How Do You Solve a Pulley System Problem with Frictionless Incline?

[vu10758]

Two masses are connected as shown at http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1423595691
It is problem 15

I drew my free body diagrams, but I can't figure out the relationship to write for the pulley since the incline is frictionless. I know that angular acceleration = (M1*g*sin(theta) - M2*g)R/ (I +M2R^2 +M1*R^2)

How do I get there though? I have to somehow get T1 and T2 out of the problem to write everything in terms of M1,M2,I,R,and theta. However, I am missing an important piece because I don't know the relationship between the forces for the pulley.

 

[OlderDan]

Write three separate equations: one for M1, one for M2, and one for the pulley. The pulley "knows nothing" about the masses at the other end of the string. All it knows is there are strings with different tensions trying to rotate it in opposite directions.

Express the angular acceleration in terms of the acceleration of the masses. There will then be three unknowns: T1, T2 and acceleration. You can solve the three equations for the three unknowns.

 

[vu10758]

Since I know that T1 is greater than T2, wouldn't the pulley have the relationship

T1-T2= Ia, where a is angular acceleration?

 

[OlderDan]

Since I know that T1 is greater than T2, wouldn't the pulley have the relationship

T1-T2= Ia, where a is angular acceleration?

Yes. And each mass will have an acceleration that depends in some way on a tension. The angular acceleration of the wheel and the linear acceleration of the masses are all related.

 

[vu10758]

Thanks, I got it now.

 

[OlderDan]

Would you check my work? I got the wrong answer but it seems like my answer is very close

For M1)
x) T1-m1*g*sin(theta)=M1*a
y) N - M1*gcos(theta) <== This is not needed since there is no friction

For M2)

T2+M2*g + M2*a <== This should be
M2*g - T2 = M2*a

Pulley

T1 - T2 = I(alpa) <== This should be
T2 - T1 = I(alpha) It is very important to keep the directions consistent. M1 is moving up the plane, M2 is going down, and the wheel is rotating clockwise.

I know that a is alpha * r <== This is OK

M1*g*sin(theta) + M1*a + M2*a - M2*g = I(alpha)
M1*g*sin(theta) + M1*(r)(alpha) + M2*(r)*alpha - M2*g = I(alpha)
M1*g*sin(theta) -M2*g = I*(alpha) - M1*r(alpha) - M2*r*alpha
alpha = (M1*g*sin(theta) - M2*g)/(I-M1*r - M2*r)

This is wrong though. The correct answer is (M1*g*sin(theta) - M2*g)R/(I + M2*R^2 + M1*R^2)

See the annotations in the quote. Make the corrections and try the Algebra again.

 

[vu10758]

See the annotations in the quote. Make the corrections and try the Algebra again.

Thank you.