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## Q. 8.7

A timing belt is used to transfer power in the exercise treadmill from the motor to the roller that supports the walking/running belt (Figure 8.33). The pitch diameters of the sheaves on the two shafts are 1.0 in. and 2.9 in., and the roller has a diameter of 1.75 in. For a running pace of 7 min/mi, at what speed should the motor be set? Express your answer in the units of rpm.

Approach
To find the appropriate speed of the motor, we first find the roller’s angular velocity and then relate this to the motor speed. The speed of the walking/running belt is related to the roller’s angular velocity by Equation (8.2).

$v=r\omega$                              (8.2)

Since the timing belt’s speed is the same whether it is in contact with the sheave on the roller or with the sheave on the motor, the angular velocities of those two shafts are related by Equation (8.23).

$VR=\frac{output \ speed}{input \ speed}=\frac{\omega _{output}}{\omega _{input}}=\frac{d_{input}}{d_{output}}$                          (8.23)

## Verified Solution

In dimensionally consistent units, the speed of the walking/running belt is

$v=\left(\frac{1 \ mi}{7 \ min} \right) \left(5280 \frac{ft}{mi} \right) \left(12 \frac{in.}{ft} \right)\left(\frac{1}{60} \frac{min}{S} \right)$

$=150.9 \left(\frac{\cancel{mi}}{\cancel{min}} \right) \left(\frac{\cancel{ft}}{\cancel{mi}} \right) \left(\frac{in.}{\cancel{ft}} \right)\left(\frac{\cancel{min}}{S} \right)$

$= 150.9 \frac{in.}{S}$

The angular velocity of the roller is

$\omega _{roller}=\frac{150.9 in./s}{(1.75 in.)/2} \longleftarrow \left[v=r\omega \right]$

$= 172.4 \left(\frac{\cancel{in.}}{S} \right) \left(\frac{1}{\cancel{in.}} \right)$

$= 172.4 \frac{rad}{S}$

where the dimensionless radian unit has been used for the roller’s angle. In the engineering units of rpm, the roller’s speed is

$\omega _{roller}= \left(172.4 \frac{rad}{S} \right) \left(9.549 \frac{rpm}{rad/S} \right)$

$= 1646 \left(\cancel{\frac{rad}{S} }\right) \left(\frac{rpm}{\cancel{rad/S}} \right)$

$=1646 rpm$

Since the 2.9-in.-diameter sheave turns at the same speed as the roller, the speed of the motor’s shaft is

$\omega _{motor}= \left(\frac{2.9 in.}{1.0 in.} \right)(1646 rpm)\longleftarrow \left[VR=\frac{d_{input}}{d_{output}} \right]$

$= 4774 \left(\frac{\cancel{in.}}{\cancel{in.}} \right)(rpm)$

$= 4774 rpm$

Discussion
This belt drive is similar to a compound geartrain in the sense that two belts (the timing belt and the walking/running belt) are in contact with sheaves on the same shaft. Because the roller and the 2.9-in. sheave have different diameters, the two belts move at different speeds v and are assumed not to slip. Many treadmill motors are rated for a maximum of 4000–5000 rpm. So our answer, though on the high end of the range, is certainly reasonable for the speed of the runner.

$\omega _{motor}= 4774 rpm$