Solving Physics Problem 9-72: Angular Speed of Pulley

[Dominique19]

1. The problem states:

Problem 9-72a:
The system shown in the figure below consists of a m1 = 4.24-kg block resting on a frictionless horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to a hanging m2 = 2.12-kg block.
http://loncapa.mines.edu/res/whfreeman/tipler/Physics_for_Scientists_and_Engineers_6e/Chap09/graphics/tipler9-68.gif

The pulley is a uniform disk of radius 8.19 cm and mass 0.565 kg. Calculate the speed of the m2 = 2.12-kg block after it is released from rest and falls a distance of 2.23 m.

Problem 9-72b:
What is the angular speed of the pulley at this instant?

Homework Equations

v=w(R)
K=1/2mv^2

The Attempt at a Solution

I set my system to be both the masses and the pulley, therefore the only external force would be the force of gravity. I think I'm supposed to set that equal to the translational and rotational energies of the system, translational for the masses and rotational for the pulley. But i don't know what they equations for the translational and rotational energies would be. Once i figure that out i can solve for the second part of the problem. Thanks in advance![/B]

 

[NTW]

You cannot equal force with energy.

The problem can be solved by considering the conversion of one type of energy into another. You quote one relevant equation for energy, and -for this problem- you need three... Look them up in your book...

 

[_N3WTON_]

Use conservation of energy:
$K_i + U_i = K_f + U_f$
Remember that there are two types of energies in this problem (translational and rotational)

 

[_N3WTON_]

To go a step further, conservation of energy will give an equation like:
$\frac{1}{2}I\omega^{2} + \frac{1}{2}M{v_1}^2+ MgH_1 = \frac{1}{2}M{v_2}^{2} + \frac{1}{2}I\omega^{2} + MgH_2$