what is the acceleration of the masses?

Question

A mass m1 is connected by alight string that passes over a pulley of mass M to a mass m2 as shown in the figure. Both masses move vertically ans there is no slippage between the string and the pulley. The pulley has a radius of 30 cm and a moment of inertia of MR^2. If m1 is 4.0 kg, m2 is 3.0 kg, and M is 6.0 kg then what is the acceleration of the masses??

a) 0.695 m/s^2
b) 0.703
c) 0.731
d) 0.754
e) 0.805

please help, I keep getting answers that aren't a choice!!

Answer 1

If the MI of the pulley is MR^2, then all the mass M is concentrated at the rim. Thus the total moving mass is m1 + m2 + M. The force is g*(m1-m2) = g. The accel is F/Mtotal = 9.806/13 = .07543, answer D.

Answer 2

f = Ta = W + Ma/2- Ma/2 - w; where T = (M + m1 + m2) = total mass of the system, W = m1 g, a = acceleration of the system, M = pulley mass, and w = m2 g. f is the net force acting on the accelerating system (T).

It is important to note that the whole system is accelerating at the same rate (a); otherwise it would fly apart. It is also imporatant to note that the "system" mass is the sum of all the masses, including the pulley.

Since the inertial moment is MR^2, the pulley acts like an infinitely thin cylinder; so all M is in the outer rim. [See source] Which means the force on it will be f1 = Ma/2 = force pulling downward on the m1 side and f2 = Ma/2 = force pulling downward on the m2 side as the system accelerates at a m/sec^2. f1 = f2 because they are the tension on the string.

Clearly the pulley forces cancel out given they are both pulling downward, but on opposite sides of the pulley. Thus we have Ta = W - w. In other words, because we can assume all the mass of the pulley is in the rim, its inertial momentum plays no part in the system's acceleration.

To find a, rearrange Ta = W - w; so that a = g(m1 - m2)/(m1 + m2 + M); m1 = 4 kg, m2 = 3 kg, M = 6 kg, and g = 9.81 m/sec^2 on Earth's surface. You can do the math.