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 ${\bar{x}}_0=0$, ${\bar{y}}_0=2\text{m}$, and its velocity ${\bar{v}}_0=(1.2\text{m/s})i+(0.96\text{m/s})j$. 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 = 7.5 m from O. Determine (a) the velocities of A and B after the cord breaks, (b) the distance a from the y axis to the path of A.

Fig. P14.53 and P14.54

To determine

Find the velocity of A and B after the cord breaks.

Answer to Problem 14.53P

The velocity of A after the cord breaks is $\underset{\_}{2.56\text{m}/\text{s}}$.

The velocity of B after the cord breaks is $\underset{\_}{4.24\text{m}/\text{s}}$ at an angle $31.9°$.

Explanation of Solution

Given information:

The mass $\left(m_A\right)$ of disk A is $3\text{kg}$.

The mass $\left(m_B\right)$ of the disk B is $1.5\text{kg}$.

The cord length AB is $600\text{mm}$.

The rate of spin $\omega$ is $10\text{rad}/\text{s}$.

The coordinates of G, $\left({\overline{x}}_0,{\overline{y}}_0\right)$ is $\left(0,2\right)\text{m}$.

The velocity ${\overline{\text{v}}}_0$ is $\left(1.2\text{m}/\text{s}\right)\text{i}+\left(0.96\text{m}/\text{s}\right)\text{j}$.

The distance b is $7.5\text{m}$.

Calculation:

Sketch the disk A and B as shown in Figure 1.

Refer to Figure 1.

The small disks A and B of mass $3\text{kg}$ and $1.5\text{kg}$ respectively, may slide on a horizontal, frictionless surface. They are connected by a cord $600\text{mm}$ long.

At location G,

Find the total mass (m) using the relation as follows:

$m=m_A+m{B}_{}$

Substitute $3\text{kg}$ for $\left(m_A\right)$ and $1.5\text{kg}$ for $\left(m_B\right)$.

$\begin{array}{c}m=3+1.5\\ =4.5\text{kg}\end{array}$

$\frac{AG}{m_B}=\frac{BG}{m_A}=\frac{AG+GB}{m_B+m_A}=\frac{AB}{m}$

Substitute $600\text{mm}$ for $AB$ and $4.5\text{kg}$ for m in Equation.

$\begin{array}{l}\frac{AG}{m_B}=\frac{BG}{m_A}=\frac{AG+GB}{m_B+m_A}=\frac{600\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{4.5\text{kg}}\\ \frac{AG}{m_B}=\frac{BG}{m_A}=\frac{AG+GB}{m_B+m_A}=\frac{0.6}{4.5}\end{array}$

$\begin{array}{c}AG=1.5\left(\frac{0.6}{4.5}\right)\\ =0.2\text{m}\end{array}$

$\begin{array}{c}BG=3\left(\frac{0.6}{4.5}\right)\\ =0.4\text{m}\end{array}$

Find the linear momentum using the relation as follows:

${\text{L}}_0=m{\overline{\text{v}}}_0$ (1)

Substitute $4.5\text{kg}$ for m and $\left(1.2\text{m}/\text{s}\right)\text{i}+\left(0.96\text{m}/\text{s}\right)\text{j}$ for ${\overline{\text{v}}}_0$ in Equation (1).

$\begin{array}{c}{\text{L}}_0=4.5\left(1.2\text{m}/\text{s}\right)\text{i}+\left(0.96\text{m}/\text{s}\right)\text{j}\\ =\text{5}\text{.4i}+4.32\text{j}\end{array}$

Find the angular moment about G using the equation as follows:

${\left(H_G\right)}_0=\overrightarrow{GA}\times m_A{{\text{v}}^{\prime }}_A+\overrightarrow{GB}\times m_B{{\text{v}}^{\prime }}_B$ (2)

Substitute $0.2\text{m}$ for GA, $0.4\text{m}$ for GB, $\left(0.2\text{m}\times \text{10}\text{rad}/\text{s}\right)\text{k}$ for ${{\text{v}}^{\prime }}_A$, $3\text{kg}$ for $m_A$, $\left(0.4\text{m}\times \text{10}\text{rad}/\text{s}\right)\text{k}$ for ${{\text{v}}^{\prime }}_B$, and $1.5\text{kg}$ for $m_B$ in Equation (2).

$\begin{array}{c}{\left(H_G\right)}_0=\left(0.2\right)\times 3\times \left(0.2\times 10\right)\text{k}+\left(0.4\right)\times 1.5\times \left(0.4\times 10\right)\text{k}\\ \text{=}\left(\text{3}\text{.6}\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{k}\end{array}$

Find the angular moment about G using the equation as follows:

Refer problem 14.27,

${\left(H_O\right)}_0=\overrightarrow{\text{r}}\times {\text{m}\overline{\text{v}}}_0+{\left(H_G\right)}_0$ (3)

Substitute $2\text{j}$ for $\overrightarrow{\text{r}}$, $\left(\text{3}\text{.6}\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{k}$ for ${\left(H_G\right)}_0$, and $\text{5}\text{.4i}+4.32\text{j}$ for ${\text{m}\overline{\text{v}}}_0$ in Equation (3).

$\begin{array}{c}{\left(H_O\right)}_0=2\text{j}\times \left(5.4\text{i}+\text{4}\text{.32j}\right)+3.6\text{k}\\ \text{=}-\text{10}\text{.8k}+3.6\text{k}\\ \text{=}\left(-\text{7}\text{.2}\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{k}\end{array}$

Find the kinetic energy $\left(T_0\right)$ using the Equation as follows:

Refer to equation 14.29 in section 14.2A Kinetic energy of a System of particles in the textbook.

$\begin{array}{l}T=\frac{1}{2}m{\overline{v}}_0{}^2+\frac{1}{2}{\displaystyle \sum _{i=1}^n{m}_i{v^{\prime }}^2{}_i}\\ T=\frac{1}{2}m{\overline{v}}_0{}^2+\frac{1}{2}{\displaystyle \sum _{i=1}^n{m}_i{\left(\frac{1}{2}m{\overline{v}}_0^2+\frac{1}{2}{m}_A{v^{\prime }}_A^2+\frac{1}{2}{m}_B{v^{\prime }}_B^2\right)}^2}\end{array}$ (4).

Substitute $4.5\text{kg}$ for m, $\left(1.2\text{m}/\text{s}\right)\text{i}+\left(0.96\text{m}/\text{s}\right)\text{j}$ for ${\overline{v}}_0^2$, $3\text{kg}$ for $m_A$, $\left(0.2\text{m}\times \text{10}\text{rad}/\text{s}\right)\text{k}$ for ${{\text{v}}^{\prime }}_A$, $\left(0.4\text{m}\times \text{10}\text{rad}/\text{s}\right)\text{k}$ for ${{\text{v}}^{\prime }}_B$, and $1.5\text{kg}$ for $m_B$ in Equation (4).

$\begin{array}{c}{T}_0=\frac{1}{2}4.5\left[\left({1.2}^2+{0.96}^2\right)\right]+\frac{1}{2}\times 3\times {\left(0.2\times 10\right)}^2+\frac{1}{2}\times 1.5\times {\left(0.4\times 10\right)}^2\\ =5.3136+6+12\\ =23.314\text{J}\end{array}$

Sketch the system as shown in Figure 2.

Write the conservation of linear momentum as follows:

${\text{L}}_0=\text{L}$ (5)

Substitute $\text{5}\text{.4i}+4.32\text{j}$ for ${\text{L}}_0$ and $m_A{\text{v}}_A+m_B{\text{v}}_B$ for L in Equation (5).

$\text{5}\text{.4i}+4.32\text{j}=m_A{\text{v}}_A+m_B{\text{v}}_B$ (6)

Substitute $3\text{kg}$ for $m_A$, $1.5\text{kg}$ for $m_B$, $\left(v_A\text{j}\right)$ for ${\text{v}}_A$, and ${\left(v_B\right)}_z\text{i+}{\left(v_B\right)}_y\text{j}$ for ${\text{v}}_B$ in Equation (6).

$\text{5}\text{.4i}+4.32\text{j}=3\left(v_A\text{j}\right)+1.5{\left(v_B\right)}_z\text{i+}{\left(v_B\right)}_y\text{j}$

Equate the coefficient i as follows:

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

Equate the coefficient j as follows:

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

Express the conservation of energy as follows:

$\begin{array}{l}{T}_0=T\\ T_0=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2\end{array}$ (7)

Find the velocity $\left(v_A\right)$ of A after the cord breaks using the relation:

Substitute $23.314\text{J}$ for $T_0$, $3\text{kg}$ for $m_A$, $1.5\text{kg}$ for $m_B$, and ${\left(v_B\right)}_x+{\left(v_B\right)}_y$ for $v_B^2$  in Equation (7).

$23.314\text{J}=\frac{1}{2}\left(3\right)v_A^2+\frac{1}{2}\left(1.5\right)\left[{\left(v_B\right)}_x^2+{\left(v_B\right)}_y^2\right]$ (8)

Substitute $2.88-2v_A$ for ${\left(v_B\right)}_y$ and $3.6\text{m}/\text{s}$ for ${\left(v_B\right)}_x$ in Equation (8).

$\begin{array}{l}23.314\text{J}=\frac{1}{2}\left(3\right)v_A^2+\frac{1}{2}\left(1.5\right)\left[{\left(3.6\right)}^2+{\left(2.88-2v_A\right)}^2\right]\\ 23.314\text{J}=1.5v_A^2+0.75{\left(3.6\right)}^2{\left(2.88-2v_A\right)}^2\\ 4.5v_A^2-8.64v_A-7.373=0\\ v_A^2-1.92v_A-1.6389=0\end{array}$

Apply the quadratic formula as follows:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (9)

Substitute 1 for a, 1.92 for b, and $1.6389$ for c in Equation (9).

$\begin{array}{c}{v}_A=\frac{-\left(-1.92\right)\pm \sqrt{{1.92}^2-4\left(1\times -1.6389\right)}}{2\left(1\right)}\\ =\frac{1.92\pm \sqrt{10.24}}{2\left(1\right)}\\ =\frac{1.92\pm 3.2}{2\left(1\right)}\\ =0.96\pm 1.6\end{array}$

$\begin{array}{c}{v}_A=0.96+1.6=2.56\text{m}/\text{s}\\ v_A=0.96-1.6=-0.64\text{m}/\text{s}\end{array}$

$v_A=0.96-1.6=-0.64\text{m}/\text{s}$ Rejected because it’s shown directed up.

Thus, the velocity of A after the cord breaks is $\underset{\_}{2.56\text{m}/\text{s}}$.

Find the velocity ${\left(v_B\right)}_y$ as follows:

${\left(v_B\right)}_y=2.88-2v_A$

Substitute $2.56\text{m}/\text{s}$ for $v_A$.

$\begin{array}{c}{\left(v_B\right)}_y=2.88-2\left(2.56\right)\\ =2.88-5.12\\ =-2.24\text{m}/\text{s}\end{array}$

The velocity ${\left(v_B\right)}_x$ along x direction is $3.6\text{m}/\text{s}$.

Find the velocity $\left(v_B\right)$ of B after the cord breaks using the relation:

$\begin{array}{c}\left(v_B\right)=3.6\text{i}-\text{2}\text{.24j}\\ \text{=}\sqrt{{\left(3.6\right)}^2+{\left(-2.24\right)}^2}\\ =\sqrt{17.9776}\\ =4.24\text{m}/\text{s}\end{array}$

Thus, the velocity of B after the cord breaks is $\underset{\_}{4.24\text{m}/\text{s}}$ at an angle $31.9°$.

To determine

Find the distance a from the y axis to the path of A.

Answer to Problem 14.53P

The distance a from the y axis to the path of A is $\underset{\_}{2.34\text{m}}$.

Explanation of Solution

Calculation:

Find the distance a from the y axis to the path of A using the relation:

${\left(H_O\right)}_0=H_0$

Substitute $-7.2\text{k}$ for $H_O$ and $a\text{i}\times m_A{\text{v}}_A+b\text{i}\times m_B{\text{v}}_B$ for $H_0$.

$-7.2\text{k}=a\text{i}\times m_A{\text{v}}_A+b\text{i}\times m_B{\text{v}}_B$ (10)

Substitute $3\text{kg}$ for $m_A$, $2.56\text{i}$ for $b\text{i}$, $2.56\text{j}$ for ${\text{v}}_A$, $3.6\text{i}-\text{2}\text{.24j}$ for ${\text{v}}_B$, and $1.5\text{kg}$ for $m_B$ in Equation (10).

$\begin{array}{l}-7.2\text{k}=a\text{i}\times 3\left(2.56\text{j}\right)+7.5\text{i}\times 1.5\left(3.6\text{i}-\text{2}\text{.24j}\right)\\ -7.2\text{k}=7.68a\text{k}-\text{25}\text{.2k}\\ 18.3\text{k}=7.68a\text{k}\\ a=\text{2}\text{.34}\text{m}\end{array}$

Thus, the distance a from the y axis to the path of A is $\underset{\_}{2.34\text{m}}$.