A rope of negligible mass is wrapped around a uniform solid cylinder, of radius R, that can rotate freely about its axis, which is horizontal and fixed. A block hangs vertically from the rope and causes the cylinder to spin. The blocks mass, M_B is 4 times the mass, M, of the cylinder. Find an expression for the magnitude of the linear acceleration of the hanging block in terms of R and/or g. (Rotational inertia of a cylinder is I=1/2*MR^2) I know the motion of the block is Fnet=M_B*a or M_B*g - T (tension) = M_B*a (btw I'm taking up to be negative y direction and down positive y direction) I also know that Torque= I * alpha. and also Torque = R T(tension)sin(theta) where theta =1 so TR = 1/2*MR^2*alpha (from Torque=I*alpha) solving for T i have T=1/2*M*R*alpha Now if i plug this equation for Tension into F_net = M_B*a i get: M_B*g - 1/2*M*R*alpha = M_B*a **(also i know that alpha = a/r) solving for a; i found two answers. by computing the algebra differently for the two answers and im not sure which is correct...if either of them even are correct. a = g(1+(2M_B/M)) AND a = g/(1+(M/2M_B)) Any insight is appreciated! thanks!