An atwood machine consists of a mass of 3.5 kg connected by a light string to a mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 kg m2 and a radius of 12.5 cm. if the system is released from rest, what is the speed of the masses after they have moved through 1.25 m if the string does not slip on the pulley?
note: the professor has told us that the correct answer is 2.3 m/s. how does one arrive at this answer?

Explanation:

For the first mass we have , and for the second . The pulley has a moment of intertia and a radius .

We solve this with conservation of energy.  The initial and final states in this case, where no mechanical energy is lost, must comply that:

Where K is the kinetic energy and U the gravitational potential energy.

We can write this as:

Initially we depart from rest so , while in the final state we will have both masses moving at velocity v and the tangential velocity of the pully will be also v since it's all connected by the string, so we have:

where we have used the rotational kinetic energy formula and that

For the gravitational potential energy part we will have:

We don't know the final and initial heights of the masses, but since the heavier, , will go down and the lighter, , up, both by the same magnitude h=1.25m (since they are connected) we know that and , so we can write:

Putting all together we have:

Which means:

d) physical means can be used to separate the components of air.

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