A string that passes over a pulley has a 0.331g mass attached to one end and a 0.625g mass attached to the other end. The pulley, which is a disk of radius 9.50cm , has friction in its axle.

Question

What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium?

Answer 1

Difference in mass, Δm
= 0.625-0.331 g
= 0.294/1000 kg
= 0.000294 kg

Radius, r
= 9.5 cm
= 0.095 m

Torque required
=Δm * r Nm

Answer 2

F=mv^2/r v=u^2+2as