91Ó°ÊÓ

We have the given data:

The radius of the disk =30.0 cm.

Mass of the disk =10.0 Kg.

Distance traveled by the stone (height) =12.6 m.

At rest, the velocity , time

We have the moment of inertia of the uniform disk about an axis through its center is,

$I=\frac{2}{5}M{R}^2$

Then the torque is given by,

$\begin{array}{r}∑{\mathrm{Ï„}}_0=I\mathrm{Î\pm }\\ ⇒N\left(0\right)+Mg\left(0\right)+TR=\left(\frac{1}{2}M{R}^2\right)\mathrm{Î\pm }\\ ⇒T=\frac{MR\mathrm{Î\pm }}{2}⋯⋯\left(1\right)\end{array}$

Where, is the tension in the wire.

is the reaction force from the bearing on the pulley.

is the weight of the pulley acting on its center of mass.

The acceleration of the stone is equal to the tangential acceleration of any point in the rim of the disk.

Thus, $a_y=R\mathrm{Î\pm }⇒\mathrm{Î\pm }=\frac{a_y}{R}$

Putting this value in , we get,

$\begin{array}{r}T=\frac{1}{2}MR\left(\frac{a_y}{R}\right)\\ T=\frac{M{a}_y}{2}⋯⋯\left(2\right)\end{array}$

Now,

$\begin{array}{r}∑F_y=m{a}_y\\ ⇒mg−T=m{a}_y\\ ⇒T=m\left(g−a_y\right)\\ ⇒m=\frac{T}{\left(g−a_y\right)}\end{array}$

Here, is the tension in the wire

is the weight of the stone.

Now, let us find the value of $a_y$.

Stone has the constant acceleration downward.

From the given question we have the vertical displacement at the initial position and positive direction downward.

Therefore,

$\begin{array}{r}y−y_1=v_{1}t+\frac{1}{2}{a}_y{t}^2\\ ⇒12.6−0=\left(0\right)â‹\dots \left(3.00\right)+\frac{1}{2}{a}_y(3.00{)}^2\\ ⇒a_y=2.80{\mathrm{ms}}^2\end{array}$

Using this value in , we get,

$$\begin{gathered} T=\frac{1}{2}\left(10.0\right)\left(2.80\right) \\ T=14.0N \end{gathered}$$

Now, substituting above value in , we get,

$$\begin{gathered} \mathrm{m}=\frac{14}{9.80-2.80} \\ \mathrm{m}=2.00\mathrm{Kg} \end{gathered}$$

Hence, the mass of the stone is 2.00 Kg.

Stone has the constant acceleration downward.

From the given question we have the vertical displacement at the initial position and positive direction downward.

Therefore,

$\begin{array}{r}y−y_1=v_{1}t+\frac{1}{2}{a}_y{t}^2\\ ⇒12.6−0=\left(0\right)â‹\dots \left(3.00\right)+\frac{1}{2}{a}_y(3.00{)}^2\\ ⇒a_y=2.80{\mathrm{ms}}^2\end{array}$

Using this value in , we get,

$\begin{array}{r}T=\frac{1}{2}\left(10.0\right)\left(2.80\right)\\ T=14.0\mathrm{N}\end{array}$

Hence, the tension in the wire is, 14.0 N.