Chapter 18, Problem 18.147RP

To determine

The distance of weight from disc $B$ to balance the system dynamically.

The distance of weight from disc $C$ to balance the system dynamically.

Answer to Problem 18.147RP

The distance of weight from disc $B$ to balance the system dynamically is $\frac{25}{4}\text{in}$.

The distance of weight from disc $C$ to balance the system dynamically is $\frac{25}{8}\text{in}$.

Explanation of Solution

Write the expression for the force.

$F=mr{\omega }^2$

Here, the mass is $m$, the speed is $\omega$ and the radius is $r$.

Write the expression for the force on disk $A$.

$F_A=m{r}_A{\omega }^2$

Here, the radius of disc $A$ is $r_A$.

Write the expression for the force on disk $B$.

$F_B=m^{\prime }{r}_B{\omega }^2$

Here, the radius of disc $B$ is $r_B$ and the balancing mass is $m^{\prime }$.

Write the expression for the force on disk $C$.

$F_C=m^{\prime }{r}_C{\omega }^2$

Here, the radius of disc $C$ is $r_C$.

Write the expression for the sum of effective forces.

$F_A+F_B+F_C=0$ ...... (I)

Here, the force on the rotor is $A$ is $F_A$, the force on the rotor is $B$ is $F_B$, the force on the rotor is $C$ is $F_C$.

Write the expression for the moment balance.

$a{F}_A+b{F}_B+c{F}_C=0$ ...... (II)

Conclusion:

Substitute $-m{r}_A{\omega }^2$ for $F_A$, $m^{\prime }{r}_B{\omega }^2$ for $F_B$ and $m^{\prime }{r}_C{\omega }^2$ for $F_C$ in Equation (I).

$-m{r}_A{\omega }^2+m^{\prime }{r}_B{\omega }^2+m^{\prime }{r}_C{\omega }^2=0$ ...... (III)

Substitute $-m{r}_A{\omega }^2$ for $F_A$, $m^{\prime }{r}_B{\omega }^2$ for $F_B$ and $m^{\prime }{r}_C{\omega }^2$ for $F_C$ in Equation (I).

$-am{r}_A{\omega }^2+b{m}^{\prime }{r}_B{\omega }^2+c{m}^{\prime }{r}_C{\omega }^2=0$ ...... (IV)

Substitute $\frac{1}{4}\text{in}$ for $r_A$, $25\text{lb}$ for $m$ and $2\text{lb}$ for $m^{\prime }$.

$\begin{array}{l}\left(-25\text{lb}\times \frac{1}{4}\text{in}\right){\omega }^2+\left(2\text{lb}\times r_B\right){\omega }^2+\left(2\text{lb}\times r_C\right){\omega }^2=0\\ r_B+r_C=\frac{25\text{lb}}{2\text{lb}}\left(\frac{1}{4}\text{in}\right)\\ r_B+r_C=3.125\text{in}\end{array}$ ...... (V)

Substitute $4\text{in}$ for $a$, $10\text{in}$ for b, $16\text{in}$ for $c$, $\frac{1}{4}\text{in}$ for $r_A$, $25\text{lb}$ for $m$ and $2\text{lb}$ for $m^{\prime }$ .in Equation (IV).

$\begin{array}{l}-4\text{in}\times 25\text{lb}\times \frac{1}{4}\text{in}\times {\omega }^2+10\text{in}\times 2\text{lb}\times r_B{\omega }^2+16\text{in}\times 2\text{lb}\times r_C{\omega }^2=0\\ 25{\text{lbin}}^2{\omega }^2+20\text{l}{\text{bin}}^2{r}_B{\omega }^2+32{\text{lbin}}^2{r}_C{\omega }^2=0\\ 10r_B+16r_C=12.5\text{in}\end{array}$ ...... (VI)

Solve Equation (V) and Equation (VI).

$r_C=\frac{25}{8}\text{in}$

$r_B=\frac{25}{4}\text{in}$

The distance of weight from disc $B$ to balance the system dynamically is $\frac{25}{4}\text{in}$.

The distance of weight from disc $C$ to balance the system dynamically is $\frac{25}{8}\text{in}$.