Atwood Problem

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1. Oct 19, 2014

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 [Broken]

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?

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

3. 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!

Last edited by a moderator: May 7, 2017
2. Oct 19, 2014

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...

3. Oct 19, 2014

_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)

4. Oct 19, 2014

_N3WTON_

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

Last edited: Oct 19, 2014