Chapter 10, Problem 86P

(a)

To determine

ToCalculate: The angular speed of the pulley.

Answer to Problem 86P

  $\omega =27\text{rad/s}$

Explanation of Solution

Given information :

The circumference of the pulley $=1.2\text{m}$

Mass of the pulley $2.2\text{kg}$

Length of the rope $=8.0\text{m}$

Mass of the rope $=4.8\text{kg}$

The difference in height of the two ends of the rope when the pulley system is at rest $=0.6\text{m}$

Formula used :

Rotational kinetic energy $=\frac{1}{2}I{\omega }^2$

Where, Iis the moment of inertia, $\omega$ is the angular speed.

Linear kinetic energy $=\frac{1}{2}M{v}^2$

Where, M is the mass and v is the speed.

Calculation:

Let the co-ordinate mechanism start at the center of the pulley, with the positive upward direction.

The conservation of energy to relate the final kinetic energy of the system $\Delta K+\Delta U=0$

Initial kinetic energy is zero, $K_i=0$

Final kinetic energy = K

Use the law of conservation of mechanical energy to find the angular velocity of the pulley when the difference in height between the two ends of the rope is 7.2 m.

By applying conservation of energy to relate the final kinetic energy of the system to the change in potential energy, can get,

  $K+\Delta U=0$

The change in potential energy of the system is

  $\begin{array}{l}\Delta U=U_f-U_i\\ \Delta U=-\frac{1}{2}\left(L_{{}_{1f}}^2+L_{{}_{2f}}^2\right)\lambda g+\frac{1}{2}\left(L_{{}_{1i}}^2+L_{{}_{2i}}^2\right)\lambda g\\ \Delta U=-\frac{1}{2}\lambda g\left[\left(L_{{}_{1f}}^2+L_{{}_{2f}}^2\right)-\left(L_{{}_{1i}}^2+L_{{}_{2i}}^2\right)\right]\end{array}$

Here,

  $\lambda =$ The linear density of the rope (Mass per unit length)

  $L_1\text{and }{L}_2$ are the lengths of the hanging parts of the rope.

Because $L_1+L_2=7.4\text{m}$

  $\begin{array}{l}{L}_{1i}=3.4\text{m}\\ L_{2i}=4.0\text{m}\\ L_{1f}=0.1\text{m}\\ L_{2f}=7.3\text{m}\end{array}$

  $\begin{array}{l}{L}_{2i}-L_{1i}=0.6\text{m}\\ L_{2f}-L_{1f}=7.2\text{m}\end{array}$

Substitute numerical values and evaluate $\Delta U$

  $\begin{array}{l}\Delta U=-\frac{1}{2}\left(0.6\text{kg/m}\right)\left(9.81{\text{ms}}^{\text{-2}}\right)\left[{\left(0.1\text{m}\right)}^2+{\left(7.3\text{m}\right)}^2-{\left(3.4\text{m}\right)}^2-{\left(4.0\text{m}\right)}^2\right]\\ \Delta U=-75.68\text{J}\end{array}$

The kinetic energy of the system when the difference in height between the two ends of the rope is $7.2\text{m}$ .

  $\begin{array}{l}K=\frac{1}{2}{I}_p{\omega }^2+\frac{1}{2}M{v}^2\\ K=\frac{1}{2}\left(\frac{1}{2}{M}_p{R}^2\right){\omega }^2+\frac{1}{2}M{R}^2{\omega }^2\\ K=\frac{1}{2}\left(\frac{1}{2}{M}_p+M\right)R^2{\omega }^2\end{array}$

  $\begin{array}{l}K=\frac{1}{2}\left[\frac{1}{2}\left(2.2\text{kg}\right)+4.8\text{kg}\right]{\left(\frac{1.2\text{m}}{2\pi }\right)}^2{\omega }^2\\ K=\left(0.1077{\text{kgm}}^{\text{2}}\right){\omega }^2\end{array}$

Substitute values in equation:

  $\begin{array}{l}\left(0.1077\text{kg}\cdot {\text{m}}^{\text{2}}\right){\omega }^2-75.68\text{J=0}\\ \omega \text{=}\sqrt{\frac{75.68\text{J}}{0.1077\text{kg}\cdot {\text{m}}^{\text{2}}}}\\ \omega \text{=}26.5\text{rad/s}\end{array}$

Conclusion:

The angular speed of the pulley when the difference in height between the two ends of the rope is $7.2\text{m}$ is $26.5\text{rad/s}$ .

(b)

To determine

ToFind: An expression for the angular momentum of the system as a function of time while neither end of the rope is above the center of the pulley.

Answer to Problem 86P

  $L=\left(0.30\text{kg}{\text{.m}}^{\text{2}}/s\right)e^{\left(1.41{\text{s}}^{\text{-1}}\right)t}$

Explanation of Solution

Given information :

The circumference of the pulley $=1.2\text{m}$

Mass of the pulley $2.2\text{kg}$

Length of the rope $=8.0\text{m}$

Mass of the rope $=4.8\text{kg}$

The difference in height of the two ends of the rope when the pulley system is at rest $=0.6\text{m}$

Formula Used :

Angular momentum:

  $L=I\omega$

Where, Iis the moment of inertia and $\omega$ is the angular velocity.

For the pulley

  $I=\frac{1}{2}{M}_p{R}^2$

For the rope

  $I=M_r{R}^2$

Calculation:

The total angular momentum of the system is,

  $\begin{array}{l}L=L_p+L_r\\ L=I_p\omega +M_r{R}^2\omega \\ L=\left(\frac{1}{2}{M}_p{R}^2+M_r{R}^2\right)\omega \\ L=\left(\frac{1}{2}{M}_p+M_r\right)R^2\omega \end{array}$

Letting $\theta$ be the angle through which the pulley has turned, express $U\left(\theta \right)$

  $U\left(\theta \right)=-\frac{1}{2}\left[{\left(L_{1i}-R\theta \right)}^2+{\left(L_{2i}+R\theta \right)}^2\right]\lambda g$

  $\begin{array}{l}\Delta U=U_f-U_i\\ \Delta U=U\left(\theta \right)-U\left(0\right)\\ \Delta U=-\frac{1}{2}+\left[{\left(L_{1i}-R\theta \right)}^2+{\left(L_{2i}+R\theta \right)}^2\right]\lambda g+\frac{1}{2}\left(L_{1i}^2+L_{2i}^2\right)\lambda g\\ \Delta U=-R^2{\theta }^2\lambda g\left(L_{1i}-L_{2i}\right)R\theta \lambda g\end{array}$

Assuming at $t=0,L_{1i}\approx L_{2i}$

  $\Delta U\approx -R^2{\theta }^2\lambda g$

Substitute for $K$ and $\Delta U$ to obtain

  $\left(0.1076{\text{kgm}}^{\text{2}}\right){\omega }^2-R^2{\theta }^2\lambda g=0$

Solving for $\omega$ yields:

  $\omega =\sqrt{\frac{R^2{\theta }^2\lambda g}{0.1076{\text{kgm}}^{\text{2}}}}$

  $\begin{array}{l}\omega =\sqrt{\frac{{\left(\raisebox{1ex}{$1.2\text{m}$}\!\left/ \!\raisebox{-1ex}{$2\pi $}\right.\right)}^2\left(0.6\text{kg/m}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)}{0.1076kg{m}^2}}\theta \\ \omega =\left(1.41s^{-1}\right)\theta \end{array}$

Express $\omega$ as the rate of change of $\theta$ :

  $\begin{array}{l}\frac{d\theta }{dt}=\left(1.41{\text{s}}^{\text{-1}}\right)t\\ \frac{d\theta }{dt}=\left(1.41{\text{s}}^{\text{-1}}\right)dt\end{array}$

Integrate $\theta$ from $0$ to $\theta$ to obtain:

  $\ln \theta =\left(1.41{\text{s}}^{\text{-1}}\right)t$

Transform from logarithmic to exponential form to obtain:

  $\theta \left(t\right)=e^{\left(1.41{\text{s}}^{\text{-1}}\right)t}$

Differentiate to express $\omega$ as a function of time:

  $\begin{array}{l}\omega \left(t\right)=\frac{d\theta }{dt}\\ \omega \left(t\right)=\left(1.41{\text{s}}^{\text{-1}}\right)e^{\left(1.41{\text{s}}^{\text{-1}}\right)t}\end{array}$

Substitute for $\omega$ in equation to obtain;

  $L=\left(\frac{1}{2}{M}_p+M_r\right)R^2\left(1.41{\text{s}}^{\text{-1}}\right)e^{\left(1.41{\text{s}}^{\text{-1}}\right)t}$

  $\begin{array}{l}L=\left[\frac{1}{2}\left(2.2\text{kg}\right)+\left(4.8\text{kg}\right)\right]{\left(\frac{1.2m}{2\pi }\right)}^2\left[\left(1.41s^{-1}\right)e^{\left(1.41{\text{s}}^{\text{-1}}\right)t}\right]\\ L=\left(0.30\text{kg}{\text{.m}}^{\text{2}}/s\right)e^{\left(1.41{\text{s}}^{\text{-1}}\right)t}\end{array}$

Conclusion:

An expression for the angular momentum of the system as a function of time while neither end of the rope is above the center of the pulley is:

  $L=\left(0.30\text{kg}{\text{.m}}^{\text{2}}/s\right)e^{\left(1.41{\text{s}}^{\text{-1}}\right)t}$