Question

A YO-YO is a toy in which a string is wound round a central shaft as shown in the figure. The string unwinds and rewinds itself alternatively making the YO-YO rise and fall. The ratio of the tension in the string during descent and ascent is:

& \text{A) 1:1} \\\ & \text{B) }{{r}_{2}}:{{r}_{1}} \\\ & \text{C) }{{r}_{1}}:{{r}_{2}} \\\ & \text{D) }{{r}_{1}}:1 \\\ \end{aligned}$$

Answer 1

We know that the YO-YO toys are a direct application of the mass-tension relation. We need to understand its working and need to consider any chance of relation with the radii of the shafts used in the toy as per the figure in the problem.

Complete answer:
The YO-YO toy uses a wire wound around its inner circle to descent and ascent. It descends due to the gravitational force due to the mass of the shaft and ascents back to the initial situation due to the tension caused on the wire during its descent.

We can understand that the toy doesn’t require any external force of momentum as the gravitational force by virtue of its mass and height has enough potential energy which converts to kinetic energy during descent and again converts back to potential energy on ascent.
The tension on the string attached to the shaft is offered by the gravitational force which is given as the weight of the shaft. The tension on the string can be given as –
$T=mg$
From the relation, we understand that the radii of the two shafts of the toy are not related to the tension on the string at any instant.
The ratio between the tension during ascent and descent is therefore, $\text{1:1}$.

The correct answer is option A.

Note:
The YO-YO toy’s functioning is greatly dependent on the mass of the shaft used in the toy. The force due to gravitational force increases with mass and the body descents faster, also this increases the tension on the string making a faster ascent.