Thread: Rubber band
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zwtipp05
#4
Aug11-05, 02:49 PM
P: 107
Rubber band

I found delta L to be .002 m as well.
Here's what I did.

Fc=m*(dL+1)*w^2
Fs=k*dL

.5 N=k*.01 m
k=50 N/m

Fc=Fs
m*(dL+1)*w^2=k*dL
m*w^2=k*dL - m*w^2*dL
m*w^2=dL*(k-m*w^2)
dL=(m*w^2)/(k-m*w^2)

When you plug in the values (m=.1 kg; w=1 rps; k=50 N/m) the answer is dL=.1/49.9
So dL = .020 m