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## Q. 8.6

Computer Scanner

The mechanism in a desktop computer scanner converts the rotation of a motor’s shaft into the side-to-side motion of the scanning head (Figure 8.32). During a portion of a scan operation, the drive motor turns at 180 rpm. The gear attached to the motor’s shaft has 20 teeth, and the timing belt has 20 teeth/in. The two other gears are sized as
indicated. In the units of in./s, calculate the speed at which the scanning head moves.

Approach
To find the speed of the scanning head, we must first determine the speed of the transfer shaft by applying Equation (8.9).

$VR=\frac{output speed}{input speed}=\frac{\omega _{g}}{\omega _{p}}=\frac{N_{p}}{N_{g}}$            (8.9)

Equation (8.2)

$v=r\omega$                              (8.2)

can then be used to relate the rotational speed of the transfer shaft to the belt’s speed.

## Verified Solution

The angular velocity of the transfer shaft is

$\omega = \left(\frac{20}{80} \right)(180 rpm) \longleftarrow \left[\omega _{g}=\left(\frac{N_{P}}{N_{g}} \right )\omega _{P} \right]$

$=45 rpm$

Because the 30-tooth and 80-tooth gears are mounted on the same shaft, the 30-tooth gear also rotates at 45 rpm. With each rotation of the transfer shaft, 30 teeth of the timing belt mesh with the gear, and the timing belt advances by the distance

$x=\frac{30 teeth/rev}{20 teeth/in.}$

$= 1.5 \left(\frac{\cancel{teeth}}{rev} \right) \left(\frac{in.}{\cancel{teeth}} \right)$

$=1.5 \frac{in.}{rev}$

The speed v of the timing belt is the product of the shaft’s speed ω and the amount x by which the belt advances with each revolution. The scanning head’s speed becomes

$v=\left(1.5\frac{in.}{rev} \right) (45 rpm)$

$= 67.5\left(\frac{\cancel{rev}}{min} \right) \left(\frac{in.}{\cancel{rev}} \right)$

$=67.5\frac{in.}{min}$

In the units of inches per second, the velocity is

$v=\left( 67.5\frac{in.}{min} \right) \left(\frac{1}{60}\frac{min}{s} \right)$

$= 1.125\left(\frac{in.}{\cancel{min}} \right) \left(\frac{\cancel{min}}{s} \right)$

$=1.125\frac{in.}{s}$

Discussion
This mechanism achieves a speed reduction and a conversion between rotational and straight-line motion in two stages. First, the (input) angular velocity of the motor’s shaft is reduced to the speed of the transfer shaft. Second, the timing belt converts the transfer shaft’s rotational motion to the (output) straight-line motion of the scanning head.

$v=1.125\frac{in.}{s}$