Question

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?

Short answer

a)Time required to reach at the end of the string is$0.89sec$

b) Thetotal kinetic energy is.$9.4J$

c)Thelinearspeed is.$1.41m/s$

d)Thetranslation kinetic energyis.$0.12J$

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

f) Therotational kinetic energy is$9.2J$

Step-by-step solution

Listing the given quantities

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

Understanding the concept of motion

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.

 Calculation of theperiod

(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

Calculation of thetotal kinetic energy

(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.$9.4J$

Calculation of thelinear speed

(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

Calculation of thetranslation kinetic energy

(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.$0.12J$

Calculation of theangular velocity at that moment

(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$4.4\times {10}^{2}rad/s$

 Calculation of therotational kinetic energy

(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$9.2J$