Solve Pulley Problem: Speed of Blocks After Moving Distance d

AI Thread Summary
The discussion revolves around calculating the speed of two blocks connected by a pulley after they have moved a distance d. The initial energy is set based on the gravitational potential energy of the blocks, and the kinetic energy is derived from the motion of the blocks and the pulley. The user initially calculates the speed as v = √[(8gd)/3], but the textbook states the correct answer is v = √[(4gd)/5]. The confusion arises from the incorrect application of gravitational potential energy, as only one block falls a distance d, not both. The final consensus emphasizes the importance of correctly accounting for the energy contributions of the pulley and the blocks in the system.
Thorskur
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Homework Statement




Sorry guys but uploading the image didnt work so i post it on imageshack. Here is the link: http://imageshack.us/f/208/apib.jpg/

So the quastion is:

In the image the pulley is a uniform cylindrical disk of mass m and radius r. The strings are massless and there is no friction. If the system is initially at rest, find the speed of the blocks after they have moved a distance d.


Homework Equations



E = E + W

K = 1/2Iω^2

The Attempt at a Solution



Let's say that the potential energy is set at 0 when the blockmoves from y=d --> y=0.

Then the starting energy becomes (m1+m2)gh = 2mgd

Translating to kinetic energy, ΔU =(1/2)(2m)v2 - (1/2)Iω2

Because our pulley also has a mass m, and I =(1/2)mr2

ΔU = mv2 -(1/4)m(r2ω2) = mv2-(1/4)mv2 (Using v =rω)

So ΔU = (3mv2)/4 = 2mgd

And my answer is:

v =√[(8gd)/3]

My book say the answer is : v =√[(4gd)/5]. So I am wondering what I am doing wrong?
 
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Only one block is falling through distance d; Check your change in gravitational PE.
 
gneill said:
Only one block is falling through distance d; Check your change in gravitational PE.

So there is no 2mgh? Just mgh?

I still can't figure out how i can get 5 there.
 
find the speed of the blocks after they have moved a distance d.
.......
From top,
T1=ma ...(1)
mg-T2=ma ...(2)
(T2-T1)r=Iα=1/2(mr2) a/r ...(3)

From rest,
v2=2ad

=======
Work done by mother Earth =ΔE
mgd=2(1/2 mv2) + 1/2 Iω2
mgd=mv2+ 1/2 (1/2 m r2) (v/r)2

mgd=mv2+ 1/4 mv2

Add: Using conservation of energy
 
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