Falling Mass on a Pulley - Rotational Energy

[hitspace]

Homework Statement

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius of .2 m. A 1.50 kg mass is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest. a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy. b) What percent of the total kinetic energy does the pulley have?

Homework Equations

KE_Rotational = (1/2)Iw^2
I_Disk = (1/2)Mr^2
v=rw
Vf^2=Vi^2 + 2gΔy

The Attempt at a Solution

Plugging in I_Disk into KE Rotational, I had a single unknown variable, w, which I solved for and got
KE = .5 [ .5(2.5)(.2)^2 ] w^2 w = 13.41 rad/s

V= rw so V=.2(13.41) = 2.683 m/s which is the linear velocity of the falling mass connected in tandem with the pulley.

My solutions manual uses some sort of conservation of energy approach at this point. I tried using
kinematics and the formula Vf^2=Vi^2 + 2gΔy as such.

2.68^2 = 0 + 2(9.8)(y) , and Delta Y = .367 m. I'm posting because my book says this answer is wrong; it says .673m. I understand that their was a energy approach to this problem, but I'm confused why my answer is not the same despite the differing approach. I didn't see any road blocks in my way, and it's making me wonder what I didn't account for trying this with kinematics.

Any help would be appreciated.

 

[haruspex]

2.68^2 = 0 + 2(9.8)(y)

I don't think the mass will be accelerating downwards at the same rate as if it were disconnected from the string.

 

[hitspace]

Because the disk's rotational inertia will resist the acceleration you mean? Meaning acceleration would be less but unknown... which would be the roadblock to using kinematics?

So if I did this the way of energy conservation, would it go like this?

KE_rot_i + KE_i + Ug_i = KE_rot_f + KE_f + Ug_f
4.5 joule was given at the beginning of the problem.

0 + 0 + mg(h1) = (4.5 J) + .5 (1.5) (2.683)^2 + mg (h2)

h= 1.7 m ... still doesn't match. I'm guessing I screwed up again?

 

[TomHart]

0 + 0 + mg(h1) = (4.5 J) + .5 (1.5) (2.683)^2 + mg (h2)

If you assume that h2=0, then it looks like h1 comes out to match the book's answer . . . unless my math is wrong.

 

[hitspace]

No, no, you are absolutely right... though I am confused why it only works if we assume h2 = 0.

 

[TomHart]

Well, you can also solve for the delta, h1 - h2, which is how far it moved. It is just convenient to assign the final height to be 0.

 

[haruspex]

0 + 0 + mg(h1) = (4.5 J) + .5 (1.5) (2.683)^2 + mg (h2)

It's often worth doing a mental approximation to check for finger trouble on the calculator.
2.7^2 is about 7; 7x.5x1.5 about 5; +4.5 =9.5; /g=1; /1.5=2/3=.67.

 

[haruspex]

acceleration would be less but unknown... which would be the roadblock to using kinematics?

It's not a roadblock. You can calculate the acceleration.

 

[hitspace]

Well, you can also solve for the delta, h1 - h2, which is how far it moved. It is just convenient to assign the final height to be 0.

That is actually what I did previously - got 1.7 , and it turns out, it was calculator error. Thank you haruspex!

It's not a roadblock. You can calculate the acceleration.

With regards to this, I would be interested in the general approach. Does it have to do with torque? I haven't yet reviewed that, but I'm starting that chapter now.

 

[haruspex]

With regards to this, I would be interested in the general approach. Does it have to do with torque? I haven't yet reviewed that, but I'm starting that chapter now.

Yes.