Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 2.19

Transfer Function—Two Equations of Motion

PROBLEM: Find the transfer function, $\theta _{2}(s)/T(s)$, for the rotational system shown in Figure 2.22(a). The rod is supported by bearings at either end and is undergoing torsion. A torque is applied at the left, and the displacement is measured at the right.

## Verified Solution

First, obtain the schematic from the physical system. Even though torsion
occurs throughout the rod in Figure 2.22(a),$\overset{9}{}$ we approximate the system by assuming that the torsion acts like a spring concentrated at one particular point in the rod, with an
inertia $J_{1}$ to the left and an inertia $J_{2}$ to the right.$\overset{10}{}$ We also assume that the damping inside the flexible shaft is negligible. The schematic is shown in Figure 2.22(b). There are two degrees of freedom, since each inertia can be rotated while the other is held still .Hence, it will take two simultaneous equations to solve the system.

Next, draw a free-body diagram of $J_{1}$ using superposition. Figure 2.23(a) shows the torques on $J_{1}$ if $J_{2}$ is held still and $J_{1}$ rotated. Figure 2.23(b) shows the torques on $J_{1}$ if $J_{1}$ is held still and $J_{2}$ rotated. Finally, the sum of Figures 2.23(a) and 2.23(b) is shown in Figure 2.23(c), the final free-body diagram for $J_{1}$ . The same process is repeated in
Figure 2.24 for $J_{2}$ .

Summing torques, respectively, from Figures 2.23(c) and 2.24(c) we obtain the equations of motion,

$(J_{1}s^{2}+D_{1}s+K )\theta _{1} (s)$          $-K\theta _{2}(s)=T(s)$

$-K\theta _{1} (s)+(J_{2}s^{2}+D_{2}s+K)\theta _{2}(s)=0$

from which the required transfer function is found to be

$\frac{\theta _{2}(s) }{T(s)} =\frac{K}{\Delta }$

as shown in Figure 2.22(c), where

$\Delta =\left|\begin{matrix} (J_{1}s^{2}+D_{1}s+K ) & -K \\ -K &(J_{2}s^{2}+D_{2}s+K ) \end{matrix} \right|$

Notice that Eqs. (2.127) have that now well-known form

$\left[\begin{matrix} Sum of \\ impedances \\ connected \\ to the motion \\ at \theta _{1} \end{matrix} \right] \theta _{1}(s)-\left[\begin{matrix} Sum of \\ impedances \\ between \\ \theta _{1} and \theta _{2} \end{matrix} \right] \theta _{2}(s)=\left [\begin{matrix} Sum of \\ applied torques \\ at \theta _{1} \end{matrix} \right]$

$-\left[\begin{matrix} Sum of \\ impedances \\ between \\ \theta _{1} and \theta _{2} \end{matrix} \right] \theta _{1}(s)+\left [\begin{matrix} Sum of \\ impedances \\ connected \\ to the motion \\ at \theta _{1} \end{matrix} \right] \theta _{2}(s)=\left[\begin{matrix} Sum of \\ applied torques \\ at \theta _{2} \end{matrix} \right]$