo is determined by the net external force. (The tension is dependent on the friction at the yo-yo string interface, for example whether or not there is pure rolling or slipping.)

When the yo-yo is moving downwards the force of gravity exceeds the tension so the CM moves downward. Without friction the yo-yo would slide down the string, but friction opposes the sliding and causes the yo-yo to rotate counter-clockwise about the CM; that is, the yo-yo rotates down the string.

For the yo-yo to "bounce back" upwards the tension must exceed the force of gravity for the CM to move upwards. This increase in tension is caused by the user flicking the string upwards. Without friction the yo-yo would slide up the string, but friction opposes the sliding and causes the yo-yo to rotate clockwise about the CM; that is, the yo-yo rotates up the string.

A similar situation is a fixed pully with a rope, with a mass hanging vertically from one end of the rope and a person pulling vertically downwards at the other end of the rope. If the person does not pull hard the tension in the rope is less than the force of gravity on the mass and the mass moves downward. If the person pulls hard the tension in the rope is greater than the force of gravity on the mass and the mass moves upward. Due to friction between the rope and the pully the pully rotates in one direction when the mass moves downward and rotates in the opposite direction when the mass moves upward. In basics physics this problem is evaluated assuming no slip at the rope pully interface. For a detailed discussion of tension and friction see
https://www.researchgate.net/publication/318107848 Force and torque of a string on a pulley.

John Darby
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