Question

In 1980, over San Francisco Bay, a large yo-yo was released from a crane. The 116kg yo-yo consisted of two uniform disks of radius 32cmconnected by an axle of radius 3.2cm(a) What was the magnitude of the acceleration of the yo-yo during its fall ? (b) What was the magnitude of the acceleration of the yo-yo during its rise? (c) What was the tension in the cord on which it rolled? (d) Was that tension near the cord’s limit of 52kN? Suppose you build a scaled-up version of the yo-yo (same shape and materials but larger). (e) Will the magnitude of your yo-yo’s acceleration as it falls be greater than, less than, or the same as that of the San Francisco yo-yo? (f) How about the tension in the cord?

Short answer

1. Acceleration during its fall is 0.19m/s²
2. Acceleration during its rise is 0.19m/s²
3. Tension in cord $1.1\times {10}^{3}N$
4. The tension is not near limit of 52kN
5. Magnitude of acceleration for scaled-up version of yo-yo is 0.19m/s²
6. Tension will be same.

Step-by-step solution

Identification of given data

Mass of yo-yo is 116kg,

Radius of disk is 32cm

Radius of axel is 3.2cm.

To understand the concept

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.

(a) Determining the magnitude of acceleration during fall and rise.

Referring to figure,

$f_s-Mg\sin \theta =Ma$

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

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

Therefore,

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

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,

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

Using$\theta ={90}^o$, 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

(b) Determining the magnitude of the acceleration of the yo-yo during its rise

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

(c) Determining the tension in cord

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

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

(d) Find out if the tension is near the cord’s limit of 25kN

This is not near to limit of 52kN

(e) Find out if the magnitude of your yo-yo’s acceleration as it falls will be greater than, less than, or the same as that of the San Francisco yo-yo

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,$a_{com}=0.19m/s^2$

(f) Find out if the magnitude of the tension as the yo-yo falls will be greater than, less than, or the same as that of the San Francisco yo-yo.

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.