A Pulley problem

  • Thread starter Thorskur
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
gneill
Mentor
20,946
2,892
Only one block is falling through distance d; Check your change in gravitational PE.
 
  • #3
Thorskur
2
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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.
 
  • #4
azizlwl
1,065
10
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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