Solving Chain on Pulley: Find Velocity

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echao12
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[SOLVED] Chain On Pulley

Homework Statement


Given: g = 9.8m/s^2 . A uniform flexible chain whose mass is 7 kg and length is 6 m.
Given: A small frictionless pulley whose circumference is negligible compared to the length of the chain.
Initially the chain is hung over the pulley with nearly equal lengths on both sides, but just unequal enough so that the unstable equilibrium condition will let the chain start to move.
After some time, the longer end of the chain is a distance ℓ = 4.2 m down from the pulley’s axle.

1. Find the acceleration a of the chain when the chain is at this position. Answer in units of m/s2. (I already solved this part, a = 3.92)

2. Find the velocity v of the chain when ℓ = 4.2 m. Answer in units of m/s. (This is the part I need help with)

Homework Equations


Center of mass?
Conservation of energy? U=mgh, KE=(1/2)mv^2

The Attempt at a Solution


mgh = (1/2)mv^2 + mgh

Masses cancel, so:
gh = (1/2)v^2 + gh

initial height of center of mass = 4.2 m
final height of center of mass = 3 m

Plugin values:
9.8 * 4.2 = (1/2)v^2 + 9.8 * 3

Solving for v, I get 4.849742261 m/s
However, this is wrong. Any ideas? Thanks in advance.
 
Attachments
  • Diagram of a uniform chain hanging over a small pulley, with one side longer than the other; labels show 4.2 m from the pulley axle to the longer end, 6 m total chain length, and acceleration a marked by arrows on both sides.
    HW #2.jpg
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Last edited:
Look again at your energy balance.

Let the zero of potential energy be at the level o the pulley, so the initial CM is 1.5 m below the pulley, and V1 (initial PE) is m*g*(-1.5).

In the second condition, the CM of the chain is lower, ycm=-(1.8*(M*1.8/L)/2+4.2*(M*4.2/L)/2)/M

Then refigure your energy balance.
 
Thank you very much! apparently I completely misunderstood center of mass.