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 13 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 22 cm such that the combination does not rotate?

Answer 1

Answer:

$F_2 = 29.54 N$

Explanation:

As we know that the combination is maintained at rest position

So we will take net torque on the system to be ZERO

so we know that

$\tau = \vec r \times \vec F$

here we will have

$\vec r_1 \times F_1 = \vec r_2 \times F_2$

so we have

$13 \times 50 = 22 \times F_2$

so we have

$F_2 = \frac{13 \times 50}{22}$

$F_2 = 29.54 N$