Question

Two​ pulleys, one with radius r 1r1 and the other with radius r 2r2​, are connected by a belt. The pulley with radius r 1r1 rotates at omega 1ω1 revolutions per​ minute, whereas the pulley with radius r 2r2 rotates at omega 2ω2 revolutions per minute. Show that StartFraction r 1 Over r 2 EndFraction equals StartFraction omega 2 Over omega 1 EndFraction r1 r2

Answer 1

Answer:

r1/r2 = 2ω2/ω1

Step-by-step explanation:

The velocity of each pulley is expressed as v = ωr where;

v is the linear velocity of the pulley

ω is the angular velocity of the pulley

r is the radius of the pulley.

For the two pulleys, the velocity I'd both pulleys are the same.

v1 = v2

v1 is the linear velocity of first pulley

v2 is the linear velocity of the second pulley.

v1 = ω1r1

v2 = 2ω2(r2)

r1 and r2 is the radius of pulley 1 and pulley 2 respectively.

Since v1 = v2

ω1r1 = 2ω2(r2)

Divide both sides by r2

ω1r1/r2 = 2ω2(r2)/r2

ω1r1/r2 = 2ω2

Divide both sides by ω1

ω1r1/r2/ω1= 2ω2/ω1

r1/r2 = 2ω2/ω1