Question

In the system shown in Fig. 9.17, a 12.0-kg mass is releasedfrom rest and falls, causing the uniform 10.0-kg cylinder of diameter 30.0 cm to turn about a frictionless axle through its center. How far will the mass have to descend to give the cylinder 480 J of kinetic energy?

Short answer

The height descended by the mass is 12.23m

Step-by-step solution

Identification of the given data.

Given in the question,

Mass of the cylinder $M=10\text{\hspace{0.17em}kg}$

The diameter of the cylinder, $d=\text{30\hspace{0.17em}\hspace{0.17em}cm}$

Therefore, the radius of the cylinder,$R=\text{15\hspace{0.17em}\hspace{0.17em}cm}$

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

The kinetic energy of the cylinder, $K_c=480\text{\hspace{0.17em}\hspace{0.17em}J}$

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

Concept used to solve the question

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.

Finding the height descended by the mass.

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

$\begin{array}{l}{U}_i=mgh\\ U_i=\left(12\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\left(9.8\text{\hspace{0.17em}\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right)\left(h\right)\end{array}$

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

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

Where I is inertia and $\omega$is the angular velocity

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

$K{E}_i=0$

Final kinetic energy

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

Substituting the values into equation (i)

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

Since the cylinder is rotating therefore

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

$\begin{array}{l}{K}_r=\frac{1}{2}I{\omega }^2\\ 480\text{\hspace{0.17em}\hspace{0.17em}J=}\frac{1}{2}I{\omega }^2\end{array}$

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

$480\text{\hspace{0.17em}\hspace{0.17em}J=}\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2$

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

Therefore, we can use the formula

$v=R\omega$

Substituting the values

$\begin{array}{c}480\text{\hspace{0.17em}\hspace{0.17em}J=}\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\left(\frac{v}{R}\right)}^2\\ 480\text{\hspace{0.17em}\hspace{0.17em}J=}\frac{1}{4}M{v}^2\\ 480=\frac{1}{4}\left(12\text{\hspace{0.17em}\hspace{0.17em}kg}\right)v^2\\ v=12.64m/s\end{array}$

Hence the linear speed of the block $v=12.64m/s$.

Now to find the height using the equation (ii)

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

Substituting all the values

$\begin{array}{c}\left(12\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\left(9.8m/s^2\right)h=\frac{1}{2}\left(12\text{\hspace{0.17em}\hspace{0.17em}kg}\right){\left(12.64m/s\right)}^2+480\text{\hspace{0.17em}\hspace{0.17em}J}\\ h=\frac{\frac{1}{2}\left(12\text{\hspace{0.17em}\hspace{0.17em}kg}\right){\left(12.64m/s\right)}^2+480\text{\hspace{0.17em}\hspace{0.17em}J}}{\left(12\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\left(9.8m/s^2\right)}\\ h=12.23\text{\hspace{0.17em}m}\end{array}$

Hence the height descended by the mass is 12.23m