Solve Pulley Problem: Speed of Blocks After Moving Distance d

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Homework Help Overview

The problem involves a pulley system with a uniform cylindrical disk and two blocks, where the goal is to determine the speed of the blocks after they have moved a distance d. The setup includes considerations of gravitational potential energy and kinetic energy, with the system initially at rest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of energy approach, questioning the correct change in gravitational potential energy and the contributions of each block to the system's energy. There is a focus on the relationship between the linear speed of the blocks and the angular speed of the pulley.

Discussion Status

Some participants have pointed out potential errors in the original poster's calculations, particularly regarding the gravitational potential energy associated with the blocks. There are ongoing attempts to clarify the contributions of each block and the pulley to the overall energy balance.

Contextual Notes

Participants are navigating through the implications of the problem's constraints, such as the mass of the pulley and the assumption of massless strings. There is a noted confusion regarding the correct interpretation of the energy changes involved.

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?
 
Physics news on Phys.org
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
 
Last edited:

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