Linear acceleration using atwood machine

[physics10189]

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

Homework Equations

F=ma
=I
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 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=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

 

[nrqed]

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

Homework Equations

F=ma
=I
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

The forces producing a torque on the pulley are the tensions in the rope. You are assuming that the tensions are equal to the weights, which is clearly not the case (otherwise the masses would move at constant speed).

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

This equation can't be right since the units don't match.

You need to set up three equations : one for each mass and one for the pulley.

 

[physics10189]

Would the equation be something ike this

m1g-T1=m1a
T2-m2g=m2a

where T1=T2

Torque=I(alpha)
Torque=Tr
alpha=a/r
therefore
(a/r)=(Tr/I)

 

[nrqed]

Would the equation be something ike this

m1g-T1=m1a
T2-m2g=m2a

yes, that's a good start

where T1=T2

No, the tensions are different

Torque=I(alpha)
Torque=Tr

Yes, except that T is T1-T2

alpha=a/r

Yes, that last one is good.

So you end up with three equations for three unknowns: T1, T2 and a.

 

[physics10189]

ok i got accerlation=...

a=(g(m1-m2)+(T2-T1))/(m1+m2)=((T1-T2)*R^2)/I

um exactly how am i going to do this because I cannot get rid of either T1 or T2

 

[nrqed]

ok i got accerlation=...

a=(g(m1-m2)+(T2-T1))/(m1+m2)=((T1-T2)*R^2)/I

um exactly how am i going to do this because I cannot get rid of either T1 or T2

You can isolate T1-T2.
Once you have T1-T2, you plug that back in one of these expressions and you have the acceleration

 

[physics10189]

Ok so now i have
T1-T2=(g(m1-m2)M)/((m1+m2)+M)

should i move the T2 to the other side and plug T1 somewhere in the 3 equation if so which one
or plug (T1-T2) in the equation that you are referring to which I do not see where to plug in