Question

Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller flywheel of radius 21 cm has a cord that has a pulling force of 50 N on it. What pulling force (in N) needs to be applied to the cord connecting the larger flywheel of radius 34 cm such that the combination does not rotate

Answer 1

Answer:

The pulling force applied to the cord connecting the larger flywheel is 30.88 N

Explanation:

Given;

the radius of the smaller flywheel, r₁ = 21 cm

force on the cord of the smaller flywheel, F₁ = 50 N

the radius of the larger flywheel, r₂ = 34 cm

The torque on each flywheel is equal, since there is no rotation.

τ = Fr

where;

τ is torque on each flywheel

F is the force on the cord of each flywheel

r is the radius of each flywheel

F₁r₁ = F₂r₂

$F_2 = \frac{F_1r_1}{r_2} \\\\F_2 = \frac{50*0.21}{0.34} \\\\F_2 = 30.88 \ N$

Therefore, the pulling force applied to the cord connecting the larger flywheel is 30.88 N