What is the acceleration of the mass 4m?

[chill_factor]

Homework Statement

Consider the following system: A mass 4m is suspended with a massless string from a massless pulley wheel. On the other end of the string is another massless pulley wheel with a mass 3m suspended from a massless string on it, and a mass m suspended on the other side.

Find the acceleration of the mass 4m.

Homework Equations

Use the Lagrangian formulation.

L = T - U
d/dt (dL/dq*) = dL/dq
q* is the time derivative of q.
dL/dq* is the generalized momentum of q.
dL/dq is the generalized force of q.

Let X be the distance from the the 1st pulley wheel to the mass 4m measured down. Let X' be the distance from the 1st pulley wheel to the 2nd pulley wheel measured down. Let Y be the distance from the 2nd pulley wheel to the mass 3m measured down. Let Y' be the distance from the 2nd pulley wheel to the mass m measured down.

The Attempt at a Solution

I already know the answer is not zero. However I keep getting zero.

Constraints: length of the 1st rope is a constant K = X+X'. length of 2nd rope is a constant C = Y+Y'

X = K - X'
X* = -X'*
Y = C - Y'
Y* = Y'*

T = 1/2 (4m) (X*)^2 + 1/2(3m)(Y*)^2 + 1/2(m)(Y*)^2

Kinetic energy is the sum of the kinetic energy of the 3 masses.

U = -4mgX - 3mg(X'+Y) - mg(X'+Y')

X is the distance from 1st pulley wheel to the mass 4m measured down. As X increases, potential energy decreases as expected. Same for the mass 3m. X'+Y is the total distance of 3m from the 1st pulley wheel and as this increases the potential decreases. Same for mass m.

U = -4mgX - 3mg(K-X+Y) - mg (K-X+(C-Y)) = -4mgX - 3mgK + 3mgX - 3mgY - Kmg + mgX - Cmg + Ymg

U = 0 mgX - 4mgK - Cmg -2 mgY.

L = T + U

dL/dx = 0
d/dt(dL/dx*) = m d2/dt2(x) = 0.

This says there's no acceleration of the mass 4m which is associated with the x direction. That's incorrect. Why?

 

[TSny]

T = 1/2 (4m) (X*)^2 + 1/2(3m)(Y*)^2 + 1/2(m)(Y*)^2

Suppose Y* happened to be zero at some instant of time while X* is not zero. Would the velocity of 3m be zero at that time?

 

[chill_factor]

Hmm, that is a good question. However, I think that it doesn't matter. I'm just writing down the form of the kinetic energy. If dY/dt = 0 at a certain time, that doesn't mean that the blocks will stay still; they may very well have a non-zero acceleration at that point. So there's no contradiction.

I have an idea maybe. The kinetic energies of the smaller blocks do not only depend on dY/dt but also depend on dX/dt. However, that does not eliminate the zero in the x term in the potential energy, which is the actual problem.

Also why is this posted in introductory physics? This is not an introductory problem since it deals with Lagrangian mechanics and I clearly remember putting this in advanced physics...

 

[TSny]

The important thing is that mass 3m can still have kinetic energy even when Y* is zero due to the motion of the second pulley. So, you don't yet have the correct expression for the kinetic energy of 3m. Likewise for m. Once you fix that you can then see if you still get zero accceleration for X even if U does not contain X.

I don't know why your question was moved to Introductory Physics. I agree with you that it should be in the advanced.

 

[chill_factor]

The important thing is that mass 3m can still have kinetic energy even when Y* is zero due to the motion of the second pulley. So, you don't yet have the correct expression for the kinetic energy of 3m. Likewise for m. Once you fix that you can then see if you still get zero accceleration for X even if U does not contain X.

I don't know why your question was moved to Introductory Physics. I agree with you that it should be in the advanced.

Thank you greatly for your help. Let me try this:

the 3m and m masses have kinetic energy that depends on X. Since X + X' = K, X* = -X'*

their squares are the same. (X*)^2 = (X'*)^2.

T = 1/2 (4m) (X*)^2 + 1/2(3m)(X*^2+Y*^2) + 1/2(m)(X*^2+Y*^2)

U = U = -4mgX - 3mg(K-X+Y) - mg (K-X+(C-Y)) = -4mgX - 3mgK + 3mgX - 3mgY - Kmg + mgX - Cmg + Ymg

U = 0 mgX - 4mgK - Cmg -2 mgY.

dL/dx = 0 since there is no dependence on X (only X*) in the kinetic and potential terms.

d/dt(dL/dx*) = 8m d2/dt2(X) = 0

Doesn't seem to solve the problem. The only possible way that the acceleration can be non-zero is if there is X dependence in the potential energy. How am I doing the potential energy wrong?

 

[frogjg2003]

Why can't acceleration be zero? As far as the 4m mass is concerned, it's connected to another 4m mass on the other end of the string. The force from gravity is 4mg on both masses, cancelling. The fact that your Lagrangian has no dependence on X, even though individual parts do, testifies to that.

 

[chill_factor]

because the solution manual straight up told me that it is a nonzero number and the complete problem statement asks "Why is this acceleration non-zero?" And that's another problem in itself...

 

[frogjg2003]

In T, shouldn't it be (X*+Y*)^2? This produces a cross term in the Y equation.

 

[TSny]

T = 1/2 (4m) (X*)^2 + 1/2(3m)(X*^2+Y*^2) + 1/2(m)(X*^2+Y*^2)

Still not quite correct.

Think in terms of relative velocities: The velocity of 3m relative to the ground equals the velocity of 3m relative to the second pulley plus the velocity of the second pulley relative to the ground.

Or, you can start with positions. Take a fixed point of the Earth frame of reference, say the position of the fixed pulley. How would you express the position (relative to this point) of 3m in terms of X' (or X) and Y. Then, how would you go from position to velocity for use in T?

 

[chill_factor]

I'm trying to think about that.

The position of the 2nd pulley is going to be X'+Y = K-X+Y. the time derivative of that is going to be -X*+Y*. Then square both. We get a X*Y* cross term.

For the mass m the position is X'+Y' = K-X+C-Y. Time derivative is going to be -X*-Y*.

T = 1/2 (4m) (X*)^2 + 1/2(3m)(-X*+Y*)^2 + 1/2(m)(X*+Y*)^2

Now there's going to be cross terms of X*Y*. The problem is that these will create 2 coupled ODEs, while the X component of the potential energy is still going to be zero. That does not solve the problem.

 

[TSny]

T = 1/2 (4m) (X*)^2 + 1/2(3m)(-X*+Y*)^2 + 1/2(m)(X*+Y*)^2

Looks good.

Now there's going to be cross terms of X*Y*. The problem is that these will create 2 coupled ODEs, while the X component of the potential energy is still going to be zero. That does not solve the problem.

Try solving the coulpled equations and see what happens.

 

[chill_factor]

Looks good.

Try solving the coulpled equations and see what happens.

sigh still doesn't work. i guess I'm just going to accept the lost points.

funny enough no one else i know gets it either =(

 

[TSny]

sigh still doesn't work. i guess I'm just going to accept the lost points.

What did you get for the coupled equations and what did you get for your answer?

 

[chill_factor]

I got:

m/2( 16x** + 4y**) = 0
m/2( 8y** + 4x**) = 2mg

8y** = 4g - 4x**

divide both sides by 2: 4y** = 2g - 2x**
substitute into the equation 1: 16x** - 2g - 2x** = 0.

Oh I made an algebra mistake in my paper... its the correct answer x** = g/7.

thank you greatly. what does this qualitatively mean though?

 

[TSny]

I got:

m/2( 16x** + 4y**) = 0
m/2( 8y** + 4x**) = 2mg

8y** = 4g - 4x**

divide both sides by 2: 4y** = 2g - 2x**
substitute into the equation 1: 16x** - 2g - 2x** = 0.

Oh I made an algebra mistake in my paper... its the correct answer x** = g/7.

thank you greatly. what does this qualitatively mean though?

I got different signs in the equations than you did. (See the signs I marked in red.) But I did end up with x** = g/7. I think your equations would lead to x** = -g/7.

In order to understand why the 4m of mass on the left of the first pulley is not balance by the total of 4m of mass on the right side of the first pulley, you should think about the tensions in the strings. (You can use the answers for x** and y**, along with Newton's second law applied to the individual masses, to calculate the tensions.)