Asked by dksee
A belt drives two flywheels whose diameters are 30cm and 45cm, respectively. If the smaller wheel turns through 310 rev/min, find the speed of the belt in meters per second and the regular velocity of the larger wheel in rev/min.
Answers
Answered by
mathhelper
long way:
perimeter of small wheel = 2π(30) cm = 60π cm
so at 310 rotations/min, it has gone 310(60π) = 18600π cm
perimeter of larger wheel = 2π(45) = 90π cm
velocity of larger wheel = 18600π/90π = 206 2/3 rev/min
short way:
by proportion,
30/45 = R/310
R = 310(30)/45 = 206 2/3 rev/min
perimeter of small wheel = 2π(30) cm = 60π cm
so at 310 rotations/min, it has gone 310(60π) = 18600π cm
perimeter of larger wheel = 2π(45) = 90π cm
velocity of larger wheel = 18600π/90π = 206 2/3 rev/min
short way:
by proportion,
30/45 = R/310
R = 310(30)/45 = 206 2/3 rev/min
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