Products

## Holooly Rewards

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

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

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

## Holooly Ads. Manager

Advertise your business, and reach millions of students around the world.

## 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 Sources

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.EX.4

Flexibility: Flexible Read/Write for a Disk Drive

Assume that there is some flexibility between the read head and the drive motor in Fig. 2.11. Find the equations of motion relating the motion of the read head to a torque applied to the base.

## Verified Solution

The dynamic model for this situation is shown schematically in Fig. 2.12. This model is dynamically similar to the resonant system shown in Fig. 2.5 and results in equations of motion that are similar in form to Eqs. (2.10) and (2.11). The moments on each body are shown in the free-body diagrams in Fig. 2.13. The discussion of the moments on each body is essentially the same as the discussion for Example 2.2, except that the springs and damper in that case produced forces, instead of moments that act on each inertia, as in this case. When the moments are summed, equated to the accelerations according to Eq. (2.14), and rearranged, the result is

$\ddot{x} + \frac{b}{m_1} (\dot{x} – \dot{y}) + \frac{k_s}{m_1} (x-y) +\frac{k_w}{m_1} x = \frac{k_w}{m_1}r ,$         (2.10)

$\ddot{y} + \frac{b}{m_2} (\dot{y} – \dot{x}) + \frac{k_s}{m_2} (y-x) =0.$             (2.11)

$M = I \alpha$             (2.14)

$I_1 \ddot{θ}_1 + b (\dot{θ}_1 – \dot{θ}_2 ) + k (θ_1 – θ_2) = M_c + M_D ,$             (2.17)

$I_2 \ddot{θ}_2 +b(\dot{θ}_2 – \dot{θ}_1) + k(θ_2 – θ_1) =0.$       (2.18)

Ignoring the disturbance torque $M_D$ and the damping b for simplicity, we find the transfer function from the applied torque $M_c$ to the read head motion to be

$\frac{\Theta _2 (s)}{M_c (s)} = \frac{k}{I_1 I_2 s^2 \left(s^2 + \frac{k}{I_1} +\frac{k}{I_2} \right) } .$        (2.19)

It might also be possible to sense the motion of the inertia where the torque is applied, $θ_1,$ in which case the transfer function with the same simplifications would be

$\frac{\Theta _1 (s)}{M_c (s)} = \frac{I_2 s^2 + k }{I_1 I_2 s^2 \left(s^2 + \frac{k}{I_1} +\frac{k}{I_2} \right) } .$       (2.20)

These two cases are typical of many situations in which the sensor and actuator may or may not be placed in the same location in a flexible body. We refer to the situation between sensor and actuator in Eq. (2.19) as the “noncollocated” case, whereas Eq. (2.20) describes the “collocated” case. You will see in Chapter 5 that it is far more difficult to control a system when there is flexibility between the sensor and actuator (noncollocated case) than when the sensor and actuator are rigidly attached to one another (the collocated case).