Chapter 13.2, Problem 13.64P

A 1.2-kg collar can slide along the rod shown. It is attached to an elastic cord anchored at F. The cord has an undeformed length of 300 mm and a spring constant of 70 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B, (b) at E.

Fig. P13.64

To determine

Find the speed of the collar at B $\left(v_B\right)$.

Answer to Problem 13.64P

The speed of the collar at B $\left(v_B\right)$ is $\underset{\_}{2.71\text{m}/\text{s}}$.

Explanation of Solution

Given information:

The mass of the collar (m) is $1.2\text{kg}$.

The un-deformed length $\left(l_0\right)$ is $300\text{mm}$ or $\text{0}\text{.3m}$.

The spring constant (k) is $70\text{N}/\text{m}$.

The length of AB $\left(l_{AB}\right)$ is $500\text{mm}$ or $\text{0}\text{.5m}$.

The length of AE $\left(l_{AE}\right)$ is $400\text{mm}$ or $\text{0}\text{.4m}$.

The length of DF $\left(l_{DF}\right)$ is $300\text{mm}$ or $\text{0}\text{.3m}$.

The acceleration due to gravity (g) is $9.81\text{m}/{\text{s}}^2$.

Calculation:

Show the figure with the velocity as in Figure (1).

Calculate the elongated length of the elastic chord when the collar is at position A $\left(l_{AF}\right)$ using the relation:

$l_{AF}=\sqrt{{\left(l_{AB}\right)}^2+{\left(l_{AE}\right)}^2+{\left(l_{DF}\right)}^2}$

Substitute $\text{0}\text{.5m}$ for $l_{AB}$, $\text{0}\text{.4m}$ for $l_{AE}$ and $\text{0}\text{.3m}$ for $l_{DF}$.

$\begin{array}{c}{l}_{AF}=\sqrt{{\left(0.5\right)}^2+{\left(0.4\right)}^2+{\left(0.3\right)}^2}\\ =0.70711\text{m}\end{array}$

Calculate the elongated lengths of the elastic chord when the collar is at position B$\left(l_{BF}\right)$ using the relation:

$l_{BF}=\sqrt{{\left(l_{AE}\right)}^2+{\left(l_{DF}\right)}^2}$

Substitute $\text{0}\text{.4m}$ for $l_{AE}$ and $\text{0}\text{.3m}$ for $l_{DF}$.

$\begin{array}{c}{l}_{BF}=\sqrt{{\left(0.4\right)}^2+{\left(0.3\right)}^2}\\ =0.5\text{m}\end{array}$

Calculate the elongation of the elastic cord when the collar is at position A $\left(\Delta L_{AF}\right)$

$\Delta L_{AF}=l_{AF}-l_0$

Substitute $\text{0}\text{.70711}\text{m}$ for $l_{AF}$ and $0.3\text{m}$ for $l_0$.

$\begin{array}{c}\Delta L_{AF}=0.70711\text{m}-0.3\text{m}\\ =0.40711\text{m}\end{array}$

Calculate the elongation of the elastic cord when collar is in position B $\left(\Delta L_{BF}\right)$ using the relation:

$\Delta L_{BF}=l_{BF}-l_0$

Substitute $0.5\text{m}$ for $l_{AF}$ and $0.3\text{m}$ for $l_0$.

$\begin{array}{c}\Delta L_{BF}=0.5\text{m}-0.30\text{m}\\ =0.2\text{m}\end{array}$

Calculate the potential energy stored in the cord due to elongation ${\left(V_A\right)}_e$ using the formula:

${\left(V_A\right)}_e=\frac{1}{2}k\left(\Delta L_{AF}^2\right)$

Substitute $70\text{N}/\text{m}$ for $k$ and $0.40711\text{m}$ for $\Delta L_{AF}$.

$\begin{array}{c}{\left(V_A\right)}_e=\frac{1}{2}\times 70\times {\left(0.40711\right)}^2\\ =5.800\text{J}\end{array}$

Calculate the potential energy stored in the collar due to gravitation ${\left(V_A\right)}_g$ using the formula:

${\left(V_A\right)}_g=Wh$

Substitute $mg$ for W and $l_{AE}$ for h.

${\left(V_A\right)}_g=mg{l}_{AE}$

Substitute $1.2\text{kg}$ for m, $9.81\text{m}/{\text{s}}^2$ for g and $\text{0}\text{.4m}$ for $l_{AE}$.

$\begin{array}{c}{\left(V_A\right)}_g=1.2\times 9.8\times 0.4\\ =4.71\text{J}\end{array}$

Calculate the potential energy stored in the elastic cord when collar is at position A $\left(V_A\right)$ using the relation:

$V_A={\left(V_A\right)}_e+{\left(V_A\right)}_g$

Substitute $5.800\text{J}$ for ${\left(V_A\right)}_e$ and $4.71\text{J}$ for ${\left(V_A\right)}_g$.

$\begin{array}{c}{V}_A=5.800\text{J}+4.71\text{J}\\ \text{=10}\text{.51J}\end{array}$

Calculate the potential energy stored in the cord due to elongation ${\left(V_B\right)}_e$ using the formula:

${\left(V_B\right)}_e=\frac{1}{2}k\left(\Delta L_{BF}^2\right)$

Substitute $70\text{N}/\text{m}$ for $k$ and $0.2\text{m}$ for $\Delta L_{BF}$.

$\begin{array}{c}{\left(V_B\right)}_e=\frac{1}{2}\times 70\times {\left(0.2\right)}^2\\ =1.4\text{J}\end{array}$

Calculate the potential energy stored in the collar due to gravitation ${\left(V_B\right)}_g$ using the formula:

${\left(V_B\right)}_g=Wh$

Substitute $mg$ for W and $l_{AE}$ for h.

${\left(V_B\right)}_g=mg{l}_{AE}$

Substitute $1.2\text{kg}$ for m, $9.81\text{m}/{\text{s}}^2$ for g and $\text{0}\text{.4m}$ for $l_{AE}$.

$\begin{array}{c}{\left(V_B\right)}_g=1.2\times 9.8\times 0.4\\ =4.71\text{J}\end{array}$

Calculate the potential energy stored in the elastic cord when collar is at position B $\left(V_B\right)$ using the relation:

$V_B={\left(V_B\right)}_e+{\left(V_B\right)}_g$

Substitute $1.4\text{J}$ for ${\left(V_B\right)}_e$ and $4.71\text{J}$ for ${\left(V_B\right)}_g$.

$\begin{array}{c}{V}_B=1.4+4.71\\ =6.11\text{J}\end{array}$

Here, the collar is at rest at position A. Hence, $T_A=0$.

Calculate the kinetic energy of the collar at position B $\left(T_B\right)$ using the formula:

$T_B=\frac{1}{2}m{v}_B^2$

Here, velocity of the collar at position B is $v_B$.

Substitute $1.2\text{kg}$ for $m$.

$\begin{array}{c}{T}_B=\frac{1}{2}\left(1.2\right)v_B^2\\ =0.6v_B^2\end{array}$

Use the principle of conservation of energy to collar at position A and position B, to calculate the velocity of collar at position B. The principle of conservation of energy states that sum of the kinetic and potential energy of a particle remains constant.

Find the speed of the collar at B $\left(v_B\right)$:

$T_A+V_A=T_B+V_B$

Substitute 0 for $T_A$, $\text{10}\text{.51J}$ for $V_A$, $0.6v_B^2$ for $T_B$, and $6.11\text{J}$ for $V_B$.

$\begin{array}{l}0+\text{10}\text{.51J}=0.6v_B^2+6.11\text{J}\\ 0.6v_B^2=10.51-6.11\\ 0.6v_B^2=4.4\\ v_B=\sqrt{\frac{4.4}{0.6}}\\ v_B=2.71\text{m}/\text{s}\end{array}$

Therefore, the speed of the collar at B $\left(v_B\right)$ is $\underset{\_}{2.71\text{m}/\text{s}}$.

To determine

Find the speed of the collar at E $\left(v_E\right)$.

Answer to Problem 13.64P

The speed of the collar at E $\left(v_E\right)$ is $\underset{\_}{3.6\text{m}/\text{s}}$.

Explanation of Solution

Given information:

The mass of the collar (m) is $1.2\text{kg}$.

The un-deformed length $\left(l_0\right)$ is $300\text{mm}$ or $\text{0}\text{.3m}$.

The spring constant (k) is $70\text{N}/\text{m}$.

The length of AB $\left(l_{AB}\right)$ is $500\text{mm}$ or $\text{0}\text{.5m}$.

The length of AE $\left(l_{AE}\right)$ is $400\text{mm}$ or $\text{0}\text{.4m}$.

The length of DF $\left(l_{DF}\right)$ is $300\text{mm}$ or $\text{0}\text{.3m}$.

The acceleration due to gravity (g) is $9.81\text{m}/{\text{s}}^2$.

Calculation:

Calculate the elongated length of the elastic chord when the collar is at position E$\left(l_{EF}\right)$ using the relation:

$l_{EF}=\sqrt{{\left(l_{AB}\right)}^2+{\left(l_{DF}\right)}^2}$

Substitute $\text{0}\text{.5m}$ for $l_{AB}$ and $\text{0}\text{.3m}$ for $l_{DF}$.

$\begin{array}{c}{l}_{EF}=\sqrt{{\left(0.5\right)}^2+{\left(0.3\right)}^2}\\ =0.58309\text{m}\end{array}$

Calculate the elongation of the elastic cord when collar is in position E $\left(\Delta L_{EF}\right)$ using the relation:

$\Delta L_{EF}=l_{EF}-l_0$

Substitute $0.58309\text{m}$ for $l_{AF}$ and $0.3\text{m}$ for $l_0$.

$\begin{array}{c}\Delta L_{EF}=0.58309\text{m}-0.3\text{m}\\ =0.28309\text{m}\end{array}$

Calculate the potential energy stored in the cord due to elongation ${\left(V_E\right)}_e$ using the formula:

${\left(V_E\right)}_e=\frac{1}{2}k\left(\Delta L_{EF}^2\right)$

Substitute $70\text{N}/\text{m}$ for $k$ and $0.28309\text{m}$ for $\Delta L_{EF}$.

$\begin{array}{c}{\left(V_E\right)}_e=\frac{1}{2}\times 70\times {\left(0.28309\right)}^2\\ =2.805\text{J}\end{array}$

Calculate the potential energy stored in the collar due to gravitation ${\left(V_E\right)}_g$ using the formula:

${\left(V_E\right)}_g=Wh$

Substitute $mg$ for W and 0 for h.

$\begin{array}{c}{\left(V_E\right)}_g=mg\left(0\right)\\ =0\end{array}$

Calculate the potential energy stored in the elastic cord when collar is at position E $\left(V_E\right)$ using the relation:

$V_E={\left(V_E\right)}_e+{\left(V_E\right)}_g$

Substitute $2.805\text{J}$ for ${\left(V_E\right)}_e$ and 0 for ${\left(V_E\right)}_g$.

$\begin{array}{c}{V}_E=2.805\text{J}+0\\ =2.805\text{J}\end{array}$

Calculate the kinetic energy of the collar at position E $\left(T_E\right)$ using the formula:

$T_E=\frac{1}{2}m{v}_E^2$

Here, velocity of the collar at position E is $v_E$.

Substitute $1.2\text{kg}$ for $m$.

$\begin{array}{c}{T}_E=\frac{1}{2}\left(1.2\right)v_E^2\\ =0.6v_E^2\end{array}$

Apply the principle of conservation of energy to collar at position A and position E to calculate the velocity of the collar at position E.

Find the speed of the collar at E $\left(v_E\right)$:

$T_A+V_A=T_E+V_E$

Substitute 0 for $T_A$, $\text{10}\text{.51J}$ for $V_A$, $0.6v_E^2$ for $T_E$, and $2.805\text{J}$ for $V_E$.

$\begin{array}{l}0+10.51=0.6v_E^2+2.805\text{J}\\ 0.6v_E^2=10.51-2.805\\ 0.6v_E^2=7.705\end{array}$

$\begin{array}{l}{v}_E=\sqrt{\frac{7.705}{0.6}}\\ v_E=3.6\text{m}/\text{s}\end{array}$

Therefore, the speed of the collar at E $\left(v_E\right)$ is $\underset{\_}{3.6\text{m}/\text{s}}$.