91Ó°ÊÓ

Mass of yo-yo is 116kg,

Radius of disk is 32cm

Radius of axel is 3.2cm.

This problem involves the concept of different applied forces on the object that is a yo-yo while it’s rolling takes place after being released from a crane. Using Newton's laws subjected to the matter that the object is in rolling motion, we check the applied forces on it. Here, the normal force due to the tension on the string and the gravitational force act along the object's body as it is released from the rest. Thus, using the free body diagram concept, the equation of the forces is found that is used for calculating the acceleration and the tension values.

Formulae:

The acceleration of the center of mass, $a_{com}=-\frac{g}{1+\left({\displaystyle \frac{l_{com}}{M{R}_0^2}}\right)}$

where, g is the acceleration due to gravity, l_com is the moment of inertia of the center of mass of the object, R₀ is the radius of the object, M is the mass of the object.

Referring to figure,

$f_s-Mg\sin \mathrm{θ}=Ma$

Consider torque caused by$F_N,F_g\sin \mathrm{θ}and{f}_g$about an axis passing through the center of mass of the sphere.

We have, torque caused by $F_{N}and{F}_g\sin \mathrm{θ}$as zero because these forces are acting at the center of mass of the sphere and so the position vector would be zero.

Therefore,

$\begin{array}{rcl}{R}_0{f}_s& =& l_{com}\mathrm{Î\pm }\\ F_s& =& \frac{l_{com}\mathrm{Î\pm }}{R_0}\\ \mathrm{Î\pm }& =& \frac{a}{R_0}\end{array}$

Negative sign arises because of the fact that the acceleration is along the negative x-axis,

$f_s=-\frac{l_{com}a}{R_0^2}$

Therefore, we have,

$\begin{array}{rcl}-\frac{l_{com}a}{R_0^2}-Mg\sin \mathrm{θ}& =& Ma\\ -Mg\sin \mathrm{θ}& =& Ma+\frac{l_{com}a}{R_0^2}-Mg\sin \mathrm{θ}=a\left(\frac{M{R}^2+l_{com}}{R_0^2}\right)\\ a& =& -\frac{Mg\sin \mathrm{θ}}{\left(\frac{M{R}^2+l_{com}}{R_0^2}\right)}\\ a& =& -\frac{Mg\sin \mathrm{θ}}{M\left(1+\frac{l_{com}}{M{R}_0^2}\right)}\\ & =& -\frac{g\sin \mathrm{θ}}{\left(1+\frac{l_{com}}{M{R}_0^2}\right)}\end{array}$

Using$\mathrm{θ}={90}^o$, we get

$a_{com}=-\frac{g}{1+\left({\displaystyle \frac{l_{com}}{M{R}_0^2}}\right)}$

Use $l_{com}=\frac{M{R}^2}{2},R=0.32m,R_o=0.032m,M=116kg$

So,

$\begin{array}{rcl}{a}_{com}& =& -\frac{g}{1+\left({\displaystyle \frac{\left({\displaystyle \frac{M{R}^2}{2}}\right)}{M{R}_0^2}}\right)}\\ & =& -\frac{g}{1+{\displaystyle \frac{{\left({\displaystyle \frac{R}{R_0}}\right)}^2}{2}}}\\ & =& -\frac{9.8m/s^2}{1+{\displaystyle \frac{\left({\displaystyle \frac{0.32m}{0.032m}}\right)}{2}}}\\ & =& -0.19m/s^2\end{array}$

Therefore, magnitude is 0.19m/s²

Acceleration during its rise is same as during fall that is a_com = 0.19m/s²

If we apply Newton’s second law to above disgram we get

$\begin{array}{rcl}T-Mg& =& Ma\\ T& =& Ma+Mg\\ & =& M\left(a+g\right)\\ & =& 116kg\left(-0.19m/s^2+9.8m/s^2\right)\\ & =& 1.1×{10}^{3}N\end{array}$

This is not near to limit of 52kN

From the above derived acceleration equation, it can be said that the acceleration depends on ratio of radius of disk and axel i.e. $\left(\frac{R}{R_0}\right)$.So even if we have scaled up version, this ratio would remain same and that would result in the same acceleration,$a_{com}=0.19m/s^2$

Tension depends on mass of object, so larger the mass larger the tension.

With same materials in larger shape, our yo-yo will have larger mass.

Thus, the magnitude of the tension as the yo-yo falls will be greater than that of the San Francisco yo-yo.