Chapter 18, Problem 18.147RP

Three 25-lb rotor disks are attached to a shaft that rotates at 720 rpm. Disk A is attached eccentrically so that its mass center is $\frac{1}{4}\text{in}$. from the axis of rotation, while disks B and C are attached so that their mass centers coincide with the axis of rotation. Where should 2-lb weights be bolted to disks B and C to balance the system dynamically?

Fig. P18.147

To determine

Where should 2 lb weights be bolted to disks B and C to balance the system dynamically.

Answer to Problem 18.147RP

The 2 lb weights be bolted to disks B and C to balance the system dynamically are $\underset{\_}{6\frac{1}{4}\text{in}\text{.}\text{}\left(\text{below}\text{shaft}\right)}$ and $\underset{\_}{3\frac{1}{8}\text{in}\text{.}\text{}\left(\text{above}\text{shaft}\right)}$.

Explanation of Solution

Given Information:

The weight (W) of the disk is 25 lb.

The shaft rotates at 720 rpm.

The mass center of the disk A is $\frac{1}{4}\text{in}\text{.}$

Calculation:

Draw the free body diagram of the system as shown in Figure (1).

The angular velocity of the shaft is constant. So, the rate of change of angular momentum is zero.

Express the effective force $\left(F_A\right)$ of disk A:

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

Here,$\omega$ is angular velocity.

Express the effective force $\left(F_B\right)$ of disk B:

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

Express the effective force $\left(F_C\right)$ of disk C:

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

The sum of the effective forces of the discs should be equal to zero for the system to be in dynamic balance.

$F_A+F_B+F_C=0$

Substitute $-m{r}_A{\omega }^{2}j$ for $F_A$, $m^{\prime }{r}_B{\omega }^{2}j$ for $F_B$, and $m^{\prime }{r}_C{\omega }^{2}j$ for $F_C$.

$\begin{array}{l}-m{r}_A{\omega }^{2}j+m^{\prime }{r}_B{\omega }^{2}j+m^{\prime }{r}_C{\omega }^{2}j=0\\ m^{\prime }{r}_B{\omega }^{2}j+m^{\prime }{r}_C{\omega }^{2}j=m{r}_A{\omega }^{2}j\\ m^{\prime }{\omega }^2\left(r_B+r_C\right)=m{r}_A{\omega }^{2}j\\ r_B+r_C=\frac{m}{m^{\prime }}{r}_A\end{array}$

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

$\begin{array}{c}{r}_B+r_C=\frac{25\text{lb}}{2\text{lb}}\left(\frac{1}{4}\text{in}\text{.}\right)\\ r_B+r_C=\frac{25}{8}\\ r_B+r_C=3.125\text{in}\text{.}\end{array}$ (1)

According to principle of impulse momentum, the total momentum is zero.

Express the moment about O:

$a\left(-m{r}_A{\omega }^{2}j\right)+b\left(m^{\prime }{r}_B{\omega }^{2}j\right)+c\left(m^{\prime }{r}_C{\omega }^{2}j\right)=0$

Here, $a$ is distance between O and A, b is distance between O and B and c distance between O and C .

Substitute $4i$ for a, $10i$ for b, and $16i$ for c.

$\begin{array}{l}4i\left(-m{r}_A{\omega }^{2}j\right)+10i\left(m^{\prime }{r}_B{\omega }^{2}j\right)+16i\left(m^{\prime }{r}_C{\omega }^{2}j\right)=0\\ 10i\left(m^{\prime }{r}_B{\omega }^{2}j\right)+16i\left(m^{\prime }{r}_C{\omega }^{2}j\right)=4i\left(m{r}_A{\omega }^{2}j\right)\\ m^{\prime }{\omega }^2\left(10r_B+16r_C\right)=4\left(m{r}_A{\omega }^2\right)\\ 10r_B+16r_C=4\frac{m}{m^{\prime }}{r}_A\end{array}$

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

$\begin{array}{c}10r_B+16r_C=4\frac{25\text{lb}}{2\text{lb}}\left(\frac{1}{4}\text{in}\text{.}\right)\\ 10r_B+16r_C=12.5\text{in}\text{.}\end{array}$ (2)

Multiply Equation (1) by $-10$ and add to Equation (2).

$\begin{array}{l}-10r_B-10r_C+10r_B+16r_C=12.5\text{in}\text{.}-31.25\text{in}\text{.}\\ 6r_C=-18.75\\ r_C=-\frac{18.75}{6}\\ r_C=-3.125\text{in}\text{.}\\ r_C=-\frac{25}{8}\text{in}\text{.}\\ r_C=-3\frac{1}{8}\text{in}\text{.}\end{array}$

The 2 lb weight is be placed above the shaft for disk C.

Substitute $3.125\text{in}\text{.}$ for $r_C$ in Equation (1).

$\begin{array}{l}{r}_B-3.125\text{in}.=3.125\text{in}\text{.}\\ r_B=6.25\text{in}\text{.}\\ r_B=\frac{25}{4}\text{in}\text{.}\\ r_B=6\frac{1}{4}\text{in}\text{.}\end{array}$

The 2 lb weight is be placed below the shaft for disk B.

Thus, the 2 lb weights be bolted to disks B and C to balance the system dynamically are $\underset{\_}{6\frac{1}{4}\text{in}\text{.}\text{}\left(\text{below}\text{shaft}\right)}$ and $\underset{\_}{3\frac{1}{8}\text{in}\text{.}\text{}\left(\text{above}\text{shaft}\right)}$.