Chapter 19.4, Problem 19.107P

To determine

(a)

Magnitude of maximum angular velocity of bar $AB$.

Answer to Problem 19.107P

The maximum angular velocity of bar AB is $0.45\text{ rad/s}$.

Explanation of Solution

Given:

Mass of attached sphere is $2\text{kg}$.

Spring constant is $3.6\text{ kN/m}$.

Vertical deflection is ${\delta }_m\sin \left({\omega }_{f}t\right)$.

Maximum amplitude of deflection is $3\text{ mm}$.

Frequency of rotation is $15\text{ rad/s}$.

Concept used:

Draw the FBD for bar AB in equilibrium state.

After giving small deflection to bar AB, Draw FBD of bar AB

Draw the equivalent kinetic diagram for bar AB as.

Write the expression for displacement of point C in terms of angular displacement.

$x_c=L\theta$ ...... (1)

Here, $x_c$ is the displacement of point C, $L$ is the length of AC and $\theta$ is small angular displacement.

Write the expression for displacement of point B in terms of angular displacement.

$x_B=2L\theta$ ...... (2)

Here, $x_B$ is the displacement of point B and $2L$ is the length of AB.

Substitute $x_c$ for $L\theta$ in equation (2).

$x_B=2x_c$ ...... (3)

Take moment about point A for equilibrium position.

$\sum M_A=mg\left(2L\right)-k\Delta L$

For static equilibrium substitute $0$ for $\sum M_A$ in above expression,

$0=2mgL-k\Delta L$

Simplify the above expression.

$2mg=k\Delta$ ...... (4)

Here, $\Delta$ is the deflection in spring and $g$ is the acceleration due to gravity.

Take moment about point A after giving the angular displacement.

$\sum M_A={\left(\sum M\right)}_{\text{kinetic}}$

Substitute $k\left(x_c-\Delta -\delta \right)\left(L\right)+mg\left(2L\right)$ for $\sum M_A$ and $-ma\left(2L\right)$ for ${\left(\sum M\right)}_{\text{kinetic}}$ in above expression.

$\begin{array}{l}k\left(x_c-\Delta -\delta \right)\left(L\right)+mg\left(2L\right)=-ma\left(2L\right)\\ k\left(x_c-\delta \right)L-k\Delta L+2mgL=-ma2L\end{array}$

Substitute $2mg$ for $k\Delta$ in above expression.

$\begin{array}{l}k\left(x_c-\delta \right)L=-2maL\\ k\left(x_c-\delta \right)=-2ma\end{array}$

Substitute ${\ddot{x}}_B$ for $a$ in above expression.

$\begin{array}{l}-2m{\ddot{x}}_B=k\left(x_c-\delta \right)\\ -2m{\ddot{x}}_B=k{x}_c-k\delta \end{array}$

Substitute $\frac{x_B}{2}$ for $x_c$ in above expression.

$\begin{array}{l}-2m{\ddot{x}}_B=\frac{k\left(x_B\right)L}{2}-k\delta \\ -4m{\ddot{x}}_B=k\left(x_B\right)L-2k\delta \end{array}$

Substitute ${\delta }_m\sin \left({\omega }_{f}t\right)$ for $\delta$ in above expression.

$-4m{\ddot{x}}_B=k\left(x_B\right)L-2k\left({\delta }_m\sin \left({\omega }_{f}t\right)\right)$

Rearrange the above expression as.

$4m{\ddot{x}}_B+k\left(x_B\right)=2k{\delta }_m\sin \left({\omega }_{f}t\right)$ ....... (5)

Here, ${\ddot{x}}_B$ is the acceleration of point B, $t$ is time and ${\delta }_m$ is the Maximum amplitude of deflection.

Write the standard equation of motion as.

$m_{eq}\ddot{x}+k_{eq}x=F$ ...... (6)

Compare the coefficient of $\ddot{x}$ in equation (5) and (6).

$m_{eq}=4m$ ...... (7)

Compare the coefficient of $x$ in equation (5) and (6).

$k_{eq}=k$ ...... (8)

Write the expression for maximum amplitude of force.

$P_m=2k{\delta }_m$ ...... (9)

Here, $P_m$ is the maximum amplitude of force.

Write the expression for natural frequency of system.

${\omega }_n=\sqrt{\frac{k_{eq}}{m_{eq}}}$ ...... (10)

Here, ${\omega }_n$ is the natural frequency of system, $k_{eq}$ is the equivalent stiffness of system and $m_{eq}$ is the equivalent mass of the system.

Write the expression for the amplitude of vibration of sphere B.

$x_{B\max }=\left(\frac{P_m}{k}\right)\left(\frac{1}{1-\left(\frac{{\omega }_f^2}{{\omega }_n^2}\right)}\right)$ ...... (11)

Here, $x_{B\max }$ is the maximum amplitude of vibration at point B.

Write the expression for magnitude of maximum angular velocity of bar AB.

${\omega }_{\max }=\frac{x_{B\max }{\omega }_f}{2L}$ ...... (12)

Here, ${\omega }_{\max }$ is the maximum angular velocity of sphere.

Calculation:

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

$\begin{array}{c}{m}_{eq}=\left(4\right)\left(2\right)\\ =8\text{ kg}\end{array}$

Substitute $3.6\text{ kN/m}$ for $k$ in equation (8).

$k_{eq}=3.6\text{ kN/m}$

Substitute $3.6\text{ kN/m}$ for $k$ and $3\text{ mm}$ for ${\delta }_m$ in equation (9).

$\begin{array}{c}{P}_m=2\left(3.6\text{ kN/m}\left(\frac{1000\text{ N}}{1\text{ kN}}\right)\right)\left(3\text{ mm}\left(\frac{1\text{ m}}{1000\text{mm}}\right)\right)\\ =21.6\text{N}\end{array}$

Substitute $3.6\text{ kN/m}$ for $k$ and $8\text{ kg}$ for $m$ in equation (10).

$\begin{array}{c}{\omega }_n=\sqrt{\frac{\left(3.6\text{ kN/m}\left(\frac{1000\text{ N}}{1\text{ kN}}\right)\right)}{8}}\\ =21.213\text{ rad/s}\end{array}$

Substitute $21.6\text{ N}$ for $P_m$, $3.6\text{ kN/m}$ for $k$, $21.213\text{ rad/s}$ for ${\omega }_n$ and $15\text{ rad/s}$ for ${\omega }_f$ in equation (11).

$\begin{array}{c}{x}_{B\max }=\frac{21.6}{\left(3.6\text{ kN/m}\left(\frac{1000\text{ N}}{1\text{ kN}}\right)\right)}\left(\frac{1}{\left(1-\frac{{15}^2}{{21.213}^2}\right)}\right)\\ =0.012\text{ m}\end{array}$

Substitute $0.012\text{ m}$ for $x_{B\max }$, $15\text{ rad/s}$ for ${\omega }_f$ and $0.2\text{ m}$ for $L$ in equation (12).

$\begin{array}{c}{\omega }_{\max }=\frac{\left(0.012\right)\left(15\right)}{2\left(0.2\right)}\\ =0.45\text{ rad/s}\end{array}$

The maximum angular velocity of bar AB is $0.45\text{ rad/s}$.

Conclusion:

Thus, the maximum angular velocity of bar AB is $0.45\text{ rad/s}$.

To determine

(b)

Magnitude of maximum acceleration of sphere $B$.

Answer to Problem 19.107P

The maximum acceleration of sphere is $2.7{\text{ m/s}}^2$.

Explanation of Solution

Concept used:

Write the expression for maximum acceleration of sphere.

$a_{B\max }=x_{B\max }{\omega }_f^2$ ...... (13)

Here, $a_{B\max }$ is the maximum acceleration of sphere.

Calculation:

Substitute $0.012\text{m}$ for $x_{B\max }$ and $15\text{ rad/s}$ for ${\omega }_f$ in equation (13).

$\begin{array}{c}{a}_{B\max }=0.012{\left(15\right)}^2\\ =2.7{\text{ m/s}}^2\end{array}$

The maximum acceleration of sphere is $2.7{\text{ m/s}}^2$.

Conclusion:

Thus, the maximum acceleration of sphere is $2.7{\text{ m/s}}^2$.