91Ó°ÊÓ

The total of a system's final energies is equal to the sum of its initial energies plus any external forces' work on the system.

${\mathit{E}}_{\mathbf{i}}\mathbf{+}\mathit{W}\mathbf{=}{\mathit{E}}_{\mathbf{f}}$ (1)

The transitional kinetic energy of an object is

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$ (2)

Whereis mass of object andis speed relative to given coordinate system.

${\mathit{K}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}{\mathit{Ó\neg }}^{\mathbf{2}}$ (3)

The gravitational potential energy of an object system is,

$U_g=mgy$ (4)

If a particle of massexperience a nonzero net force, its acceleration is related to the net force by Newton’s second law:

$∑F=T−mg=ma$ (5)

Here we have given that diameter of flywheel is$D=35.6\text{ c³¾}$

So, radius of wheel is$r=0.178\text{″¾}$

Mass of block is $m=5.60\text{ k²µ}$

The initial angular velocity is${\mathrm{Ó\neg }}_i=0$

The initial translational speed of the block is$v_i=0$

Also, the equation for the graph shown in given figure that gives good fit to data point is$d=\left(165{\text{ c³¾/²õ}}^2\right)t^2$

Here we have, the equation for the graph shown in given figure that gives good fit to the data point is$d=\left(165{\text{ c³¾/²õ}}^2\right)t^2$

We know that acceleration is double derivative of distance .

$$\begin{gathered} ⇒a=\frac{d^2}{d{t}^2}\left(d\right)=\frac{d^2}{d{t}^2}\left[\left(165{\text{ c³¾/²õ}}^2\right)t^2\right] \\ ⇒a=330{\text{ c³¾/²õ}}^2 \end{gathered}$$

Hence, acceleration is constant

Here we have, the equation for the graph shown in given figure that gives good fit to the data point is

$d=\left(165{\text{ c³¾/²õ}}^2\right)t^2$ (6)

We know that speed is derivative of distance .

$$\begin{gathered} ⇒v=\frac{d}{dt}\left(d\right)=\frac{d}{dt}\left[\left(165{\text{ c³¾/²õ}}^2\right)t^2\right] \\ ⇒v=\left(330{\text{ c³¾/²õ}}^2\right)t\left(6\right) \end{gathered}$$

Now,$d=\left(165{\text{ c³¾/²õ}}^2\right)t^2⇒t=\sqrt{\frac{d}{165{\text{ c³¾/²õ}}^2}}$

Here block descended$1.50\text{″¾}=150\text{ c³¾}$

$\begin{array}{l}⇒t=\sqrt{\frac{150\text{ c³¾}}{165{\text{ c³¾/²õ}}^2}}\\ ⇒t=0.954\text{ s}\end{array}$

Now, substitute value of in equation (6)

$\begin{array}{l}⇒v=\left(330{\text{ c³¾/²õ}}^2\right)\left(0.954\text{ s}\right)\\ ⇒v=315\text{ }c\text{m/s}\end{array}$

Hence, speed of the block when it has descended $\text{1.50″¾}$ is$315cm/s$ .

In given system there are no external force is apply. So, equation (1) reduces to

$E_i=E_f$

Now, by mechanical energy equation

$⇒K_i+U_{gi}=K_f+U_{gf}$ (7)

Here total initial kinetic energy$K_i=0$

Also, the final potential energy$U_{gf}=0$

Now, from equation (2), (3), (4), and (7),

$$\begin{gathered} \begin{array}{l}⇒0+mgd=\frac{1}{2}I{\mathrm{Ó\neg }}^2+\frac{1}{2}m{v}^2+0\\ ⇒mgd=\frac{1}{2}I{\mathrm{Ó\neg }}^2+\frac{1}{2}m{v}^2\\ ⇒\frac{1}{2}I{\mathrm{Ó\neg }}^2=mgd−\frac{1}{2}m{v}^2\\ ⇒I=\frac{2mgd−m{v}^2}{{\mathrm{Ó\neg }}^2}\\ ⇒I=\frac{2mgd−m{v}^2}{{\left(\frac{v}{r}\right)}^2}\end{array} \\ \begin{array}{l}⇒I=\frac{2\left(5.60\text{ k²µ}\right)\left(9.8{\text{″¾/s}}^2\right)\left(1.50\text{″¾}\right)−\left(5.60\text{ k²µ}\right){\left(3.15\text{″¾/s}\right)}^2}{{\left(\frac{3.15\text{″¾/s}}{0.178\text{″¾}}\right)}^2}\\ ⇒I=0.348\text{ k²µ}â‹\dots {\text{m}}^2\end{array} \end{gathered}$$

Hence, the moment of inertia of the flywheel is$I=0.348\text{ k²µ}â‹\dots {\text{m}}^2$

Here by equation (5) we have,

$\begin{array}{l}∑F=ma=T−mg\\ ⇒T=m\left(g−a\right)\end{array}$

Substitute the values in above equation,

$\begin{array}{l}⇒T=\left(5.60\text{ k²µ}\right)\left(9.8{\text{″¾/s}}^2−3.30{\text{″¾/s}}^2\right)\\ ⇒T=36.4\text{ N}\end{array}$

Hence, the tension in the line is$T=36.4\text{ N}$