91Ó°ÊÓ

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:

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

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:

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

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

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

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

Energy conservation leads to

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

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

$\begin{array}{rcl}\mathrm{Ó\neg }& =& \frac{v^{2}com}{R0}\\ & =& \frac{-1.41m/s}{0.0032 m}\\ & =& 441rad/s\\ & ≈& 4.4×{10}^{2}rad/s\\ & & \end{array}$

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

(f)

The rotational kinetic energy is

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

Therotational kinetic energy is$9.2J$