Suppose that the yo-yo in Problem 17, instead of rolling from rest, is thrown so that its initial speed down the string is$\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$.(a) How long does the yo-yo take to reach the end of the string? As it reaches the end of the string, what are its (b) total kinetic energy, (c) linear speed, (d) translational kinetic energy, (e) angular speed, and (f) rotational kinetic energy?

Initial speed down the string is$1.3m/s$

Here,the concept of the period, velocity is used

Formula:

The acceleration of the center of the mass,

$acom=\frac{g}{1+\frac{Icom}{MR{0}^2}}$

where,gis the acceleration due to gravity,$\mathit{I}\mathit{c}\mathit{o}\mathit{m}$is the moment of inertia of center of mass,is the mass of the object,is the distance of the central axis to th$R_0$e surface of the object.

(a)

The acceleration is given by

$acom=\frac{g}{1+\frac{Icom}{MR{0}^2}}$

Where upward is the positive translational direction. Taking the coordinate origin at the initial position.

$$\begin{gathered} y_{com}=v_{com,0}t+\frac{1}{2}{a}_{com}{t}^2 \\ =v_{com,0}t-\frac{\frac{1}{2}g{t}^2}{1+\frac{Icom}{MR{0}^2}} \\ g{t}^2-2t\left(1+\frac{Icom}{MR{0}^2}\right)v_{com,0}+2y_{com}\left(1+\frac{Icom}{MR{0}^2}\right)=0 \end{gathered}$$

The roots of the above quadratic equation can be given as follows:

$\begin{array}{rcl}a& =& g,b=-2\left(1+\frac{Icom}{MR{0}^2}\right)v_{com,0},c=2y_{com}\left(1+\frac{Icom}{MR{0}^2}\right)\\ t& =& \frac{-\left(-2\left(1+\frac{Icom}{MR{0}^2}\right)v_{com,0}\right)\pm \sqrt{{\left(-2\left(1+\frac{Icom}{MR{0}^2}\right)v_{com,0}\right)}^2-4\left(g\right)\left(2y_{com}\left(1+\frac{Icom}{MR{0}^2}\right)\right)}}{2g}\\ & =& \frac{\left(1+\frac{Icom}{MR{0}^2}\right)v_{com,0}\pm \sqrt{{\left(\left(1+\frac{Icom}{MR{0}^2}\right)v_{com,0}\right)}^2-\left(g\right)\left(2y_{com}\left(1+\frac{Icom}{MR{0}^2}\right)\right)}}{g}\\ & =& \frac{\left(1+\frac{Icom}{MR{0}^2}\right)\left(v_{com,0}\pm \sqrt{{\left(v_{com,0}\right)}^2-\frac{2g{y}_{com}}{\left(1+\frac{Icom}{MR{0}^2}\right)}}\right)}{g}\\ & & \end{array}$

Where$ycom=-1.2m$,and$vcom,0=-1.3m/s$

Substituting,$Icom=0.000095kg.m^2,$$M=0.12kg$

$R0=0.0032m,andg=9.8m/s^2$

We use the quadratic formula and we find the following values as:

$\begin{array}{rcl}t& =& \frac{\left(1+\frac{Icom}{MR{0}^2}\right)\left(v_{com,0}\pm \sqrt{{\left(v_{com,0}\right)}^2-\frac{2g{y}_{com}}{\left(1+\frac{Icom}{MR{0}^2}\right)}}\right)}{g}\\ & =& \frac{\left(1+\frac{0.000095}{(0.12){(0.0032)}^2}\right)\left(-1.3\pm \sqrt{{\left(-1.3\right)}^2-\frac{2(9.8)(-1.2)}{\left(1+\frac{0.000095}{(0.12){(0.0032)}^2}\right)}}\right)}{9.8}\\ & =& -21.7 or 0.885\\ & & \end{array}$

We choose t = 0.89 s as the answer

(b)

Initial potential energy is$Ui=mgh$and$h=1.2m$(using the bottom as the reference level for computing.

The initial kinetic energy is

$\left(K=\frac{1}{2}{I}_{com}{\omega }^2+\frac{1}{2}M{v}_{com}^2\right)$. . . Eq.11 -5

Where the initial angular and linear speeds are related by. . . Eq.11-2

Energy conservation leads to

$$\begin{gathered} \begin{array}{rcl}Kf& =& Kf+Uf\\ & =& \frac{1}{2}m{v}^{2}com,0 +\frac{1}{2}I{\left(\frac{vcom,0 }{R0}\right)}^2+Mgh\\ & =& \left\{\left[\frac{1}{2}(0.12kg)+{(1.3m/s)}^2\right] +\left[\frac{1}{2}(9.5\times {10}^{-5}kg.m^2){\left(\frac{1.3m/s }{0.0032 m}\right)}^2\right]+ \\ \left[(0.12kg)(9.8m/s^2)(1.2 m)\right] \\ \right\}\\ & =& 9.4J\\ & & \end{array} \end{gathered}$$

Thetotal kinetic energy is.$9.4J$

(c)

As it reaches the end ofthe string, its center ofmass velocity is given by Eq.11-2

$\begin{array}{rcl}vcom& =& vcom,0+acomt\\ & =& vcom,0-\frac{gt}{1+\frac{Icom}{MR{0}^2}}\\ & =& -1.3m/s-\frac{(9.8m/s^2)(0.885 s)}{1+\frac{0.000095kg.m^2}{(0.12kg){(0.0032m)}^2}}\\ & =& -1.41m/s\\ & & \end{array}$

Thus, we obtain

So, its linear speed at that moment is approximately 1.4 m

(d)

The translational kinetic energy is

$\begin{array}{rcl}\frac{1}{2}{v}^{2}com& =& \frac{1}{2}(0.12kg)(-1.41m/s{)}^2\\ & =& 0.12J\\ & & \end{array}$

The translation kinetic energy is.$0.12J$

(e)

The angular velocity at that moment is given by

$\begin{array}{rcl}\omega & =& \frac{v^{2}com}{R0}\\ & =& \frac{-1.41m/s}{0.0032 m}\\ & =& 441rad/s\\ & \approx & 4.4\times {10}^{2}rad/s\\ & & \end{array}$

The angular speed is$4.4\times {10}^{2}rad/s$

(f)

The rotational kinetic energy is

$\begin{array}{rcl}\frac{1}{2}Icom{\omega }^2& =& \frac{1}{2}(9.50\times {10}^{-5} kg.m^2)(441rad/s{)}^2\\ & =& 9.2J\\ & & \end{array}$

Therotational kinetic energy is$9.2J$