you are rebuilding a 1965 Chevrolet. To decide whether to replace the flywheel with a newer, lighter-weight one, you want to determine the moment of inertia of the original,35.6 cmdiameter flywheel. It is not a uniform disk, so you can’t use$\mathit{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{M}{\mathit{R}}^{\mathbf{2}}$to calculate the moment of inertia. You remove the flywheel from the car and use low-friction bearings to mount it on a horizontal, stationary rod that passes through the center of the flywheel, which can then rotate freely (aboutabove the ground). After gluing one end of a long piece of flexible fishing line to the rim of the flywheel, you wrap the line a number of turns around the rim and suspend a$\mathbf{5}\mathbf{.60}\mathbf{\text{\hspace{0.17em}m}}$metal block from the free end of the line. When you release the block from rest, it descends as the flywheel rotates. With high-speed photography, you measure the distance d the block has moved downward as a function of the time since it was released. The equation for the graph shown in the given figure that gives a good fit to the data point is$\mathit{d}\mathbf{=}\mathbf{\left(}\mathbf{165}{\mathbf{\text{\hspace{0.17em}cm/s}}}^{\mathbf{2}}\mathbf{\right)}{\mathit{t}}^{\mathbf{2}}$.

(a) Based on the graph, does the block fall with constant acceleration? Explain

(b) Use the graph to calculate the speed of the block when it has descended 1.50m

(c) Apply conservation of mechanical energy to the system of flywheel and block to calculate the moment of inertia of the flywheel

(d) You are relieved that the fishing line doesn’t break. Apply Newton’s second law to the block to find the tension in the line as the block descended.

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.

${\mathit{K}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}{\mathit{\omega }}^{\mathbf{2}}$ (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:

$\sum F=T-mg=ma$ (5)

Here we have given that diameter of flywheel is$D=35.6\text{\hspace{0.17em}cm}$

So, radius of wheel is$r=0.178\text{\hspace{0.17em}m}$

Mass of block is $m=5.60\text{\hspace{0.17em}kg}$

The initial angular velocity is${\omega }_i=0$

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$d=\left(165{\text{\hspace{0.17em}cm/s}}^2\right)t^2$

Here we have, the equation for the graph shown in given figure that gives good fit to the data point is$d=\left(165{\text{\hspace{0.17em}cm/s}}^2\right)t^2$

We know that acceleration is double derivative of distance .

$$\begin{gathered} \Rightarrow a=\frac{d^2}{d{t}^2}\left(d\right)=\frac{d^2}{d{t}^2}\left[\left(165{\text{\hspace{0.17em}cm/s}}^2\right)t^2\right] \\ \Rightarrow a=330{\text{\hspace{0.17em}cm/s}}^2 \end{gathered}$$

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

$d=\left(165{\text{\hspace{0.17em}cm/s}}^2\right)t^2$ (6)

We know that speed is derivative of distance .

$$\begin{gathered} \Rightarrow v=\frac{d}{dt}\left(d\right)=\frac{d}{dt}\left[\left(165{\text{\hspace{0.17em}cm/s}}^2\right)t^2\right] \\ \Rightarrow v=\left(330{\text{\hspace{0.17em}cm/s}}^2\right)t\left(6\right) \end{gathered}$$

Now,$d=\left(165{\text{\hspace{0.17em}cm/s}}^2\right)t^2\Rightarrow t=\sqrt{\frac{d}{165{\text{\hspace{0.17em}cm/s}}^2}}$

Here block descended$1.50\text{\hspace{0.17em}m}=150\text{\hspace{0.17em}cm}$

$\begin{array}{l}\Rightarrow t=\sqrt{\frac{150\text{\hspace{0.17em}cm}}{165{\text{\hspace{0.17em}cm/s}}^2}}\\ \Rightarrow t=0.954\text{\hspace{0.17em}s}\end{array}$

Now, substitute value of in equation (6)

$\begin{array}{l}\Rightarrow v=\left(330{\text{\hspace{0.17em}cm/s}}^2\right)\left(0.954\text{\hspace{0.17em}s}\right)\\ \Rightarrow v=315c\text{m/s}\end{array}$

Hence, speed of the block when it has descended $\text{1.50\hspace{0.17em}m}$ 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

$\Rightarrow K_i+U_{gi}=K_f+U_{gf}$ (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),

$$\begin{gathered} \begin{array}{l}\Rightarrow 0+mgd=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2+0\\ \Rightarrow mgd=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2\\ \Rightarrow \frac{1}{2}I{\omega }^2=mgd-\frac{1}{2}m{v}^2\\ \Rightarrow I=\frac{2mgd-m{v}^2}{{\omega }^2}\\ \Rightarrow I=\frac{2mgd-m{v}^2}{{\left(\frac{v}{r}\right)}^2}\end{array} \\ \begin{array}{l}\Rightarrow I=\frac{2\left(5.60\text{\hspace{0.17em}kg}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^2\right)\left(1.50\text{\hspace{0.17em}m}\right)-\left(5.60\text{\hspace{0.17em}kg}\right){\left(3.15\text{\hspace{0.17em}m/s}\right)}^2}{{\left(\frac{3.15\text{\hspace{0.17em}m/s}}{0.178\text{\hspace{0.17em}m}}\right)}^2}\\ \Rightarrow I=0.348\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\end{array} \end{gathered}$$

Hence, the moment of inertia of the flywheel is$I=0.348\text{\hspace{0.17em}kg}\cdot {\text{m}}^2$

Here by equation (5) we have,

$\begin{array}{l}\sum F=ma=T-mg\\ \Rightarrow T=m\left(g-a\right)\end{array}$

Substitute the values in above equation,

$\begin{array}{l}\Rightarrow T=\left(5.60\text{\hspace{0.17em}kg}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^2-3.30{\text{\hspace{0.17em}m/s}}^2\right)\\ \Rightarrow T=36.4\text{\hspace{0.17em}N}\end{array}$

Hence, the tension in the line is$T=36.4\text{\hspace{0.17em}N}$