91影视

Given in the question,

Mass of the cylinder $M=10\text{鈥塳驳}$

The diameter of the cylinder, $d=\text{30鈥夆赌塩尘}$

Therefore, the radius of the cylinder,$R=\text{15鈥夆赌塩尘}$

The mass of the block,$m=\text{12Kg}$

The kinetic energy of the cylinder, $K_c=480\text{鈥夆赌塉}$

The height descended by the mass, $h=?$

Law of conservation of energy

According to the conservation of energy, the energy of a systemcan neither be created nor destroyed it can only be converted from one form of energy to another.

$\begin{array}{c}{E}_F=E_I\\ U_i+K{E}_i=U_f+K{E}_f\end{array}$

Where,

$U_i$ is initial potential energy,$U_f$is final potential energy, $K{E}_i$is initial kinetic energy, $K{E}_f$is final kinetic energy.

From the conservation of energy,

$U_i+K{E}_i=U_f+K{E}_f$鈥�(颈)

Potential energy can be given as

$U=mgh$

Where Uis potential energy, m is mass and h is the height.

The mass m is initially at height h, therefore initial potential energy of the system can be given as

$\begin{array}{l}{U}_i=mgh\\ U_i=\left(12\text{鈥夆赌塳驳}\right)\left(9.8\text{鈥夆赌尘}/{\text{s}}^{\text{2}}\right)\left(h\right)\end{array}$

Since finally the height of the mass is zero

Final potential energy

$U_f=0$

The final kinetic energy will be the sum of the translational kinetic energy of the mass and the rotational kinetic energy of the cylinder.

$K_t=\frac{1}{2}m{v}^2$

The formula of transitional kinetic energy is

Where m is mass and v is the linear speed

The formula of rotational kinetic energy

$k_r=\frac{1}{2}I{蝇}^2$

Where I is inertia and $蝇$is the angular velocity

Initially, the system was at rest therefore initial kinetic energy.

$K{E}_i=0$

Final kinetic energy

$\begin{array}{c}K{E}_f=K_t+K_r\\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{蝇}^2\end{array}$

Substituting the values into equation (i)

$$\begin{gathered} mgh+0=0+\frac{1}{2}m{v}^2+\frac{1}{2}I{蝇}^2 \\ mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{蝇}^2\left(ii\right) \end{gathered}$$

Since the cylinder is rotating therefore

The kinetic energy of the cylinder is equal to its rotational kinetic energy

$\begin{array}{l}{K}_r=\frac{1}{2}I{蝇}^2\\ 480\text{鈥夆赌塉=}\frac{1}{2}I{蝇}^2\end{array}$

Moment of inertia of the cylinder

$I=\frac{1}{2}M{R}^2$

Where M is mass and R is the radius of the cylinder

Hence

$480\text{鈥夆赌塉=}\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){蝇}^2$

We know, the mass and cylinder are connected so they will move with the same linear speed

Therefore, we can use the formula

$v=R蝇$

Substituting the values

$\begin{array}{c}480\text{鈥夆赌塉=}\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\left(\frac{v}{R}\right)}^2\\ 480\text{鈥夆赌塉=}\frac{1}{4}M{v}^2\\ 480=\frac{1}{4}\left(12\text{鈥夆赌塳驳}\right)v^2\\ v=12.64\text{鈥夆赌�}m/s\end{array}$

Hence the linear speed of the block $v=12.64\text{鈥夆赌�}m/s$.

Now to find the height using the equation (ii)

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{蝇}^2$

Substituting all the values

$\begin{array}{c}\left(12\text{鈥夆赌塳驳}\right)\left(9.8\text{鈥夆赌�}m/s^2\right)h=\frac{1}{2}\left(12\text{鈥夆赌塳驳}\right){\left(12.64\text{鈥夆赌�}m/s\right)}^2+480\text{鈥夆赌塉}\\ h=\frac{\frac{1}{2}\left(12\text{鈥夆赌塳驳}\right){\left(12.64\text{鈥夆赌�}m/s\right)}^2+480\text{鈥夆赌塉}}{\left(12\text{鈥夆赌塳驳}\right)\left(9.8\text{鈥夆赌�}m/s^2\right)}\\ h=12.23\text{鈥尘}\end{array}$

Hence the height descended by the mass is 12.23m