Consider two masses of 3.3 kg and 7.4 kg
connected by a string passing over a pulley
having a moment of inertia 12 g · m2
about its axis of rotation, as in the ﬁgure below. The
string does not slip on the pulley, and the
system is released from rest. The radius of
the pulley is 0.35 m.
Find the linear speed of the masses after
the 7.4 kg mass descends through a distance
21 cm. Assume mechanical energy is conserved during the motion. The acceleration of
gravity is 9.8 m/s^2.
Answer in units of m/s
delta PE = KE
KE = K(translational) + K(rotational)
The Attempt at a Solution
Found that change in potential energy should be equal to change in potential energy of the heavier mass (where PE is lost) subtracted by the change in potential energy of the lighter mass (where some PE is gained).
Therefore: 7.4*g*(21/100) - 3.3 * g * (21/100) = delta PE
I then set total change in KE to the delta PE. I determined rotational KE to be 1/2 I * omega^2
where I is given to me and omega is (v/r)^2 and r is given to me.
K translational = 1/2 * Mtotal * v^2. I factored out V^2, and set that equal to delta PE / rest of that mess
So in the end: v = sqrt ( delta PE / (6/r^2 + 1/2M)).
However, it wasn't the right answer :( so any ideas?