In Fig. P9.80, the cylinder and pulley turn without frictionabout stationary horizontal axles that pass through their centers.A light rope is wrapped around the cylinder, passes over thepulley, and has a 3.00-kg box suspended from its free end. Thereis no slipping between the rope and the pulley surface. The uniformcylinder has mass 5.00 kg and radius 40.0 cm. The pulley isa uniform disk with mass 2.00 kg and radius 20.0 cm. The box isreleased from rest and descends as the rope unwraps from thecylinder. Find the speed of the box when it has fallen 2.50 m.

Given in the question,

The mass of the block,$m=3.00\text{\hspace{0.17em}\hspace{0.17em}kg}$

The Mass of the cylinder,$m_c=5.00\text{\hspace{0.17em}kg}$

The radius of the cylinder, $r_c=4\text{0\hspace{0.17em}\hspace{0.17em}cm}$

The Mass of the cylinder$m_p=2.00\text{\hspace{0.17em}kg}$

The radius of the cylinder, $r_p=2\text{0\hspace{0.17em}\hspace{0.17em}cm}$

The height descended by the mass, $h=2.50\text{\hspace{0.17em}\hspace{0.17em}m}$

Law of conservation of energy

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

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.

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,

$U_i+K{E}_i=U_f+K{E}_f$…(i)

Potential energy can be given as

$U=mgh$

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

The block is initially at height $h=2.50m$, therefore initial potential energy of the system can be given as

$\begin{array}{l}{U}_i=m_{b}gh\\ U_i=\left(3\text{\hspace{0.17em}kg}\right)\left(9.8\text{\hspace{0.17em}\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right)\left(2.5\text{\hspace{0.17em}\hspace{0.17em}m}\right)\\ U_i=73.5\text{\hspace{0.17em}\hspace{0.17em}J}\end{array}$

Since finally the height of the block is zero

Final potential energy $U_f=0$

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

The formula of transitional kinetic energy is

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

Where m is mass and v is the linear speed

The formula of rotational kinetic energy

$k_r=\frac{1}{2}I{\omega }^2$

Where I is inertia and is the angular velocity

$K{E}_i=0$

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

Final kinetic energy

$\begin{array}{c}K{E}_f=K_t+K_{rc}+K_{rp}\\ =\frac{1}{2}m{v_b}^2+\frac{1}{2}{I}_c{{\omega }_c}^2+\frac{1}{2}{I}_p{{\omega }_p}^2\end{array}$

The moment of inertia can be calculated using the formula.

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

Where M is mass and R is the radius

$\begin{array}{c}{I}_c=\frac{1}{2}{m}_c{r_c}^2\\ =\frac{1}{2}\left(5.00\text{\hspace{0.17em}kg}\right){\left(0.4\text{0\hspace{0.17em}\hspace{0.17em}m}\right)}^2\\ =0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\end{array}$

So, the moment of inertia of the cylinder

$\begin{array}{c}{I}_c=\frac{1}{2}{m}_c{r_c}^2\\ =\frac{1}{2}\left(5.00\text{\hspace{0.17em}kg}\right){\left(0.4\text{0\hspace{0.17em}\hspace{0.17em}m}\right)}^2\\ =0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\end{array}$

So, the moment of inertia of the pully

$\begin{array}{c}{I}_p=\frac{1}{2}{m}_p{r_p}^2\\ =\frac{1}{2}\left(2.00\text{\hspace{0.17em}kg}\right){\left(0.2\text{0\hspace{0.17em}\hspace{0.17em}m}\right)}^2\\ =0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\end{array}$

Substituting the values into equation (i)

$73.5\text{\hspace{0.17em}\hspace{0.17em}J+0=0+}\frac{1}{2}\left(3.00\text{\hspace{0.17em}\hspace{0.17em}kg}\right){v_b}^2+\frac{1}{2}\left(0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\right){{\omega }_c}^2+\frac{1}{2}\left(0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\right){{\omega }_p}^2$--------(ii)

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

$v_b=v_p=v_c$

Now using the formula

$v=r\omega$

Where v is linear speed, r is radius and is the angular speed

Therefore,

$\begin{array}{c}{v}_b=v_p\\ v_b=r_p{\omega }_p\\ {\omega }_p=\frac{v_b}{r_p}\end{array}$

And

$\begin{array}{c}{v}_b=v_p\\ v_b=r_c{\omega }_c\\ {\omega }_c=\frac{v_b}{r_c}\end{array}$

Substituting the values into the equation (ii)

$\begin{array}{l}73.5\text{\hspace{0.17em}\hspace{0.17em}J+0=0+}\frac{1}{2}\left(3.00\text{\hspace{0.17em}\hspace{0.17em}kg}\right){v_b}^2+\frac{1}{2}\left(0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\right){\left(\frac{v_b}{r_c}\right)}^2+\frac{1}{2}\left(0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\right){\left(\frac{v_b}{r_p}\right)}^2\\ 73.5\text{\hspace{0.17em}\hspace{0.17em}J=}\frac{1}{2}\left(3.00\text{\hspace{0.17em}\hspace{0.17em}kg}\right){v_b}^2+\frac{1}{2}\left(0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\right){\left(\frac{v_b}{0.4\text{0\hspace{0.17em}\hspace{0.17em}m}}\right)}^2+\frac{1}{2}\left(0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\right){\left(\frac{v_b}{0.2\text{0\hspace{0.17em}\hspace{0.17em}m}}\right)}^2\\ 73.5\text{\hspace{0.17em}\hspace{0.17em}J=}\left[\frac{1}{2}\left(3.00\text{\hspace{0.17em}\hspace{0.17em}kg}\right){v_b}^2+\frac{1}{2}\frac{\left(0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\right)}{0.4\text{0\hspace{0.17em}\hspace{0.17em}m}}+\frac{1}{2}\frac{\left(0.40{\text{\hspace{0.17em}\hspace{0.17em}Kg.m}}^{\text{2}}\right)}{0.2\text{0\hspace{0.17em}\hspace{0.17em}m}}\right]{v_b}^2\\ v_b=4.76\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s}\end{array}$

The speed of the block when it has fallen 2.50m is 4.76m/s .