Find Magnitude of Angular Acceleration?

Question

A cable passes over a pulley. Because the cable grips the pulley and the pulley has zero mass, the tension in the cable is not the same on opposite sides of the pulley. Then force on one side is 167 N, and the force on the other side is 64 N. Assuming the pulley is a uniform disk of mass 1.56 kg and radius 0.334m, determine the magnitude of its angular acceleration.

Answer 1

The difference in force due to moment of inertia

T=aI
I=.5 mR^2

a=(F2-F1)/I=
a=(F2-F1)/ (0.5 mR^2)
a=2(F2-F1)/( mR^2)
a=2(167-64)/( 1.56 (0.334)^2)=
a=1183.7 rad/s^2

Answer 2

the torque utilized may be t = r F sin(theta) so that you'll make certain the torque that is .3487 N.m the instantaneous of inertia of the roller is I = m ok r^2 the position m is the mass of the roller and ok is the inertia consistent of the article. for that reason, the roller is a cylinder, so ok is a million/2 I = 2.4kg * (a million/2) * (.038 m)^2 = .0017328 Kg.m^2 The angular acceleration is torque over second of inertia (alpha) = .3487 / .0017 Angular acceleration is 201.25 radians in retaining with 2d.