A solid cylinder (I = 1/2 mr^2) is being pulled by a force applied at 40 degrees above the horizontal to the axle and directed through the center of the roller. Roller moves across a flat, horizontal surface. Cylinder has radius of 0.60 m, and mass of 250 kg. Assume roller rolls with just enough friction to keep from slipping, determine the amount of force needed to give the roller a horizontal acceleration of 0.50 m/s^2
Angular Acc. = a / r
Torque = I (Ang. Acc.) : F Cos(40) (r) = (1/2 mr^2) a/r
The Attempt at a Solution
Is the last equation I wrote sufficient to solve problem? Or do you need to do sum of forces in x direction equal to horizontal acceleration times mass? If so, does static friction point in direction opposite of motion or in same direction of motion (i.e. to prevent the rolling disk from slipping in place)?