Chapter 13.2, Problem 13.64P

A 2-kg collar is attached to a spring and slides without friction in a vertical plane along the curved rod ABC. The spring is undeformed when the collar is at C and its constant is 600 N/m. if the collar is released at A with no initial velocity, determine its velocity (a) as it passes through B, (b) as it reaches C.

    

To determine

(a)

Velocity of collar as it passes through $B$.

Answer to Problem 13.64P

Velocity of collar as it passes through $B$ is $2.485\text{ m/s}$.

Explanation of Solution

Given:

Mass of collar is $2\text{ kg}$.

Spring constant is $600\text{N/m}$.

Initial velocity at $A$ is $0\text{ m/s}$.

Concept used:

Refer to Fig.P13.64;

Calculate the elongation of the spring when collar is at point $A$.

$e_A=l_{OA}-l_{OC}$   ...... (1)

Here, $e_A$ is the elongation in spring at point $A$, $l_{OA}$ is the length $OA$ and $l_{OC}$ is the length $OC$.

Calculate the elongation of the spring when collar is at point $B$.

$e_B=l_{OB}-l_{OC}$   ...... (2)

Here, $e_B$ is the elongation in spring at point $B$, $l_{OB}$ is the length $OB$ and $l_{OC}$ is the length $OC$.

Calculate the elongation of the spring when collar is at point $C$.

$e_C=0$

Here, $e_C$ is the elongation of spring at $C$.

Write the expression for Potential energy of spring at point $A$.

$V_A=\frac{1}{2}k{e}_A{}^2$   ...... (3)

Here, $V_A$ is the potential energy at point $A$ and $k$ is the spring constant.

Write the expression for Potential energy of spring at point $B$.

$V_B=\frac{1}{2}k{e}_B{}^2$   ...... (4)

Here, $V_B$ is the potential energy at point $B$.

Write the expression for Potential energy of spring at point $A$.

$V_C=\frac{1}{2}k{e}_C{}^2$   ...... (5)

Here, $V_C$ is the potential energy at point $C$.

Write the expression for the kinetic energy of collar at point $A$.

$T_A=\frac{1}{2}m{v}_A{}^2$   ...... (6)

Here, $T_A$ is the kinetic energy of collar at point $A$, $m$ is the mass of collar and $v_A$ is the velocity of collar at point $A$.

Write the expression for the kinetic energy of collar at point $B$.

$T_B=\frac{1}{2}m{v}_B{}^2$   ...... (7)

Here, $T_B$ is the kinetic energy of collar at point $B$ and $v_B$ is the velocity of collar at point $C$.

Write the expression for the kinetic energy of collar at point $C$.

$T_C=\frac{1}{2}m{v}_C{}^2$   ...... (8)

Here, $T_C$ is the kinetic energy of collar at point $C$ and $v_C$ is the velocity of collar at point $C$.

Consider the datum to be the surface.

Write the expression for the gravitational potential energy at point $B$.

$V_{Bg}=-mg\left(l_{OB}\right)$   ...... (9)

Here, $V_{Bg}$ is the gravitational potential energy at point $B$ and $g$ is acceleration due to gravity.

The gravitational potential energy at point $A$ and $C$ will be zero as they are on datum.

Write the expression of conservation of energy for the system at point $B$.

$T_A+V_A+V_{Ag}=T_B+V_{B\text{}}+V_{Bg}$   ...... (10)

Calculation:

Substitute $250\text{ mm}$ for $l_{OA}$ and $150\text{ mm}$ for $l_{OC}$ in equation (1).

$\begin{array}{c}{e}_A=250-150\\ =100\text{ mm}\end{array}$

Substitute $200\text{ mm}$ for $l_{OB}$ and $150\text{ mm}$ for $l_{OC}$ in equation (2).

$\begin{array}{c}{e}_B=200-150\\ =50\text{mm}\end{array}$

Substitute $600\text{N/m}$ for $k$ and $100\text{ mm}$ for $e_A$ in equation (3).

$\begin{array}{c}{V}_A=\frac{1}{2}\left(600\right){\left(100\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)}^2\\ =3\text{ J}\end{array}$

Substitute $600\text{N/m}$ for $k$ and $50\text{ mm}$ for $e_B$ in equation (4).

$\begin{array}{c}{V}_B=\frac{1}{2}\left(600\right){\left(50\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)}^2\\ =0.75\text{ J}\end{array}$

Substitute $600\text{N/m}$ for $k$ and $0\text{ mm}$ for $e_C$ in equation (5).

$\begin{array}{c}{V}_C=\frac{1}{2}\left(600\right){\left(0\text{ mm}\right)}^2\\ =0\text{ J}\end{array}$

Substitute $2\text{ kg}$ for $m$ and $0\text{ m/s}$ for $v_A$ in equation (6).

$\begin{array}{c}{T}_A=\frac{1}{2}\left(2\text{ kg}\right)\left(0\text{ m/s}\right)\\ =0\text{J}\end{array}$

Substitute $2\text{ kg}$ for $m$ in equation (7).

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

Substitute $2\text{ kg}$ for $m$, $200\text{ mm}$ for $l_{OB}$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (9).

$\begin{array}{c}{V}_{Bg}=-\left(2\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\left(200\text{mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)\\ =-3.924\text{J}\end{array}$

Substitute $0$ for $T_A$, $3\text{ J}$ for $V_A$, $0$ for $V_{Ag}$, $v_B{}^2$ for $T_B$, $0.75\text{ J}$ for $V_B$ and $-3.924\text{ J}$ for $V_{Bg}$ in equation (10).

$0+3+0=v_B{}^2+0.75-3.924$

Simplify the above expression for $v_B$.

$v_B=2.485\text{ m/s}$

Velocity of collar as it passes through $B$ is $2.485\text{ m/s}$.

Conclusion:

Thus, Velocity of collar as it passes through $B$ is $2.485\text{ m/s}$.

To determine

(b)

Velocity of collar as it passes through $C$.

Answer to Problem 13.64P

Velocity of collar as it passes through $C$ is $1.732\text{ m/s}$.

Explanation of Solution

Write the expression of conservation of energy for the system at point $C$.

$T_A+V_A+V_{Ag}=T_C+V_{C\text{}}+V_{Cg}$   ...... (11)

Calculation:

Substitute $2\text{ kg}$ for $m$ in equation (8).

$\begin{array}{c}{T}_A=\frac{1}{2}\left(2\text{ kg}\right)v_c{}^2\\ =v_c{}^2\end{array}$

Substitute $0$ for $T_A$, $3\text{ J}$ for $V_A$, $0$ for $V_{Ag}$, $v_c{}^2$ for $T_C$, $0\text{ J}$ for $V_C$ and $0\text{ J}$ for $V_{Cg}$ in equation (11).

$0+3+0=v_c{}^2+0+0$

Simplify the above expression for $v_c$.

$v_c=1.732\text{ m/s}$

Velocity of collar as it passes through $C$ is $1.732\text{ m/s}$.

Conclusion:

Thus, Velocity of collar as it passes through $C$ is $1.732\text{ m/s}$.