Half atwood machine with accelerating pulley

[krackers]

Homework Statement

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Consider a half-atwood machine on a cart as below, with mass [itex]m_2[/itex] attached to [itex]M[/itex] via a frictionless track that keeps it pinned to M but allows it to move vertically. All surfaces (except between the wheels/ground) are frictionless, and the pulley and rope are massless.

If the system is released from rest, what will the acceleration of [itex]M[/itex] be?

Homework Equations

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Not sure if the below are right —

[tex]T=m_{1}\left( a-a_{M} \right)[/tex]
[tex]T-m_{2}\text{g}=-m_{2}a[/tex]

Where the acceleration of [itex]m_1[/itex] is [itex]a - a_M[/itex] (where [itex]a[/itex] is the magnitude of the acceleration of [itex]m_2[/itex]) since [itex] m_1 [/itex] moves right while [itex]M[/itex] moves left.

Also possibly:
[tex]T = a_M\left(M + m_2 \right)[/tex]

since the tension must be enough to accelerate the combined mass of [itex]M[/itex] and [itex]m_2[/itex]

The Attempt at a Solution

The force responsible for accelerating [itex]M[/itex] has to be the reaction to the tension in the [itex]x[/itex] direction. I tried to calculate the tension in the string while the cart and came up with two possible sets of equations. However, the solution to the first set does not match the second set so one is clearly not properly accounting for all forces.

Once the tension is found, the normal force between [itex]M[/itex] and [itex]m_2[/itex] also has to be accounted for — how is this expressed?

 

[andrewkirk]

I would have thought ##a_M## should not come into the calculation of the tension in the string. The acceleration of M comes from the horizontal component of the string's push on the pulley.

m2 is being pulled down by gravity with a force of ##gm_2##, which accelerates the ##m_1,m_2## complex along the line of the string at rate ##a##. So the acceleration of the two blocks along that line should be ##a=\frac{m_2g}{m_1+m_2}##. Horizontal momentum must be conserved, so the leftwards momentum of ##M+m_2## must equal the rightwards momentum of ##m_1##. Given that, and the rightwards acceleration of ##m_1## from above, the leftwards acceleration of ##M+m_2## can be calculated by proportioning on masses.

This is a little bit hand-wavy. If you want to be more rigorous, you could either use d'Alembert or Lagrangians with generalized coordinates, or perhaps just write and then solve the equations for conservation of energy and momentum.

Once the tension is found, the normal force between M and m_2 also has to be accounted for

Why? I don't think that's necessary in order to answer the question.

 

[krackers]

@andrewkirk Doesn't the fact that the pulley is moving along with M to the left decreases the tension string along the x-direction though? If you look at it from the reference frame of [itex]m_1[/itex] which is non-inertial, there will be a "pseudo-force" acting towards the right, essentially reducing the net tension force acting on [itex]m_1[/itex]

 

[andrewkirk]

I don't think so. As far as ##m_1## is concerned there's no difference whether the string is going over a pulley towards a downwards gravitational pull or is being pulled by someone standing on top of M, with their boots glued to the surface of M, or even pulled by somebody standing on a stepladder away from the M-m1-m2 block, at the correct height to just pull rightwards (in the latter case the effect on m1 is unchanged, but the effect on M is not).

 

[krackers]

That seems reasonable — so the tension on the pulley would then be [itex]\frac{g m_1 m_2}{m_1+m_2}[/itex], which is the same magnitude of force that accelerates the [itex]M + m_2[/itex] system.

So is the acceleration of [itex]M[/itex] just [itex]\frac{g m_1 m_2}{\left(m_1+m_2\right) \left(M+M_2\right)}[/itex]?

 

[andrewkirk]

Hmmm. I did a quick calc based on conservation of energy and momentum and got the acceleration of M to be ##\frac{gm_2}{M+m_2}##, which is different from what I got using the above argument.

The energy-momentum calc is more rigorous, but I haven't had time to check whether I've made any mistakes. So I can't be sure whether the discrepancy is because of a calc error in the energy-mom calc, or an invalidity of the above hand-wave.

I suggest you do an energy-momentum calc and see what you get. As well as giving a confident answer, it may give some insight into the other considerations.

 

[krackers]

If it helps, here are the specific answers for when m1 = 5, m2 = 10, and M = 15:

T = 28.9 N
Acc. of M = -1.15 ms^-2
Acc. of m1 = 5.78 ms^-2

Assuming the answers are correct, I don't believe your calculated acceleration as found through momentum conservation is valid...

Here are some of the valid equations that fit the above:

[tex]T=a_1 m_1[/tex]
[tex]-T = a_M\left(M + m_2 \right)[/tex]

It seems that one more equations relating [itex]a_1[/itex] and [itex]a_M[/itex] is needed.

EDIT: I think I found it, though I'm not sure why it's valid:

[tex]a_2=a_M-a_1[/tex]

which gives you the third equation:

[tex]T-m_2g=m_2 \left(a_M-a_1\right)[/tex]

I have no idea why the acceleration of the second block is equal to [itex]a_M-a_1[/itex] (note that since [itex]a_M[/itex] and [itex]a_1[/itex] are in different directions, they have opposite signs. The right direction is taken to be positive so [itex]a_1[/itex] is positive while [itex]a_M[/itex] is negative — so essentially this is conveying that the magnitude of the acceleration of [itex]m_2[/itex] is equal to the magnitude of the acceleration of [itex]M[/itex] plus the magnitude of the acceleration of [itex]m_2[/itex], though I have no idea why this is true.)

 

[andrewkirk]

I think I can explain why ##a_2=a_M-a_1##. The mass ##m_1## is accelerating rightwards at rate ##a_1## relative to the ground, and the trolley, with pulley, is accelerating rightwards at ##a_M## (which will be negative, as it's accelerating leftwards). So ##m_1## is accelerating towards the pulley, at rate ##a_1-a_M## in the rest frame of the pulley. Hence ##m_2## must also be accelerating at that rate downwards, which means ##a_2=-(a_1-a_M)## since the positive vertical direction is up.

I haven't had time to check any of my calcs yet.

 

[Mister T]

If it helps, here are the specific answers for when m1 = 5, m2 = 10, and M = 15:

T = 28.9 N
Acc. of M = -1.15 ms^-2
Acc. of m1 = 5.78 ms^-2

Where did this hint come from? Was it given along with the statement of the problem?

I drew three free-body diagrams. One for ##m_1##, which, when looking at the net force in the horizontal direction, gave me one equation.

And one for ##M## gave me one equation, again looking at only the horizontal direction.

And one for ##m_2##, which gave me two equations, one for the horizontal direction and for the vertical.

Hint: There's been no mention so far of the effect of the rails!

 

[Mister T]

That seems reasonable — so the tension on the pulley would then be [itex]\frac{g m_1 m_2}{m_1+m_2}[/itex], which is the same magnitude of force that accelerates the [itex]M + m_2[/itex] system.

So is the acceleration of [itex]M[/itex] just [itex]\frac{g m_1 m_2}{\left(m_1+m_2\right) \left(M+M_2\right)}[/itex]?

I reached the same conclusion, but note that for the specific example you gave it gives an acceleration of ##\frac{2g}{15}## which is about 1.31 m/s² instead of the 1.15 m/s².

 

[krackers]

@Mister T

This problem was posed to me by a friend, and he gave me those answer values to check.

I'm guessing the reason why the acceleration is not [itex]\frac{g m_1 m_2}{m_1 + m_2}[/itex] is that the acceleration of both blocks are not the same, as explained above.

The effect of the rails is to keep [itex]m_2[/itex] fixed onto [itex]M[/itex], as otherwise it would lie at an angle due to the acceleration. I'm not sure how this would affect the calculated acceleration though — @andrewkirk could you try to expand on how the free body diagram would be altered if [itex]m_2[/itex] were no longer pinned to the side?

 

[krackers]

Also, how did you attempt to solve it via conservation of energy / momentum? Because the final velocities are not given, how was it possible to write an expression for momentum/kinetic energy conservation?

Also, I wonder if this problem can be solved by looking at the x-coordinate of the center of mass and considering that it needs to remain constant. You might be able to set up an expression for the position of [itex]m_1[/itex] and [itex]M[/itex] at any instant, whose second derivative would give you the acceleration.

 

[andrewkirk]

@andrewkirk could you try to expand on how the free body diagram would be altered if

m_2 were no longer pinned to the side?

Sorry for the hit and run. I'm at work and shouldn't really be looking at this site at all right now. But a quick visit can't hurt. If m2 weren't pinned to the side, it would swing rightwards away from M as M accelerated leftwards. I expect that would complicate things quite a lot because the string from pulley to m2 would no longer be vertical. The acceleration of M would no longer be constant because m2 would not be part of the immediate acceleration, but would become increasingly involved in it as the string moved away from the vertical and hence had a horizontal component to its pull on m2.

Energy momentum equations are, assuming constant acceleration, of which I'm pretty confident (need to think that through a bit more later), so that ##v_1=a_1t, v_M=a_Mt## are velocities of ##m_1,M## at time ##t##:

$$m_1v_1=m_1a_1t=-(M+m_2)v_M=-(M+m_2)a_Mt$$
$$0.5m_1(a_1t)^2+0.5M(a_Mt)^2
+0.5m_2\left[(a_Mt)^2+(a_Mt-a_1t)^2\right]
=m_2g(x_1-x_2)=0.5m_2g(a_1-a_M)t^2$$

Then divide the first equation by ##t## and the second by ##t^2## to get rid of ##t##, and solve for ##a_1,a_M##

EDIT: In my haste I omitted the KE from the vertical component of the motion of ##m_2## from the second equation. I have now added it in, in the last term on the LHS. That will change the answer.

 

[Mister T]

Looking back at the thread I see that the forces exerted between ##M## and ##m_2## were discussed. So, if I treat ##M## and ##m_2## as a single object I can ignore those internal forces and I get

##T=(M+m_2)a##.

Now I need only two more equations. One from ##m_1## in horizontal direction and the other from ##m_2## in vertical direction.

I get the same answer as before.

 

[Mister T]

— so essentially this is conveying that the magnitude of the acceleration of [itex]m_2[/itex] is equal to the magnitude of the acceleration of [itex]M[/itex] plus the magnitude of the acceleration of [itex]m_2[/itex], though I have no idea why this is true.)

Me, either. But if I make that assumption I get an expression that, when I plug in your numbers, gives me a value of ##\frac{2}{17}g## for the magnitude of the acceleration of ##M##. That's about ##1.153##m/s².

 

[krackers]

@Mister T

Yeah 1.153 is the correct answer. Andrewkirk explained the reasoning above.

 

[Mister T]

Andrewkirk explained the reasoning above.

Oh, yeah. I missed that somehow. But I was thinking about it, too. The magnitude of m₁'s acceleration relative to the floor is a₁, but relative to M it's a₁ plus the magnitude of M's acceleration because M is accelerating in the opposite direction. And it's the magnitude of m₁'s acceleration relative to M that's equal to the magnitude of m₂'s acceleration (i.e. the acceleration of the string relative to M).

So can you post for us the expression you got for the acceleration of M?

 

[krackers]

@Mister T

Yeah that seems essentially analogous to andrewkirk's.

Here are the solutions I got for the three variables (found with Mathematica because I'm lazy):

[tex]\left\{\left\{T\to \frac{g m_1 m_2 \left(m_2+M\right)}{m_2 \left(m_2+M\right)+m_1 \left(2 m_2+M\right)},a_M\to -\frac{g m_1 m_2}{m_2 \left(m_2+M\right)+m_1 \left(2 m_2+M\right)},a_1\to \frac{g m_2 \left(m_2+M\right)}{m_2 \left(m_2+M\right)+m_1 \left(2 m_2+M\right)}\right\}\right\}[/tex]

 

[Mister T]

@Mister T
Here are the solutions I got for the three variables (found with Mathematica because I'm lazy):

But then you miss out on all the fun!

I used the three equations I got from the three free-body diagrams. This is what I got when I solved for the magnitude of M's acceleration:

##\left(\frac{m_1m_2}{(M+m_2)(m_1+m_2)+m_1m_2}\right)g##

which is equivalent to what you got.

I don't suppose anyone slugged through Andrew's energy-momentum pair of equations.

 

[andrewkirk]

I don't suppose anyone slugged through Andrew's energy-momentum pair of equations.

I did , and I got the same formula for ##a_M## as you show in that latest post, which gives the numeric result quoted above of about 1.15 ##ms^{-2}##.