- #1
celeramo
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Homework Statement
An Atwood machine is constructed using a
hoop with spokes of negligible mass. The
2.4 kg mass of the pulley is concentrated on
its rim, which is a distance 20.9 cm from the
axle. The mass on the right is 1.36 kg and on
the left is 1.79 kg.
What is the magnitude of the linear accel-
eration of the hanging masses? The accelera-
tion of gravity is 9.8 m/s^2
Answer in units of m/s^2
Homework Equations
F=ma
[tex]\tau[/tex]=I [tex]\alpha[/tex]
[tex]\alpha[/tex] R=v
The Attempt at a Solution
Ok I found the net force action on the system as the difference of the two weight forces so:
F1=1.79*9.8=17.542N F2=1.36*9.8=13.328N
Fnet=4.214N
Now I said that F=ma, but since the wheel has a mass, and thus a moment of inertia I said:
F=(m1+m2)a+I[tex]\alpha[/tex] where m1 and m2 are the weights and I is the moment of inertia and alpha is the angular acceleration of the wheel
I think this is probably where I am mistaken...but continuing from here
I=mr^2=2.4*(.209^2)=.104834
and I say a=[tex]\alpha[/tex]r
4.214=(3.15 kg)a+I*a/r
4.214=3.15a+.501598a => 4.214=(3.6516)a
a=1.15402 m/s^2
and needless to say this is incorrect :(
Any help would be wonderful! Thanks a lot