Recent content by Richard Dumfry

Huh? Aren't they the same thing?

So... 2/3 angular acceleration, and 99 Inertia (Since 1/2 * 22 * 3^2 = 99), then torsion = Inertia * acceleration = 99 * 2/3 = 66?

Right, but I thought the tangential acceleration you'd actually have to apply would be 11.81, since gravity is pulling 9.81 in the opposite direction.

Err...should be 11.81 / 3m, since angular acceleration is just tangential acceleration / radius, right? So 3.93? I tried something like that, and then plugging into torque = I * a, but couldn't get the answer from that either.

Homework Statement A 16 kg mass is suspended from the end of a rope around a 22 kg cylinder with a 6.0 m diameter. What torque applied to the drive axle of the cylinder is required to accelerate the 16 kg mass 2.0 m/s^2 upwards?Homework Equations Torque = Radius cross Force The Attempt at a...