Recent content by vivitribal

Finally thank you everyone for your help.

In general how do you demonstrate that a given Lagrangian equation provides the correct equation of motion?

Here we go again T1=M1*a T2=-M2*a+M2*g (T2-T1)*r=1/2*m*r^2*alpha T2-T1=1/2*m*a <--r^2 cancels from alpha=a/r and for the r on the other side substituting we get -M2*a+M2*g-M1*a=1/2*m*a M2*g=a*(1/2*m+M2+M1) a=M2*g/(1/2*m+M2+M1) So hopefully for the last time is this correct?

In reference to the first one T2=-M2*a-M2*g thanks for pointing that out. For the second one are you saying that (T2-T1)*r=I*alpha?

Can someone confirm or deny this?

Ok I think I have it. T1=M1*a T2=M2*a+M2*g T2-T1=1/2*m*r^2*alpha=1/2*m*r*a combining we get M2*a+M2*g-M1*a=1/2*m*r*a solving for a we get a=2*M2*g/(m*r) Is that right or am I still missing something?

Can someone help please?

Yes I realized that about the tensions and the pullies which you can use to find the pulley's angular accel which has to be the same as the systems accel but how do you find tensions?

* M1-------o * ^ * +y * ____T1-> | * | * -x +x * _________ | * T2 * -y * _________ | * _________ M2 Wouldn't T1-T2 be the acceleration?

OK so ignoring the pulley for a second the equation would be T1-T2+M2*g=(M1+M2)*a is that correct at least?

Let's try this again sum of the forces = m*a so M1*T1+M2*g+I*(T2-T1)-M2*T2= (M1+M2+I)*a and solve for a. Is that right or am I still missing something

Ok so the force down from the hanging block is M*g which has to pull the pulley and the other block which would be (m+M)*a? or would it be m*alpha +M*a?

Homework Statement I have 2 blocks of mass M one on a frictionless horizontal surface the other hanging over the edge of the surface connected by a string of length l. The string goes over a pulley of mass m. Find the acceleration. Homework Equations PE = mgh KE = 1/2mv^2 The...