A frictionless pulley which can be modeled as a 082 kg solid

A frictionless pulley, which can be modeled as a 0.82 kg solid cylinder with a 0.35 m radius, has a rope going over it, as shown in the figure. The tensions in the rope are 12 N and 10 N. What is the angular acceleration of the pulley? Express your answer using two significant figures.

Solution

Torque = F * r = (12 - 10) * 0.35 = 0.70 N.m

Moment of intertia for a solid cylinder = 1/2 M R²

= 1/2 * 0.82 * 0.35² = 0.050225 kg.m²

Torque= Moment of Intertia * angular acceleration

0.70 = 0.050225 * angular acceleration

angular acceleration = 13.94 rad/s²