Re: Need soln explaination

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Regarding Tony's question:
> I got a problem out of an older physics text (one without a solutions
> manual). My students and spent a while debating the method of attaining
> the solution. Can someone explain how they would solve it -using energy.
>
> PROBLEM
> Masses of 3 kg and 2 kg are hanging at opposite ends of a cord that
> passes over a frictionless pulley. The system is held stationary for a
> while, then released. What will be the speed of the masses when the 3 kg
> mass has descended 0.5 m below its starting point? (USE ENERGY
> RELATIONSHIPS TO SOLVE.)

First calculate the total energy at the instant of release.  Since the
system has not yet had time to start moving all of this energy will be just
the total gravitational potential energy of both masses and the pulley
(no kinetic energy contribution).  Since the mass of the cord is not given
we can probably safely assume its mass is small enough to ignore for both
the kinetic and potential energy calculations.  Now since the position of
the pulley is held constant throughout the problem it is probably most
convenient to choose the zero level height for the potential energy to be
at the height of the pulley.  This means that the pulley (having a zero 
height) then can never contribute any potential energy to the problem.
Next calculate the total gravitational potential energy of both masses in
their new later configuration.  Since (if the system is assumed lossless)
the total energy is conserved this means that the total energy initially
calculated is the same as the total energy at all later times.  Thus the
initially calculated total energy must be the sum of the total kinetic
energy and the total potential energy for the later configuration of
interest.  Thus subtracting the final total potential energy from the total
energy (i.e. initial total potential energy) will give the total kinetic
energy of the masses and the pulley in the later configuration of interest.
Since both masses must move at the same speed (being connected by a fixed
length string) this means that the total kinetic energy must then be:
(1/2)*(m_1 + m_2)*v^2 + (rot. K.E. of pulley)
If we can assume that the pulley has negligible mass and/or moment of
inertia then we can drop this last term.  Now since both m_1 and m_2 are
known then it is straightforward to get the common value of |v| from the
remaining kinetic energy expression.

David Bowman
dbowman@gtc.georgetown.ky.us