Question

A mass m = 0.5 kg is hung from a pulley with moment of inertia I = 0.20kg⋅m2 and radius R = 10cm. There is friction in the pulley. The mass is dropped from rest, and after 2s it traveled 0.4m. What is the torque due to friction?

Answer 1

s=at²/2 => a=2s/t²=2•0.4/2²=0.2 m/s²
The equation of the load motion
ma=mg-T => T=m(g-a)
The equation of the pulley motion
Iε=M-M(fr)
M(fr) = M-Iε = TR-Ia/R=
=m(g-a)R – Ia/R = ...