Simple Pulley Problem: Energy Methods and Coefficient of Friction Explained

[Exuro89]

Homework Statement

In Fig. 9.2, two blocks, of masses 2 kg and 3 kg, are connected by a light string that passes over a pulley of moment of inertia 0.004 kgm^2 and radius 5 cm. The coefficient of friction for the table top is 0.30. The blocks are released from rest. Using energy methods, one can deduce that after the upper block has moved 0.6 m, its speed is:

Homework Equations

Work Energy Theorum

The Attempt at a Solution

I know the answer to the problem is 1.4m/s but I'm not understanding why. This answer fits if the pulley is ignored with the equation

-F_fk = 1/2mv^2-mgh

But when I add the pulley in with this equation

-F_fk = 1/2mv^2+1/2Iw^2-mgh

the answer is wrong.

Does the term "passes over the pulley" mean the pulley is not rotating? So I would not need to put that into the equation? Thanks

 

[Doc Al]

Does the term "passes over the pulley" mean the pulley is not rotating?

No. Assume the pulley is rotating as it normally would.

So I would not need to put that into the equation?

You'll need to consider the change in energy of the pulley.

(Your equations showed only one mass 'm', but each block must be considered separately when figuring the change in energy.)

 

[Exuro89]

That was quickly written, I meant to have those masses be different.

-F_fk = 1/2m_1v^2-m_2gh

This equation equates to 1.4m/s which is the correct answer while...

-F_fk = 1/2m_1v^2+1/2Iw^2-m_2gh

this equation which I believe is the correct equation equates to 1.13m/s.

What am I doing incorrectly?

 

[Doc Al]

That was quickly written, I meant to have those masses be different.

-F_fk = 1/2mv_1^2-m_2gh

This equation equates to 1.4m/s which is the correct answer

I agree that the correct answer is 1.4 m/s, but that equation is not correct. You cannot neglect the KE of the pulley (and m_2). Maybe it's just a coincidence that the numbers work out. (I didn't check.)

while...

-F_fk = 1/2m_1v^2+1/2Iw^2-m_2gh

this equation which I believe is the correct equation equates to 1.13m/s.

For one thing, you missed the KE of m_2. And you need the work done by friction, not just the force of friction.

Am I doing this problem incorrectly?

Fix up the last equation and you should be able to get the needed answer.

 

[ojs]

Are you including the kinetic energy of the 2 kg mass in this equation, that is are you using m_1 as 3 kg or 5 kg?

 

[Exuro89]

Ah thank you, that's what I forgot.

I'm having issues with this other problem

"A hoop is released from rest at the top of a plane inclined at 16 above horizontal. How long does it take the hoop to roll 16.4 m down the plane?"

Another energy problem.

I believe the equation is mgh=1/2mv^2+1/2mv^2

Am I on the right track? Still working on it.

 

[Doc Al]

I'm having issues with this other problem

"A hoop is released from rest at the top of a plane inclined at 16 above horizontal. How long does it take the hoop to roll 16.4 m down the plane?"

Another energy problem.

I believe the equation is mgh=1/2mv^2+1/2mv^2

Am I on the right track? Still working on it.

Yes, you're on the right track. Note that v is the final speed.