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Easy methods to Clear up a Multi-Atwood Machine Meeting

Easy methods to Clear up a Multi-Atwood Machine Meeting

2024-02-17 01:44:50

Introduction

angular motion of the top pulley

A double-double (AM-2) Atwood machine.

The determine on the suitable exhibits a “double-double” Atwood machine with three supreme pulleys and 4 plenty.  All pulleys are launched from relaxation concurrently.  Which of the alternatives under describes the angular movement of the highest pulley P after a while has elapsed and why?

  1. It rotates clockwise with rising angular velocity.
  2. It rotates counter-clockwise with rising angular velocity.
  3. It stays at relaxation.

In case your instant thought, light reader, was “idk the place to begin” or “let me draw a free physique diagram” or “let me write down the Lagrangian”, please learn on.  You will note how one can attain the proper reply in a single line on the again of an envelope. As well as, you’ll learn to discover the accelerations of all of the plenty and pulleys in a multi-Atwood machine meeting counting on nothing greater than three easy guidelines and Newton’s second regulation.

In what follows, AM stands for “Atwood machine”. It is a perfect pulley supported by a string. A great pulley is massless and modifications the course of the stress however not its magnitude. Two plenty are hooked up to the ends of a massless inextensible string that goes over the pulley.  The tensions in every string section are equal.  Multi AM assemblies are obtained by changing plenty with pulleys, every with its personal pair of plenty. The label AM-N signifies an meeting with N+1 tiers, ##2^N## plenty and ##(2^N-1)## supreme pulleys.  For instance, the one within the determine on the suitable is an AM-2; it has two tiers with pulleys and one tier with plenty.  A mass hanging from a ceiling with no pulleys is an AM-0; it has a single tier with one mass.

Three guidelines

Hanging mass equal AM-0 of an AM-1.

We quote with out derivation the well-known expression for the stress in an AM-1. $$T=frac{2m_2 m_1}{m_2+m_1}g.$$ We’re in search of a single equal mass, ##M_{textual content{eq}}##, to exchange the AM-1 proven within the determine on the suitable.  In different phrases, take away the 2 hanging plenty, block the pulley from spinning and assign mass ##M_{textual content{eq}}## to the pulley in order that the supporting rigidity ##F_0## within the string  stays the identical.  Which means that $$start{align} & F_0=2T=frac{4m_1m_2}{m_1+m_2}~g =M_{textual content{eq}}^{(1)}~g nonumber
& implies M_{textual content{eq}}^{(1)}=frac{4m_1m_2}{m_1+m_2}=frac{4}{frac{1}{m_1}+frac{1}{m_2}} nonumber
& implies frac{1}{M_{textual content{eq}}^{(1)}}=frac{1}{4}left(frac{1}{m_1}+frac{1}{m_2}proper).
finish{align}$$

Rule I

The pulley and lots more and plenty in an AM-1 could also be changed by an equal mass to type an AM-0. This equal mass is the same as 4 occasions the product of the plenty divided by their sum.  When acted upon by an exterior drive, the acceleration of the equal mass would be the identical because the acceleration of the pulley within the AM-1 meeting.

Proof
Take into account an AM-1 hanging from the ceiling of an elevator with acceleration ##A##, a signed amount.  Let ##F## be the drive appearing upon the equal mass.  Making use of Newton’s second regulation, $$start{align} M_{textual content{eq}}^{(1)}A=F-M_{textual content{eq}}^{(1)}~g.finish{align}$$ We all know that the drive appearing on the AM-1 meeting is ##2T##.  To show Rule I, we have to present that, with the definition of ##M_{textual content{eq}}^{(1)}## in equation (1), it follows that ##2T=F##.

From Newton’s second regulation, the accelerations of the plenty relative to the lab body are $$a_1=frac{T}{m_1}-(g+A)~;~~a_2=frac{T}{m_2}-(g+A).$$

The constraint that the size of the string connecting the plenty be fixed calls for that the sum of the accelerations be zero. Including the equations,$$start{align} & 0=a_1+a_2=Tleft(frac{1}{m_1}+frac{1}{m_2}proper)-2g-2A nonumber
& implies A=Tleft(frac{m_1+m_2}{2m_{1} m_{2}}proper)-g =Tfrac{1}{2}left(frac{1}{m_1}+frac{1}{m_2} proper)-g=frac{2T}{M_{textual content{eq}}^{(1)}}-gnonumber
& implies M_{textual content{eq}}^{(1)}A=2T-g.
finish{align}$$Comparability of equation (3) with equation (2), reveals that we should establish ##2T=F.~~~~## Q.E.D.

We conclude that we will substitute an AM-1 with an equal hanging mass AM-0.  Conversely, we will substitute a dangling mass ##M## with an equal pulley and two plenty to type an AM-1.  The 2 plenty should fulfill equation (1) however there isn’t a distinctive selection, e.g. ##m_1 = m_2=2M;##  ##m_1=frac{3}{10}M, m_2=frac{3}{2}M## and so forth.

Thus, successive replacements of pulleys by equal plenty can convert an AM-N  into an AM-0.  This conversion is exclusive.  Likewise,  successive replacements of plenty by equal pulleys can convert an AM-0 into an AM-N. This conversion is just not distinctive.

Rule II

The equal mass of an AM-N is given by $$start{align} mathbf{frac{1}{M_{textual content{eq}}^{(N)}}=frac{1}{4^N}sum_{i=1}^{2^N}frac{1}{m_i}.}finish{align}$$Proof

The highest two tiers of an AM-N type an AM-1.  The highest tier is Tier 0. We all know that an AM-1 has equal mass given by equation (1) $$frac{1}{M_{textual content{eq}}^{(1)}}=frac{1}{4}left( frac{1}{m_{1,1}}+frac{1}{m_{1,2}}proper).$$ The primary subscript denotes the tier quantity (beginning at 0) and the second denotes the place of the mass inside a tier, left to proper. These two plenty are the equivalents of pairs in Tier 2 under. Then $$start{align} frac{1}{M_{textual content{eq}}^{(2)}}nonumber & = frac{1}{4}left[frac{1}{4}left(frac{1}{m_{2,1}}+frac{1}{m_{2,2}}right)+frac{1}{4}left(frac{1}{m_{2,3}}+frac{1}{m_{2,4}}right)right]nonumber &=frac{1}{4^2}left(frac{1}{m_{2,1}}+frac{1}{m_{2,2}}+frac{1}{m_{2,3}}+frac{1}{m_{2,4}}proper).nonumber finish{align}$$ The equation above offers the reciprocal of the equal mass of an AM-2.

It ought to be apparent that, after we add one other tier and go from Tier ##okay## all the way down to Tier ##okay+1##, we’re changing an AM-k into an AM-(okay+1). The reciprocal of the equal mass of the AM-(okay+1) could have twice as many plenty added reciprocally multiplied by an extra issue of ##frac{1}{4}## up entrance.  Thus, beginning with equation (1) for AM-1 one will get the expression for the equal mass of AM-2. From the AM-2 one will get the equal mass for AM-3 and so forth till, after N steps, one reaches equation (4) for the equal mass of the AM-N.

Rule III

In an AM-N, the strings that assist pulleys or plenty within the k-th Tier have the identical rigidity. It is the same as the load of the equal AM-0 divided by the variety of strings within the Tier, ##mathbf{T_k=dfrac{M_{textual content{eq}}^{(N)}~g}{2^okay}.}##

Proof

Begin with an AM-0 hanging from a ceiling.  The strain within the string is ##T_0=M_{textual content{eq}}^{(N)}~g##. Sequentially remodeling this AM-0 into equal AM-1, AM-2, . . . , AM-N will, after every step, double the variety of strings and halve the magnitude of the stress in every string.   It follows that ##T_k=dfrac{M_{textual content{eq}}^{(N)}~g}{2^okay}.##

Rule implementation

With the three guidelines above, discovering the accelerations of all of the plenty in an AM meeting is easy.

  1. Establish ##N## for the meeting at hand.  It may be vital to remodel plenty into equal pulleys to maintain the bookkeeping easy.  For instance, the so referred to as double Atwood machine has an AM-1 on one facet of the highest pulley and a mass on the opposite.  The meeting turns into an AM-2 by remodeling the lone mass into an equal AM-1.
  2. Use equation (4) to search out the equal mass of the meeting.
  3. Use ##T_N=dfrac{M_{textual content{eq}}^{(N)}~g}{2^N}## to search out the stress in Tier N.
  4. Use Newton’s second regulation to search out the acceleration of every mass relative to the lab body: $$start{align} T_N-m_{N,i} g=m_{N,i} a_{N,i} implies a_{N,i}=frac{T_N}{m_{N,i}}-g~~~~(i=1,dots,2N)finish{align}$$ Be aware that the acceleration of the i-th mass is optimistic if the stress is larger than its weight and unfavourable if the stress is lower than its weight.  In different phrases,  the implicit conference is “up” is optimistic and “down” is unfavourable.
  5. If desired, repeat steps 3 and 4 to search out the acceleration of the equal plenty of pulleys within the tiers above.

We see that equal plenty in the identical tier of an AM-N have equal accelerations whatever the values of the opposite plenty.  Utilizing equation (5) and the expressions for ##T_N## and ##M_{textual content{eq}}^{(N)}##, we discover that the sum of the accelerations is zero. $$ sum_{i=1}^{2^N} a_{N,i}  =T_N sum_{i=1}^{2^N}frac{1}{m_{N,i}}- sum_{i=1}^{2^N}g=frac{M_{textual content{eq}}^{(N)}~g}{2^N} frac{4^N}{M_{textual content{eq}}^{(N)}}- 2^N~g =0.$$That is anticipated as a result of the relative accelerations of plenty coupled beneath the identical pulley add to zero and the acceleration of the equal AM-0 is zero.

Moreover, we will discover the acceleration of the middle of mass. With ##M= sumlimits_{i=1}^{2^N}m_{N,i}##, $$ start{align} &sum_{i=1}^{2^N}m_{N,i} a_{N,i}=sum_{i=1}^{2^N}T_{N,i}-sum_{i=1}^{2^N}m_{N,i} g
implies MA_{textual content{cm}}=M_{textual content{eq}}^{(N)}g-Mg nonumber &implies A_{textual content{cm}}=left(frac{M_{textual content{eq}}^{(N)}}{M}-1right)g.nonumber finish{align}$$

Examples

Though the strategy is conceptually easy, the hierarchical construction of nested equal plenty makes the specific formulation of the accelerations by way of the given plenty impractical. It’s expedient to arrange a spreadsheet by which the association of cells follows the sample of the AM-N.  Such an association gives a handy at-a-glance view of the movement.

There are N+1 tiers labeled by ##0leq kleq N.## Tier N exhibits the ##2^N## values of the given plenty. Tier N-1 exhibits the ##2^{N-1}## calculated equal plenty of the pulleys supporting pairs of plenty under. These equal plenty change into “given” plenty for Tier N-2 and so forth.  Working from the underside up, one reaches Tier 0 which exhibits the configuration’s equal mass.  This completes step one of filling within the equal plenty.

The second step is filling within the tensions supporting every mass. The content material of the only cell on the prime tier multiplied by ##g## offers the equal weight of the AM-N, ##F_0##.  Understanding that, one strikes down tier by tier, halving the stress from the tier above, till one reaches the Tier N with the given plenty.

The third step is filling within the accelerations.  For every mass, $$textual content{Acceleration = Stress divided by mass – Acceleration of gravity.}$$ Accelerations of equal plenty are, after all, the accelerations of the pulleys they’ve changed.

The fourth and remaining step is filling within the acceleration of every mass relative to its pulley. This step is beneficial for validating the setup.  Lots (or pulleys) on both facet of a pulley above them should have accelerations including to zero in accordance with the constraint that lengths of strings passing over pulleys are fixed.

AM-2

We at the moment are prepared to handle the a number of selection query within the introduction. To search out the course of the angular acceleration of pulley P, we have to examine the equal plenty of pulleys A and B. Whichever is bigger determines the course of the online torque.

##M_{textual content{eq-A}}^{(1)}=4timesdfrac{1times 2}{1+2}~textual content{kg}=dfrac{8}{3}~textual content{kg}##

See Also

##M_{textual content{eq-B}}^{(1)}=4timesdfrac{4times 0.8}{4+0.8}~textual content{kg}=4timesdfrac{3.2}{4.8}~textual content{kg}=dfrac{8}{3}~textual content{kg}##

For the reason that equal plenty are equal, they don’t speed up therefore the angular acceleration of pulley P is zero.

Masses are in cells highlighted with color. Masses with the same color belong to the same pulley.The determine on the suitable exhibits the spreadsheet output of the AM-2 within the query.

  1. Lots are in cells highlighted with shade. Lots with the identical shade belong to the identical pulley.
  2. The given plenty are in Tier 2.  Tiers 0 and 1 present equal plenty of pulleys.
  3. The strain on a mass is one cell above.
  4. The acceleration of a mass relative to the lab is one cell under.
  5. The acceleration of a mass relative to its pulley is 2 cells under.

AM-3

Maybe a extra fascinating instance is an AM-3 with plenty chosen for instance options talked about earlier. We use double subscript ##i,j## to establish a mass in Tier ##i## at place ##j## from the left.

Noteworthy options

  1. Accelerations in the identical tier add to zero.
  2. Relative accelerations of same-color plenty add to zero.
  3. Lots ##m_{3,4}## and ##m_{3,5}## are equal however not paired beneath the identical pulley. Their accelerations are the identical whereas their accelerations relative to their pulleys are usually not.
  4. Lots ##m_{3,7}## and ##m_{3,8}## are equal and paired beneath the identical pulley.  Their accelerations are equal to the acceleration of their widespread pulley ##m_{2,4}.## Which means that they transfer as one with the pulley.  It additionally confirms that if we view ##m_{2,4}## as an AM-0 ignoring ##m_{3,7}## and ##m_{3,8},## the impact on the remainder of the meeting will likely be equivalent to viewing ##m_{2,4}## because the pulley of an AM-1 with ##m_{3,7}## and ##m_{3,8}## hanging from it.
  5. A mass, on this case ##m_{3,3}##, might have a optimistic acceleration higher than ##g=10~textual content{m/s}^2##. Nevertheless, no mass can have a unfavourable acceleration relative to the lab body with magnitude higher than ##g##.  That is according to the notion that “you may pull however not push with a string.”

AM-4 and past

We are able to lengthen the AM-3 above to an AM-4 calculation on a spreadsheet fairly simply.

  1. Create Tier 4 with the 16 given plenty.
  2. Exchange the outdated given mass values in Tier 3 with the suitable equal mass formulae.
  3. Watch the spreadsheet recalculate all the things.

Calculations past AM-4 will most likely require abandoning the spreadsheet in favor of code to compact the output. We summarize the equations for an AM-N:

Equal mass $$start{align}M_{textual content{eq}}^{(N)} = frac{2^{2N}}{{sumlimits_{i=1}^{2^N} left(frac{1}{m_{N,i}}proper)}}.finish{align}$$Stress on any of the ##2^{okay}## plenty in Tier ##okay~## (##0 leq okay leq N##) $$start{align}T_k=frac{M_{textual content{eq}}^{(N)}g}{2^{okay}}=frac{2^{2N-k}}{{sumlimits_{i=1}^{2^N} left(frac{1}{m_{N,i}}proper)}}g.finish{align}$$Acceleration of ##j##th mass in Tier ##okay## $$start{align} a_{okay,j}=frac{T_k}{m_{okay,j}}-g=left[frac{M_{text{eq}}^{(N)}}{2^{k}~m_{k,j}}-1right]g=left[frac{2^{2N-k}}{m_{k,j}{sumlimits_{i=1}^{2^N} left(frac{1}{m_{N,i}}right)}}-1right]g.finish{align}$$

AM##-{infty}##

infinite Atwood machineThe so referred to as infinite Atwood machine consists of an infinite ladder of pulleys every with mass ##m## on one facet and a pulley on the opposite (see determine on the suitable).  Of curiosity are the accelerations of the plenty ##m.##

Let ##M_{textual content{eq}}^{(infty)}## be the equal mass of the complete meeting, i.e. the equal mass of the AM-0.  Take into account the AM-1 equal of the meeting. Its single pulley has mass ##m## on the left and ##M_{textual content{eq}}^{(infty)}## on the suitable.  The equal mass of this AM-1 is ##M_{textual content{eq}}^{(infty)}##.  Thus, $$M_{textual content{eq}}^{(infty)}=frac{4mM_{textual content{eq}}^{(infty)}}{m+M_{textual content{eq}}^{(infty)}}implies M_{textual content{eq}}^{(infty)}=3m.$$ Then,$$start{align} & T_1=frac{1}{2}M_{textual content{eq}}^{(infty)}~g=frac{3}{2}mgimplies a_1=frac{T_1}{m}-g=frac{3}{2}g-g=frac{1}{2}g nonumber & T_2=frac{1}{4}M_{textual content{eq}}^{(infty)}~g=frac{3}{4}mgimplies a_2=frac{T_2}{m}-g=frac{3}{4}g-g=-frac{1}{4}g nonumber &dots nonumber & T_N=frac{1}{2^N}M_{textual content{eq}}^{(infty)}~g=frac{3}{2^N}mgimplies a_N=frac{T_N}{m}-g=left(frac{3}{2^N}-1right)g nonumber &dots nonumber finish{align}$$Solely the topmost mass ##m## has acceleration up. The remainder of the accelerations are down approaching free fall as ##Nrightarrowinfty.##

Afterthoughts

There’s sufficient resemblance between the reciprocal mass addition in equation (1) and the expression for equal capacitance in a sequence mixture to make one ponder whether there may be {an electrical} analog for an AM-N.  I think there may be however, to maintain this story quick, I’ll save the concept for a separate article.

I must also be aware that Atwood machines changing hanging plenty in customary issues could make a easy state of affairs seem difficult.  For instance, connect an AM-N to a vertical spring and set it into oscillations. The oscillation frequency is, after all, ##omega=sqrt{okay/M_{textual content{eq}}^{(N)}}.## The acceleration of the equal mass has the shape ##A(t)=-omega^2x_0sin(omega t+phi).##  A separate AM-N calculation, following the strategies outlined right here, yields the acceleration of the plenty ##a_{AM,i}## relative to a set assist. By Rule I, the accelerations of the hanging plenty relative to the lab body will likely be  ##a_i(t)=a_{AM,i}-A(t).##

This text was impressed by a latest homework problem.  I hope that the simplicity of the strategy will encourage the reader to undertake the equal mass method in lieu of FBDs or Lagrangians when treating a number of Atwood machine issues.  Quoting David J.  Griffiths in his textbook Introduction to Quantum Mechanics quoting Peter Lorre within the movie Arsenic and Outdated Lace, “Do it ze kveek vay, Johnny!”

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