Atwood Machine

grabcad

The Atwood Machine is another classic physics model. As one mass goes down, the other goes up. You can model the forces acting on them by using Newton's Second Law if you call the positive direction the system going clockwise and the negative direction counterclockwise:1) Big Mass: Mg-T_1= Ma2) Small Mass: T_2-mg = mawhere M is the mass of the big mass, m is the mass of the small mass, T_1 is the tension that the big mass experiences, T_2 is the tension that the small mass experiences, and a is the acceleration of the masses.Using Newton's Second Law but for torque, we get:3) RT_1-RT_2 = Iαwhere I is the moment of inertia of the pulley, R is the radius of the pulley, and α is the angular acceleration of the pulley. If we assume the thread is rolling across the surface of the pulley without slipping, then we get the relation: 4) a = RαUsing these four equations, we can solve for the acceleration of the masses:4) --> 3) a = (R^2/I)(T_1-T_2)1) T_1 = Mg-Ma2) T_2 = ma+mg1) + 2) --> 3) a = (R^2/I)(g(M-m)-a(M+m)) a + a(R^2/I)(M+m) = g(R^2/I)(M-m) a = (R^2g(M-m))/(I+R^2(M+m))This is the result we get if we assume the pulley has some mass. If we ignore the mass of the pulley, then the result simplifies down to:a = g(M-m)/(M+m)We can check this by setting up Newton's Second Law Equations, if we ignore the mass of the pulley initially:Big Mass: Mg-T_1 = MaSmall Mass: T_2-mg = maPulley: T_1-T_2 = 0Solving for acceleration, we get:a = g(M-m)/(M+m)Wow, physics actually makes sense finally.NOTE: These calculations were done all under the assumption that m, M, R, and I are known values. Also, the video URL leads to a worked example of this problem with an MIT professor. He defines his directions a little differently, but our results are the same. You also have to use the rebuild feature to see a change for the thread.