Chapter 10, Problem 53P

(a)

To determine

The speed of the block.

Answer to Problem 53P

The speed of the block is $1.59\text{ m/s}$.

Explanation of Solution

The change in kinetic energy of the system is equal to the net work done on the system. Only friction on the block of mass $m_2$ and gravitation force on block of mass $m_1$ have non-zero work. Speed of a pulley of a hollow cylinder shape has a moment of inertia.

Write the expression for the conservation of energy.

  $\Delta K+\Delta U+\Delta E_{\text{int}}=0$                                                                                      (I)

Here, $\Delta K$ is the change in kinetic energy, $\Delta U$ is the change in potential energy and $\Delta E_{\mathrm{int}}$ is the internal kinetic energy.

Change in kinetic energy is sum of rotational kinetic energy and translational kinetic energy.

Write the expression for change in kinetic energy.

    $\Delta K=\Delta {{\rm K}}_{trans}+\Delta {{\rm K}}_{rotational}$

Here, $\Delta K_{trans}$ is the change in translational kinetic energy and $\Delta K_{rotational}$ is the rotational kinetic energy.

Write the expression for change in translational kinetic energy.

    $\Delta K_{trans}=K_f-K_i$

Here, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy.

Write the expression for the change in potential energy.

    $\Delta U=U_f-U_i$

Here, $U_f$ is the final potential energy and $U_i$ is the initial potential energy.

Substitute $\left(K_f-K_i\right)+\Delta K_{rotational}$ for $\Delta K$ and $U_f-U_i$ for  $\Delta U$ in equation (I).

    $\left(K_f-K_i\right)+\Delta K_{rotational}+\left(U_f-U_i\right)+\Delta E_{\mathrm{int}}=0$                                             (II)

Write the expression for final kinetic energy.

    $K_f=\frac{1}{2}{m}_2\left(v_f^2-v_i{}^2\right)$

Here, $m_2$ is the mass of second block and $v_f$ is the final speed and $v_i$ is initial speed for second block.

Write the expression for initial kinetic energy.

    $K_i=\frac{1}{2}{m}_1\left(v_f^2-v_i{}^2\right)$

Here, $m_1$ is the mass of first block.

Write the expression for change in rotational kinetic energy.

    $\Delta K_{rotational}=\frac{1}{2}I\left({\omega }_f^2-{\omega }_i{}^2\right)$                                                                          (III)

Here, ${\omega }_f$ is final angular speed, $I$ is the inertia of the pulley and ${\omega }_i$ is initial angular speed.

Write the expression for inertia of pulley.

    $I=\frac{1}{2}M\left(R_1^2+R_2^2\right)$

Here, $M$ is the mass of pulley, $R_1$ is the inner radius of pulley and $R_2$ is the outer radius of pulley.

Substitute $\frac{1}{2}M\left(R_1^2+R_2^2\right)$ for $I$ in equation (III).

  $\Delta K_{rotational}=\frac{1}{2}\left(\frac{1}{2}M\left(R_1^2+R_2^2\right)\right)\left({\omega }_f^2-{\omega }_i{}^2\right)$

Write the expression for change in potential energy.

    $\Delta U=m_{1}g\left(y_f-y_i\right)$

Here, $g$ is acceleration under gravity, $y_f$ is final height and $y_i$ is initial height.

Write the expression for energy of the second block.

    $\Delta E_{\text{int}}=f_{k}d$

Here, $f_k$ is frictional force and $d$ is the distance covered by the block.

Substitute $\frac{1}{2}{m}_1\left(v_f^2-v_i{}^2\right)$ for $K_f$, $\frac{1}{2}{m}_1\left(v_f^2-v_i{}^2\right)$ for $K_i$, $\frac{1}{2}\left(\frac{1}{2}M\left(R_1^2+R_2^2\right)\right)\left({\omega }_f^2-{\omega }_i{}^2\right)$ for $\Delta K_{rotational}$, $m_{1}g\left(y_f-y_i\right)$ for $U_f-U_i$ and $f_{k}d$ for $\Delta E_{\text{int}}$ in equation (II).

    $\left[\begin{array}{l}\frac{1}{2}{m}_2\left(v_f^2-v_i{}^2\right)+\frac{1}{2}{m}_1\left(v_f^2-v_i{}^2\right)+\\ \frac{1}{2}\left(\frac{1}{2}M\left(R_1^2+R_2^2\right)\right)\left({\omega }_f^2-{\omega }_i{}^2\right)+m_{1}g\left(y_f-y_i\right)+f_{k}d\end{array}\right]=0$                         (IV)

Write the expression for frictionof second block.

    $f_k={\mu }_k{m}_{2}g$

Here, ${\mu }_k$ is coefficient of friction.

Angular speed of the pulley is related to the speed of the objects.

Write the expression for the final angular speed.

  ${\omega }_f=\frac{v_f}{R_2}$

Write the expression for the initial angular speed.

  ${\omega }_i=\frac{v_i}{R_2}$

Substitute $\mu m_{2}g$ for $f_k$, $\frac{v_f}{R_2}$ for ${\omega }_f$, $\frac{v_i}{R_2}$ and $-d$ for $\left(y_f-y_i\right)$ in equation (IV).

    $\left[\begin{array}{l}\frac{1}{2}{m}_2\left(v_f^2-v_i{}^2\right)+\frac{1}{2}{m}_1\left(v_f^2-v_i{}^2\right)+\\ \frac{1}{2}\left(\frac{1}{2}M\left(R_1^2+R_2^2\right)\left({\left(\frac{v_f}{R_2}\right)}^2-{\left(\frac{v_i}{R_2}\right)}^2\right)\right)+m_{1}g\left(-d\right)+\mu m_2{g}_{k}d\end{array}\right]=0$

Rearrange the above expression for $v_f$.

    $v_f=\sqrt{v_i{}^2+\frac{4gd\left(m_1-{\mu }_k{m}_2\right)}{2\left(m_1+m_2\right)+M\left(1+\frac{R_1{}^2}{R_2{}^2}\right)}}$                                                        (V)

Conclusion:

Substitute $0.820\text{ m/s}$ for $v_i$, $9.80{\text{ m/s}}^2$ for $g$, $0.700\text{ m}$ for $d$, $0.420\text{ kg}$ for $m_1$, $0.250$ for ${\mu }_k$, $0.850\text{ kg}$ for $m_2$, $0.350\text{ kg}$ for $M$, $0.0200\text{ m}$ for $R_1$ and $0.0300\text{ m}$ for $R_2$ in equation (III).

    $\begin{array}{c}{v}_f=\sqrt{{\left(0.820\text{ m/s}\right)}^2+\frac{4\left(9.80{\text{ m/s}}^2\right)\left(0.700\text{ m}\right)\left(0.420\text{ kg}-0.250\left(0.850\text{ kg}\right)\right)}{2\left(0.420\text{ kg}+0.850\text{ kg}\right)+0.350\text{ kg}\left(1+\frac{{\left(0.0200\text{ m}\right)}^2}{{\left(0.0300\text{ m}\right)}^2}\right)}}\\ =\sqrt{0.6724{\left(\text{m/s}\right)}^2+\frac{5.6938\text{ kg}{\left(\text{m/s}\right)}^2}{3.0455\text{ kg}}}\\ =\sqrt{0.6724{\left(\text{m/s}\right)}^2+1.8695{\left(\text{m/s}\right)}^2}\\ =1.59\text{ m/s}\end{array}$

Thus, the speed of the block is $1.59\text{ m/s}$.

(b)

To determine

The angular speed of the pulley.

Answer to Problem 53P

The angular speed of the pulley is $53.1\text{ rad/s}$.

Explanation of Solution

Write the expression for the angular speed of the pulley.

    $\omega =\frac{v_f}{R_2}$                                                                                                        (VI)

Conclusion:

Substitute $1.59\text{ m/s}$ for $v$ and $0.0300\text{ m}$ for $R_2$ in equation (VI).

    $\begin{array}{c}\omega =\frac{1.59\text{ m/s}}{0.0300\text{ m}}\\ =53.1\text{ rad/s}\end{array}$

Thus, the angular speed of the pulley is $53.1\text{ rad/s}$.