Page 168 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 12] SIMPLE MACHINES 153

In any torque transmission system, the angular velocity ratio is the inverse of the IMA:
Output angular velocity ω out d in 1
= = =
Input angular velocity ω in d out IMA
An increase in torque is accompanied by a decrease in speed of rotation, and vice versa.

SOLVED PROBLEM 12.12
A 1200 rev/min motor is connected to a 12-in.-diameter circular saw blade with a V-belt and a pair of step
pulleys (Fig. 12-11). (a) If the pulley diameters are 4, 5, and 6 in. in each set, find the possible angular
velocities of the saw blade in revolutions per minute. (b) Find the corresponding linear velocities of the
saw’s teeth in feet per minute.

Fig. 12-11

2
3
(a) The three possible ratios of pulley diameters are , 1, and . Since
3 2
d in
ω out = ω in
d out
the three possible angular velocities of the saw blade are
2
ω 1 = (1200 rev/min)( ) = 800 rev/min
3
ω 2 = (1200 rev/min)(1) = 1200 rev/min
3
ω 3 = (1200 rev/min)( ) = 1800 rev/min
2
(b) The linear velocity v of the saw’s teeth is related to the radius r of the saw blade and its angular velocity ω by
the formula v = ωr, where ω must be expressed in radian measure. Since there are 2π rad in a revolution, the
above angular velocities can also be expressed as

rad
ω 1 = 2π (800 rev/min) = 5027 rad/min
rev

rad
ω 2 = 2π (1200 rev/min) = 7540 rad/min
rev

rad
ω 3 = 2π (1800 rev/min) = 11,310 rad/min
rev
The radius of the saw blade is r = 6 in. = 0.5 ft, and so the possible linear velocities of its teeth are
ν 1 = ω 1 r = 2514 ft/min ν 2 = ω 2 r = 3770 ft/min ν 3 = ω 3 r = 5655 ft/min