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:
$\frac{1}{2}I\omega^{2} + \frac{1}{2}M{v_1}^2+ MgH_1 = \frac{1}{2}M{v_2}^{2} + \frac{1}{2}I\omega^{2} + MgH_2$

 

[HalfLostAndSearching]

Greetings,

Thank you for providing the problem statement and your attempt at a solution. I would approach this problem by first identifying the relevant equations and principles involved. In this case, we can use the principle of conservation of energy and the relationship between linear and angular velocity.

For part a of the problem, we can use the equation for conservation of energy, where the initial kinetic energy of the system (at rest) is equal to the final potential energy (after the block falls) and the final kinetic energy (after the block falls). This can be written as:

K_initial = U_final + K_final

We can then substitute the equations for translational and rotational kinetic energy, which are:

K_translational = 1/2mv^2
K_rotational = 1/2Iw^2

Where m is the mass, v is the linear velocity, and I is the moment of inertia for the pulley (which can be calculated using the equation I = 1/2mr^2, where r is the radius of the pulley).

Substituting these equations into the conservation of energy equation, we can solve for the final velocity of the block (v_final).

For part b of the problem, we can use the relationship between linear and angular velocity, which is v = wR, where v is the linear velocity, w is the angular velocity, and R is the radius of the pulley. We can use the calculated value of v_final from part a to solve for the angular velocity (w) of the pulley at the instant the block reaches the ground.

I hope this helps guide you in solving the problem. Remember to always identify the relevant principles and equations and to carefully substitute and solve for the unknown variables. Good luck!