Question

the pulley to which the 5 cm radius is stuck is located on a horizontal axis.A cord of negligible mass is wrapped around the pulley, on the free end of which a load of 0.4 kg is hung.After being released the load decreased uniformly accelerated making 1.8 m for 3 s.Determine the moment of inertia of the roller

Answer 1

$$\alpha=\frac{T}{I}\\a=\frac{2s}{t^2}=\frac{Tr}{I}=\frac{mgr}{I}\\ \frac{2(1.8)}{3^2}=\frac{(0.5)(9.8)(0.05)}{I}\\ I=0.6125\ kg\cdot m^2$$