Chapter 14.2, Problem 14.53P

Two small disks A and B of mass 3 kg and 1.5 kg, respectively, may slide on a horizontal, frictionless surface. They are connected by a cord, 600 mm long, and spin counterclockwise about their mass center G at the rate of 10 rad/s. At t = 0, the coordinates of G are ${\bar{x}}_0=0,{\overline{\underset{\dot{}}{y}}}_0=2\text{m}$ , and its velocity ${\bar{v}}_0=\left(1.2\text{m/s}\right)i+\left(0.96\text{m/s}\right)\text{j}$.Shortly thereafter the cord breaks: disk, A is then observed to move along a path parallel to the y axis, and disk B moves along a path that intersects the x axis at a distance $b=7.5m$ from O. Determine (a) the velocities of.-) and B after the cord breaks, (b) the distance a from the y axis to the path of A.
'

To determine

(a)

The velocities of disks A and B after the cord breaks.

Answer to Problem 14.53P

$v_B=4.24m/s$

$v_A=2.56m/s$

Explanation of Solution

Given information:

Mass of the disk A, $m_A=3{\rm K}g.$

Mass of the disk B, $m_B=1.5{\rm K}g.$

Length of cord $AB=600mm=0.6m.$

At t=0,

Co-ordinates of G are $\overline{x_0}=0,\overline{y_0}=2m,$ and

Velocity $\overline{v_0}=\left(1.2m/s\right)i+\left(0.96m/s\right)j$

Distance $b=7.5m$

Firstly, calculate for initial condition.

The free body diagram for initial condition is as follows:

The distance G from points A and B,

$\begin{array}{l}\frac{AG}{m_B}=\frac{AB}{m_A+m_B}\\ \frac{AG}{m_B}=\frac{0.6}{3+1.5}\\ \frac{AG}{m_B}=\frac{0.6}{4.5}\\ AG=\frac{0.6}{4.5}\times 1.5\\ AG=0.2m\end{array}$

And,

$\begin{array}{l}\frac{BG}{m_A}=\frac{AB}{m_A+m_B}\\ \frac{BG}{m_A}=\frac{0.6}{3+1.5}\end{array}$

$\frac{BG}{m_A}=\frac{0.6}{4.5}$

$BG=\frac{0.6}{4.5}\times 3$

$BG=0.4m$

Now, considering linear momentum,

$\begin{array}{l}{L}_0=m\overline{v_0}\\ L_0=4.5\left(1.2i+0.96j\right)\\ L_0=5.4i+4.32j\end{array}$

After breaking the disks moves counter clockwise about their mass centre G, at the rate of 10 rad/sec.

Hence, angular momentum of both the disc about G,

$\begin{array}{l}{\left(H_G\right)}_0=AG\times m_A{v}_A^{\text{'}}+BG\times m_B{v}_B^{\text{'}}\\ {\left(H_G\right)}_0=0.2\times 3\left(0.2\times 10\right)k+0.4\times 1.5\left(0.4\times 10\right)k\\ {\left(H_G\right)}_0=1.2k+2.4k\\ {\left(H_G\right)}_0=3.6kkg.m^2/s\end{array}$

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

$\begin{array}{l}{\left(H_o\right)}_0=r\times m\overline{v_o}+{\left(H_G\right)}_0\\ {\left(H_o\right)}_0=2j\times \left(5.4i+4.32j\right)+3.6k\\ {\left(H_o\right)}_0=2j\times 5.4i+2j\times 4.32j\sin \left(0^{°}\right)+3.6k\end{array}$

Here, the property of determinant $i\times j=-k$ is used and the area of the parallelogram is representing by sin component in cross multiplication.

$\begin{array}{l}{\left(H_o\right)}_0=-10.8k+3.6k\\ {\left(H_o\right)}_0=-7.2kkg.m^2/s\end{array}$

Now, kinetic energy of the component is,

$\begin{array}{l}{T}_0=\frac{1}{2}m\overline{v_0^2}+\frac{1}{2}{m}_A{v}^{\text{'}}{}_A^2+\frac{1}{2}{m}_B{v}^{\text{'}}{}_B^2\\ T_0=\frac{1}{2}\left(4.5\right)\left[{\left(1.2\right)}^2+{\left(0.96\right)}^2\right]+\frac{1}{2}\left(3\right){\left(0.2\times 10\right)}^2+\frac{1}{2}\left(1.5\right){\left(0.4\times 10\right)}^2\end{array}$

$\begin{array}{l}{T}_0=\frac{1}{2}\left(4.5\right)\left[2.3616\right]+\frac{1}{2}\left(12\right)+\frac{1}{2}\left(24\right)\\ T_0=5.3136+6+12\\ T_0=23.3136J\end{array}$

Calculation:

After the cord is break then the free body diagram of both disc A and B.

Using conservation of mass of linear momentum,

$L_0=L$

$\begin{array}{l}5.4i+4.32j=m_A{v}_A+m_B{v}_B\\ 5.4i+4.32j=3\left(v_{A}j\right)+1.5\left[{\left(v_B\right)}_{x}i+{\left(v_B\right)}_{y}j\right]\end{array}$

Equation the coefficient of i :

$\begin{array}{l}5.4=1.5{\left(v_B\right)}_x\\ {\left(v_B\right)}_x=3.6m/s\_\_\_\_\_\_(1)\end{array}$

Equation the coefficient of j :

$\begin{array}{l}4.32=3v_A+1.5{\left(v_B\right)}_y\\ {\left(v_B\right)}_y=\frac{4.32-3v_A}{1.5}\\ {\left(v_B\right)}_y=2.88-2v_A\_\_\_\_\_\_\_(2)\end{array}$

Applying conservation of energy,

$T_0=T$

$\begin{array}{l}{T}_0=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2\\ 23.3136=\frac{1}{2}\left(3\right)v_A^2+\frac{1}{2}\left(1.5\right)\left[{\left(v_B\right)}_x^2+{\left(v_B\right)}_y^2\right]\end{array}$

Substituting the values from equation (1) and (2);

$\begin{array}{l}23.3136=\frac{1}{2}\left(3\right)v_A^2+\frac{1}{2}\left(1.5\right)\left[{\left(3.6\right)}^2+{\left(2.88-2v_A\right)}^2\right]\\ 23.3136=1.5v_A^2+0.75\left[12.96+8.3+4v_A^2-11.52v_A\right]\\ 23.3136=1.5v_A^2+16+3v_A^2-8.64v_A\\ 4.5v_A^2-8.64v_A-7.314=0\end{array}$

By solving the above equation we get;

$v_A=2.56m/sand{v}_A=-0.640m/s$

$v_A=-0.640m/s(rejected)$

Hence, velocity of A, $v_A=2.56m/s$

And velocity of B,

$\begin{array}{l}{\left(v_B\right)}_y=2.88-2\left(2.56\right)\\ {\left(v_B\right)}_y=-2.24m/s\end{array}$

$\begin{array}{l}{v}_B=3.6i-2.24jand,\\ v_B=4.24m/s\end{array}$

To determine

(b)

The distance a from the y-axis to the path of A.

Answer to Problem 14.53P

$a=2.343m$

Explanation of Solution

Given information:

Mass of the disk A, $m_A=3{\rm K}g.$

Mass of the disk B, $m_B=1.5{\rm K}g.$

Length of cord $AB=600mm=0.6m.$

At t=0,

Co-ordinates of G are $\overline{x_0}=0,\overline{y_0}=2m,$ and

Velocity, $\overline{v_0}=\left(1.2m/s\right)i+\left(0.96m/s\right)j$

Distance $b=7.5m$

Firstly, calculate for initial condition.

The free body diagram for initial condition is as follows:

The distance G from points A and B,

$\begin{array}{l}\frac{AG}{m_B}=\frac{AB}{m_A+m_B}\\ \frac{AG}{m_B}=\frac{0.6}{3+1.5}\\ \frac{AG}{m_B}=\frac{0.6}{4.5}\\ AG=\frac{0.6}{4.5}\times 1.5\\ AG=0.2m\end{array}$

And,

$\begin{array}{l}\frac{BG}{m_A}=\frac{AB}{m_A+m_B}\\ \frac{BG}{m_A}=\frac{0.6}{3+1.5}\end{array}$

$\frac{BG}{m_A}=\frac{0.6}{4.5}$

$BG=\frac{0.6}{4.5}\times 3$

$BG=0.4m$

Now, considering linear momentum,

$\begin{array}{l}{L}_0=m\overline{v_0}\\ L_0=4.5\left(1.2i+0.96j\right)\\ L_0=5.4i+4.32j\end{array}$

After breaking the disks moves counter clockwise about their mass centre G, at the rate of 10 rad/sec.

Hence, angular momentum of both the disc about G,

$\begin{array}{l}{\left(H_G\right)}_0=AG\times m_A{v}_A^{\text{'}}+BG\times m_B{v}_B^{\text{'}}\\ {\left(H_G\right)}_0=0.2\times 3\left(0.2\times 10\right)k+0.4\times 1.5\left(0.4\times 10\right)k\\ {\left(H_G\right)}_0=1.2k+2.4k\\ {\left(H_G\right)}_0=3.6kkg.m^2/s\end{array}$

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

$\begin{array}{l}{\left(H_o\right)}_0=r\times m\overline{v_o}+{\left(H_G\right)}_0\\ {\left(H_o\right)}_0=2j\times \left(5.4i+4.32j\right)+3.6k\\ {\left(H_o\right)}_0=2j\times 5.4i+2j\times 4.32j\sin \left(0^{°}\right)+3.6k\end{array}$

Here, the property of determinant $i\times j=-k$ is used and the area of the parallelogram is representing by sin component in cross multiplication.

$\begin{array}{l}{\left(H_o\right)}_0=-10.8k+3.6k\\ {\left(H_o\right)}_0=-7.2kkg.m^2/s\end{array}$

Now, kinetic energy of the component is,

$\begin{array}{l}{T}_0=\frac{1}{2}m\overline{v_0^2}+\frac{1}{2}{m}_A{v}^{\text{'}}{}_A^2+\frac{1}{2}{m}_B{v}^{\text{'}}{}_B^2\\ T_0=\frac{1}{2}\left(4.5\right)\left[{\left(1.2\right)}^2+{\left(0.96\right)}^2\right]+\frac{1}{2}\left(3\right){\left(0.2\times 10\right)}^2+\frac{1}{2}\left(1.5\right){\left(0.4\times 10\right)}^2\end{array}$

$\begin{array}{l}{T}_0=\frac{1}{2}\left(4.5\right)\left[2.3616\right]+\frac{1}{2}\left(12\right)+\frac{1}{2}\left(24\right)\\ T_0=5.3136+6+12\\ T_0=23.3136J\end{array}$

Calculation:

After the cord is break then the free body diagram of both disc A and B.

Using conservation of mass of linear momentum,

$L_0=L$

$\begin{array}{l}5.4i+4.32j=m_A{v}_A+m_B{v}_B\\ 5.4i+4.32j=3\left(v_{A}j\right)+1.5\left[{\left(v_B\right)}_{x}i+{\left(v_B\right)}_{y}j\right]\end{array}$

Equation the coefficient of i :

$\begin{array}{l}5.4=1.5{\left(v_B\right)}_x\\ {\left(v_B\right)}_x=3.6m/s\_\_\_\_\_\_(1)\end{array}$

Equation the coefficient of j :

$\begin{array}{l}4.32=3v_A+1.5{\left(v_B\right)}_y\\ {\left(v_B\right)}_y=\frac{4.32-3v_A}{1.5}\\ {\left(v_B\right)}_y=2.88-2v_A\_\_\_\_\_\_\_(2)\end{array}$

Applying conservation of energy,

$T_0=T$

$\begin{array}{l}{T}_0=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2\\ 23.3136=\frac{1}{2}\left(3\right)v_A^2+\frac{1}{2}\left(1.5\right)\left[{\left(v_B\right)}_x^2+{\left(v_B\right)}_y^2\right]\end{array}$

Substituting the values from equation (1) and (2);

$\begin{array}{l}23.3136=\frac{1}{2}\left(3\right)v_A^2+\frac{1}{2}\left(1.5\right)\left[{\left(3.6\right)}^2+{\left(2.88-2v_A\right)}^2\right]\\ 23.3136=1.5v_A^2+0.75\left[12.96+8.3+4v_A^2-11.52v_A\right]\\ 23.3136=1.5v_A^2+16+3v_A^2-8.64v_A\\ 4.5v_A^2-8.64v_A-7.314=0\end{array}$

By solving the above equation we get;

$v_A=2.56m/sand{v}_A=-0.640m/s$

$v_A=-0.640m/s(rejected)$

Hence, velocity of A, $v_A=2.56m/s$

And velocity of B,

$\begin{array}{l}{\left(v_B\right)}_y=2.88-2\left(2.56\right)\\ {\left(v_B\right)}_y=-2.24m/s\end{array}$

$\begin{array}{l}{v}_B=3.6i-2.24jand,\\ v_B=4.24m/s\end{array}$

Now, Conservation of angular momentum about O,

${\left(H_O\right)}_0=H_O$

$\begin{array}{l}-7.2k=ai\times 3\times 2.56j+7.5i\times 1.5\left(3.6i-2.24j\right)\\ -7.2k=7.68ak-25.2k\\ 7.68a=-7.2+25.2\\ 7.68a=18\\ a=2.343m\end{array}$