# Conservation of energy and string pulley problem

• gunster
In summary, given two masses of 3.3 kg and 7.4 kg connected by a string passing over a pulley with a moment of inertia of 12 g · m2, and released from rest with the system not slipping, the linear speed of the masses after the 7.4 kg mass descends 21 cm can be found by using the equations for change in potential energy and kinetic energy. However, if the moment of inertia is given in grams rather than kilograms, the resulting calculation will not yield the correct answer.

## Homework Statement

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.

## Homework Equations

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?

In what units is the moment of inertia given?

wow I see the moment of inertia is given in grams. Thank you so much that would explain why I failed XD