Chapter 14.2, Problem 14.54P

Two small disks A and B of mass 2 kg and 1 kg, respectively, may slide on a horizontal and frictionless surface. They are connected by a cord of negligible mass and spin about their mass center G. At t = 0, G is moving with the velocity ${\stackrel{\to }{v}}_0$ and its coordinates are ${\bar{x}}_0=0,{\bar{y}}_0=1.89\text{m}$ . Shortly thereafter, the cord breaks and disk A is observed to move with a velocity $v_A=\left(5\text{m/s}\right)\text{j}$ in a straight line and at a distance a = 2.56 m from the y axis, while B moves with a velocity $v_B=\left(7.2\text{m/s}\right)\text{i-}\left(4.6\text{m/s}\right)\text{j}$ along a path intersecting the x axis at a distance b = 7.48 m from the origin O. Determine (a) the initial velocity ${\bar{v}}_0$ of the mass center G of the two disks, (b) the length of the cord initially connecting the two disks, (c) the rate in rad s at which the disks were spinning about G.

To determine

(a)

The initial velocity, $V_o$ of the centre of mass G.

Answer to Problem 14.54P

$\overline{v_0}=\left(2.4m/s\right)i+\left(1.8m/s\right)j$

Explanation of Solution

Given information:

Mass of the disk A, $m_A=2{\rm K}g$

Mass of the disk B, $m_B=1{\rm K}g$

At t=0,

Co-ordinates of G are $\overline{x_0}=0,\overline{y_0}=1.89m$ and

Velocity of A, $V_A=\left(5m/s\right)j$

Velocity, $\overline{v_B}=\left(7.2m/s\right)i-\left(4.6m/s\right)j$

Distance $a=2.56m$

$b=7.48m$

Firstly, calculate for initial condition:

The free body diagram for initial condition is as follows:

The distance G from points A and B,

$\begin{array}{l}\frac{AG}{m_B}=\frac{AG+GB}{m_A+m_B}\\ \frac{AG}{m_B}=\frac{l}{2+1}\\ AG=\frac{1}{3}l\end{array}$

And,

$\begin{array}{l}\frac{BG}{m_A}=\frac{AG+GB}{m_A+m_B}\\ \frac{BG}{m_A}=\frac{l}{2+1}\\ BG=\frac{m_A}{3}l\\ BG=\frac{2}{3}l\end{array}$

Now, considering linear momentum,

$\begin{array}{l}{L}_0=m\overline{v_0}\\ L_0=3\overline{v_0}\end{array}$

Hence, angular momentum of both the disc about G,

$\begin{array}{l}{\left(H_G\right)}_0=AG\times m_A{v}_A^{\text{'}}+BG\times m_B{v}_B^{\text{'}}\\ {\left(H_G\right)}_0=\left(\frac{1}{3}l\right)\times 2\times \left(\frac{1}{3}l\omega \right)k+\left(\frac{2}{3}l\right)\times 1\times \left(\frac{2}{3}l\omega \right)k\\ {\left(H_G\right)}_0=\left(\frac{2}{9}{l}^2\omega \right)k+\left(\frac{4}{9}{l}^2\omega \right)k\\ {\left(H_G\right)}_0=\left(\frac{2}{3}{l}^2\omega \right)k\end{array}$

Now, kinetic energy of the component is,

$\begin{array}{l}{T}_0=\frac{1}{2}m\overline{v_0^2}+\frac{1}{2}{m}_A{v}^{\text{'}}{}_A^2+\frac{1}{2}{m}_B{v}^{\text{'}}{}_B^2\\ T_0=\frac{1}{2}\left(3\right)\overline{v_0^2}+\frac{1}{2}\left(2\right){\left(\frac{1}{3}l\omega \right)}^2+\frac{1}{2}\left(1\right){\left(\frac{2}{3}l\omega \right)}^2\end{array}$

$\begin{array}{l}{T}_0=\frac{3}{2}{\overline{v}}_0^2+\frac{1}{9}\left(l^2{\omega }^2\right)+\frac{2}{9}\left(l^2{\omega }^2\right)\\ T_0=\frac{3}{2}{\overline{v}}_0^2+\frac{3}{9}\left(l^2{\omega }^2\right)\\ T_0=\frac{3}{2}{\overline{v}}_0^2+\frac{1}{3}\left(l^2{\omega }^2\right)\end{array}$

Calculation:

Using conservation of mass of linear momentum,

$L_0=L$

$\begin{array}{l}m\overline{v_0}=m_A{v}_A+m_B{v}_B\\ 3\overline{v_0}=2\left(v_{A}j\right)+1\left[{\left(v_B\right)}_{x}i-{\left(v_B\right)}_{y}j\right]\\ 3\overline{v_0}=2\left(5j\right)+\left(1\right)\left[7.2i-4.6j\right]\\ 3\overline{v_0}=10j+\left[7.2i-4.6j\right]\\ 3\overline{v_0}=7.2i+5.4j\\ \overline{v_0}=\frac{7.2i+5.4j}{3}\\ \overline{v_0}=\left(2.4m/s\right)i+\left(1.8m/s\right)j\end{array}$

To determine

(b)

The length of cord initially connecting the two disks.

Answer to Problem 14.54P

$l=600mm$

Explanation of Solution

Given information:

Mass of the disk A, $m_A=2{\rm K}g$

Mass of the disk B, $m_B=1{\rm K}g$

At t=0,

Co-ordinates of G are $\overline{x_0}=0,\overline{y_0}=1.89m$ and

Velocity of A, $V_A=\left(5m/s\right)j$

Velocity, $\overline{v_B}=\left(7.2m/s\right)i-\left(4.6m/s\right)j$

Distance $a=2.56m$

$b=7.48m$

Firstly, calculate for initial condition:

The free body diagram for initial condition is as follows:

The distance G from points A and B,

$\begin{array}{l}\frac{AG}{m_B}=\frac{AG+GB}{m_A+m_B}\\ \frac{AG}{m_B}=\frac{l}{2+1}\\ AG=\frac{1}{3}l\end{array}$

And,

$\begin{array}{l}\frac{BG}{m_A}=\frac{AG+GB}{m_A+m_B}\\ \frac{BG}{m_A}=\frac{l}{2+1}\\ BG=\frac{m_A}{3}l\\ BG=\frac{2}{3}l\end{array}$

Now, considering linear momentum,

$\begin{array}{l}{L}_0=m\overline{v_0}\\ L_0=3\overline{v_0}\end{array}$

Hence, angular momentum of both the disc about G,

$\begin{array}{l}{\left(H_G\right)}_0=AG\times m_A{v}_A^{\text{'}}+BG\times m_B{v}_B^{\text{'}}\\ {\left(H_G\right)}_0=\left(\frac{1}{3}l\right)\times 2\times \left(\frac{1}{3}l\omega \right)k+\left(\frac{2}{3}l\right)\times 1\times \left(\frac{2}{3}l\omega \right)k\\ {\left(H_G\right)}_0=\left(\frac{2}{9}{l}^2\omega \right)k+\left(\frac{4}{9}{l}^2\omega \right)k\\ {\left(H_G\right)}_0=\left(\frac{2}{3}{l}^2\omega \right)k\end{array}$

Now, kinetic energy of the component is,

$\begin{array}{l}{T}_0=\frac{1}{2}m\overline{v_0^2}+\frac{1}{2}{m}_A{v}^{\text{'}}{}_A^2+\frac{1}{2}{m}_B{v}^{\text{'}}{}_B^2\\ T_0=\frac{1}{2}\left(3\right)\overline{v_0^2}+\frac{1}{2}\left(2\right){\left(\frac{1}{3}l\omega \right)}^2+\frac{1}{2}\left(1\right){\left(\frac{2}{3}l\omega \right)}^2\end{array}$

$\begin{array}{l}{T}_0=\frac{3}{2}{\overline{v}}_0^2+\frac{1}{9}\left(l^2{\omega }^2\right)+\frac{2}{9}\left(l^2{\omega }^2\right)\\ T_0=\frac{3}{2}{\overline{v}}_0^2+\frac{3}{9}\left(l^2{\omega }^2\right)\\ T_0=\frac{3}{2}{\overline{v}}_0^2+\frac{1}{3}\left(l^2{\omega }^2\right)\end{array}$

Calculation:

Using conservation of mass of linear momentum,

$L_0=L$

$\begin{array}{l}m\overline{v_0}=m_A{v}_A+m_B{v}_B\\ 3\overline{v_0}=2\left(v_{A}j\right)+1\left[{\left(v_B\right)}_{x}i-{\left(v_B\right)}_{y}j\right]\\ 3\overline{v_0}=2\left(5j\right)+\left(1\right)\left[7.2i-4.6j\right]\\ 3\overline{v_0}=10j+\left[7.2i-4.6j\right]\\ 3\overline{v_0}=7.2i+5.4j\\ \overline{v_0}=\frac{7.2i+5.4j}{3}\\ \overline{v_0}=\left(2.4m/s\right)i+\left(1.8m/s\right)j\end{array}$

Conservation of angular momentum of both the disc about point O,

$\begin{array}{l}\left(1.89j\right)\times m\overline{v_o}+{\left(H_G\right)}_0=\left(2.56i\right)m_A{v}_A+\left(7.48i\right)m_B{v}_B\\ \left(1.89j\right)\times 3\left(2.4i+1.8j\right)+\frac{2}{3}{l}^2\omega k=\left(2.56i\right)\times 2\times 5j+\left(7.48i\right)\times \left(7.2i-4.7j\right)\end{array}$

$\begin{array}{l}-13.608k+\frac{2}{3}{l}^2\omega k=25.6k-34.408k\\ \frac{2}{3}{l}^2\omega k=25.6k-34.408k+13.608k\\ \frac{2}{3}{l}^2\omega =4.80\\ l^2\omega =7.20\_\_\_\_\_\_\_\_\_(1)\end{array}$

Now, applying law of conservation of energy: $T_0=T$

$\begin{array}{l}\frac{3}{2}\overline{v_0^2}+\frac{1}{3}{l}^2{\omega }^2=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2\\ \frac{3}{2}\left[{\left(2.4\right)}^2+{\left(1.8\right)}^2\right]+\frac{1}{3}{l}^2{\omega }^2=\frac{1}{2}\left(2\right){\left(5\right)}^2+\frac{1}{2}\left(1\right)\left[{\left(7.2\right)}^2+{\left(4.6\right)}^2\right]\\ \frac{3}{2}\times 9+\frac{1}{3}{l}^2{\omega }^2=25+\frac{73}{2}\\ 13.5+\frac{1}{3}{l}^2{\omega }^2=25+36.5\\ \frac{1}{3}{l}^2{\omega }^2=48\\ l^2{\omega }^2=144\_\_\_\_\_\_\_\_\_(2)\end{array}$

Dividing equation (2) by equation (1);

$\begin{array}{l}\frac{l^2{\omega }^2}{l^2\omega }=\frac{144}{7.20}\\ \omega =20rad/s\end{array}$

Now, substituting the value of ? in equation (1);

$\begin{array}{l}{l}^2\omega =7.20\\ l^2\left(20\right)=7.20\\ l^2=\frac{7.20}{20}\end{array}$

$\begin{array}{l}{l}^2=0.36\\ l=0.6mor,\\ l=600mm\end{array}$

To determine

(c)

The rate in rad/s at which the disks were spinning about G.

Answer to Problem 14.54P

$\omega =20rad/s$

Explanation of Solution

Given information:

Mass of the disk A, $m_A=2{\rm K}g$

Mass of the disk B, $m_B=1{\rm K}g$

At t=0,

Co-ordinates of G are $\overline{x_0}=0,\overline{y_0}=1.89m$ and

Velocity of A, $V_A=\left(5m/s\right)j$

Velocity, $\overline{v_B}=\left(7.2m/s\right)i-\left(4.6m/s\right)j$

Distance $a=2.56m$

$b=7.48m$

Firstly, calculate for initial condition:

The free body diagram for initial condition is as follows:

The distance G from points A and B,

$\begin{array}{l}\frac{AG}{m_B}=\frac{AG+GB}{m_A+m_B}\\ \frac{AG}{m_B}=\frac{l}{2+1}\\ AG=\frac{1}{3}l\end{array}$

And,

$\begin{array}{l}\frac{BG}{m_A}=\frac{AG+GB}{m_A+m_B}\\ \frac{BG}{m_A}=\frac{l}{2+1}\\ BG=\frac{m_A}{3}l\\ BG=\frac{2}{3}l\end{array}$

Now, considering linear momentum,

$\begin{array}{l}{L}_0=m\overline{v_0}\\ L_0=3\overline{v_0}\end{array}$

Hence, angular momentum of both the disc about G,

$\begin{array}{l}{\left(H_G\right)}_0=AG\times m_A{v}_A^{\text{'}}+BG\times m_B{v}_B^{\text{'}}\\ {\left(H_G\right)}_0=\left(\frac{1}{3}l\right)\times 2\times \left(\frac{1}{3}l\omega \right)k+\left(\frac{2}{3}l\right)\times 1\times \left(\frac{2}{3}l\omega \right)k\\ {\left(H_G\right)}_0=\left(\frac{2}{9}{l}^2\omega \right)k+\left(\frac{4}{9}{l}^2\omega \right)k\\ {\left(H_G\right)}_0=\left(\frac{2}{3}{l}^2\omega \right)k\end{array}$

Now, kinetic energy of the component is,

$\begin{array}{l}{T}_0=\frac{1}{2}m\overline{v_0^2}+\frac{1}{2}{m}_A{v}^{\text{'}}{}_A^2+\frac{1}{2}{m}_B{v}^{\text{'}}{}_B^2\\ T_0=\frac{1}{2}\left(3\right)\overline{v_0^2}+\frac{1}{2}\left(2\right){\left(\frac{1}{3}l\omega \right)}^2+\frac{1}{2}\left(1\right){\left(\frac{2}{3}l\omega \right)}^2\end{array}$

$\begin{array}{l}{T}_0=\frac{3}{2}{\overline{v}}_0^2+\frac{1}{9}\left(l^2{\omega }^2\right)+\frac{2}{9}\left(l^2{\omega }^2\right)\\ T_0=\frac{3}{2}{\overline{v}}_0^2+\frac{3}{9}\left(l^2{\omega }^2\right)\\ T_0=\frac{3}{2}{\overline{v}}_0^2+\frac{1}{3}\left(l^2{\omega }^2\right)\end{array}$

Calculation:

Using conservation of mass of linear momentum,

$L_0=L$

$\begin{array}{l}m\overline{v_0}=m_A{v}_A+m_B{v}_B\\ 3\overline{v_0}=2\left(v_{A}j\right)+1\left[{\left(v_B\right)}_{x}i-{\left(v_B\right)}_{y}j\right]\\ 3\overline{v_0}=2\left(5j\right)+\left(1\right)\left[7.2i-4.6j\right]\\ 3\overline{v_0}=10j+\left[7.2i-4.6j\right]\\ 3\overline{v_0}=7.2i+5.4j\\ \overline{v_0}=\frac{7.2i+5.4j}{3}\\ \overline{v_0}=\left(2.4m/s\right)i+\left(1.8m/s\right)j\end{array}$

Conservation of angular momentum of both the disc about point O,

$\begin{array}{l}\left(1.89j\right)\times m\overline{v_o}+{\left(H_G\right)}_0=\left(2.56i\right)m_A{v}_A+\left(7.48i\right)m_B{v}_B\\ \left(1.89j\right)\times 3\left(2.4i+1.8j\right)+\frac{2}{3}{l}^2\omega k=\left(2.56i\right)\times 2\times 5j+\left(7.48i\right)\times \left(7.2i-4.7j\right)\end{array}$

$\begin{array}{l}-13.608k+\frac{2}{3}{l}^2\omega k=25.6k-34.408k\\ \frac{2}{3}{l}^2\omega k=25.6k-34.408k+13.608k\\ \frac{2}{3}{l}^2\omega =4.80\\ l^2\omega =7.20\_\_\_\_\_\_\_\_\_(1)\end{array}$

Now, applying law of conservation of energy: $T_0=T$

$\begin{array}{l}\frac{3}{2}\overline{v_0^2}+\frac{1}{3}{l}^2{\omega }^2=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2\\ \frac{3}{2}\left[{\left(2.4\right)}^2+{\left(1.8\right)}^2\right]+\frac{1}{3}{l}^2{\omega }^2=\frac{1}{2}\left(2\right){\left(5\right)}^2+\frac{1}{2}\left(1\right)\left[{\left(7.2\right)}^2+{\left(4.6\right)}^2\right]\\ \frac{3}{2}\times 9+\frac{1}{3}{l}^2{\omega }^2=25+\frac{73}{2}\\ 13.5+\frac{1}{3}{l}^2{\omega }^2=25+36.5\\ \frac{1}{3}{l}^2{\omega }^2=48\\ l^2{\omega }^2=144\_\_\_\_\_\_\_\_\_(2)\end{array}$

Dividing equation (2) by equation (1);

$\begin{array}{l}\frac{l^2{\omega }^2}{l^2\omega }=\frac{144}{7.20}\\ \omega =20rad/s\end{array}$