A 1.00 kg mass is attached to a string wrapped around a cylinder of radius 0.05m. Two masses are connected with a rod and attached to the end of the cylinder, perpendicular to its axis. as the mass falls from rest the rotating contraption spins through 12 pi radians in 1.2 seconds.

Question

A 1.00 kg mass is attached to a string wrapped around a cylinder of radius 0.05m. Two masses are connected with a rod and attached to the end of the cylinder, perpendicular to its axis. as the mass falls from rest the rotating contraption spins through 12 pi radians in 1.2 seconds.
A) Find the angular acceleration of the rotating object.
B) Using the falling mass, find the tension in the string.
C) Calculate the rotational inertia of the object.

Damon · Answer

12 pi = (1/2) alpha (1.2)^2 so solve for alpha a = acceleration down of mass = r alpha = .05 alpha so 1 g - T = 1 a T = 1(g-a) Torque = .05 T = I alpha solve for I