An Interesting Mechanics Problem - My Attempt

Recently, the following problem was shared in the Brilliant Community:

My attempt is as follows. Consider the diagram at an arbitrary instant as shown below.

Let:

BAC=2ϕ\angle BAC = 2\phi

Say the mass BB is situated at a distance ss along the X-axis. Then the coordinates of each mass are:

xB=sx_B = s yB=0y_B = 0

xA=s+Lsinϕx_A = s + L\sin{\phi} yA=Lcosϕy_A = L\cos{\phi}

xC=s+2Lsinϕx_C = s + 2L\sin{\phi} yC=0y_C = 0

Since no external force acts one the system along the X direction, the momentum along X is conserved. This gives:

mx˙A+mx˙B+2mx˙C=0m\dot{x}_A + m\dot{x}_B + 2m\dot{x}_C = 0

The initial momentum of the system is zero. Plugging in all variables above and simplifying gives:

s˙=54Lϕ˙cosϕ(1)\dot{s} = -\frac{5}{4}L \dot{\phi} \cos{\phi} \dots (1)

The kinetic energy of this system is:

T=m2(x˙A2+y˙A2)+m2(x˙B2+y˙B2)+2m2(x˙C2+y˙C2)\mathcal{T} = \frac{m}{2}\left(\dot{x}_A^2 + \dot{y}_A^2\right) + \frac{m}{2}\left(\dot{x}_B^2 + \dot{y}_B^2\right) + \frac{2m}{2}\left(\dot{x}_C^2 + \dot{y}_C^2\right)

Simplifying the above expression gives:

T=2ms˙2+5mLs˙ϕ˙cosϕ+mL2ϕ˙22+4mL2cos2ϕϕ˙2\mathcal{T} = 2m\dot{s}^2 + 5mL\dot{s} \dot{\phi}\cos{\phi} + \frac{mL^2\dot{\phi}^2}{2} + 4mL^2\cos^2{\phi} \dot{\phi}^2

Eliminating s˙\dot{s} by using (1) and simplifying gives:

T=7mL2ϕ˙2cos2ϕ8+mL2ϕ˙22\mathcal{T} = \frac{7mL^2\dot{\phi}^2\cos^2{\phi}}{8} + \frac{mL^2\dot{\phi}^2}{2}

The potential energy of the system is:

V=mgyA+mgyB+2mgyC\mathcal{V} = mgy_A + mgy_B + 2mgy_C V=mgLcosϕ\mathcal{V} = mgL\cos{\phi}

Initial potential energy is:

Vo=mgL\mathcal{V}_o = mgL

Initial kinetic energy is:

To=0\mathcal{T}_o = 0

ϕ(0)=0\phi(0)=0 ϕ˙(0)=0\dot{\phi}(0) = 0

Applying conservation of energy gives:

T+V=To+Vo\mathcal{T} + \mathcal{V} = \mathcal{T}_o + \mathcal{V}_o     7mL2ϕ˙2cos2ϕ8+mL2ϕ˙22+mgLcosϕ=mgL\implies \frac{7mL^2\dot{\phi}^2\cos^2{\phi}}{8} + \frac{mL^2\dot{\phi}^2}{2} + mgL\cos{\phi} = mgL

    ϕ˙=g(1cosϕ)7Lcos2ϕ8+L2\implies \dot{\phi} = \sqrt{\frac{g(1-\cos{\phi})}{\frac{7L\cos^2{\phi}}{8} + \frac{L}{2}}}

Now, when the angle between the rods is 30 degrees, then:

ϕ=π12\phi = \frac{\pi}{12}

This gives the value of ϕ˙\dot{\phi} at that instant.

Finally:

xA=s+Lsinϕx_A = s + L\sin{\phi} yA=Lcosϕy_A = L\cos{\phi}

x˙A=s˙+Lcosϕϕ˙=Lcosϕϕ˙4\dot{x}_A = \dot{s} + L\cos{\phi}\dot{\phi} = -\frac{L\cos{\phi} \dot{\phi}}{4} y˙A=Lsinϕϕ˙\dot{y}_A = -L\sin{\phi} \dot{\phi}

The velocity of point AA at this instant :

vA=x˙A i^+y˙A j^\vec{v}_A = \dot{x}_A \ \hat{i} + \dot{y}_A \ \hat{j}

The final expression for speed is:

V=gL2(1cosϕ)(1+15sin2ϕ7cos2ϕ+4)\boxed{V = \sqrt{\frac{gL}{2}\left(1 - \cos{\phi}\right)\left(\frac{1 + 15\sin^2{\phi}}{7\cos^2{\phi}+4}\right)}}

Assumptions:

  • The cylinders are treated as point masses.

Note by Karan Chatrath
1 week, 1 day ago

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1 vote

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@Lil Doug This is my attempt. Unable to tag KrishnaKarthik. Do tag him if possible

@Steven Chase @Mark Hennings Your inputs would be helpful.

Karan Chatrath - 1 week, 1 day ago

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Hey; very nice approach. So this is what momentum is used for, outside of collisions, of course. Nice.

Krishna Karthik - 1 week, 1 day ago

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Suppose that the horizontal displacement of AA from its starting position is xx, and that the angle BAC=2θ\angle BAC = 2\theta. Then A,B,CA,B,C have position vectors (xLcosθ)(xLsinθ0)(x+Lsinθ0) \binom{x}{L\cos\theta} \hspace{2cm} \binom{x - L\sin\theta}{0} \hspace{2cm} \binom{x+L\sin\theta}{0} respectively, so that the kinetic energy of the system is T=12m(x˙2+L2sin2θθ˙2)+12m(x˙Lcosθθ˙)2+m(x˙+Lcosθθ˙)2 T = \tfrac12m(\dot{x}^2 + L^2\sin^2\theta \dot{\theta}^2) + \tfrac12m(\dot{x}-L\cos\theta\dot{\theta})^2 + m(\dot{x} + L\cos\theta\dot{\theta})^2 Since the xx-coordinate of the centre of mass of the system remains constant, we have 4x+Lsinθ=04x + L\sin\theta = 0 (the system starts with x=0x=0, θ=0\theta = 0. Thus T  =  18mL2(4+7cos2θ)θ˙2 T \; = \; \tfrac18mL^2(4 + 7\cos^2\theta)\dot{\theta}^2 and so conservation of energy gives 18mL2(4+7cos2θ)θ˙2+mgLcosθ=  mgLθ˙2=  8g(1cosθ)L(4+7cos2θ) \begin{aligned} \tfrac18mL^2(4 + 7\cos^2\theta)\dot{\theta}^2 + mgL\cos\theta & = \; mgL \\ \dot{\theta}^2 & = \; \frac{8g(1 - \cos\theta)}{L(4 + 7\cos^2\theta)} \end{aligned} Thus the speed (not velocity) of AA is vv, where v2  =  x˙2+L2sin2θθ˙2  =  116L2(1+15sin2θ)θ˙2  =  gL(1cosθ)(1+15sin2θ)2(4+7cos2θ) v^2 \; = \; \dot{x}^2 + L^2\sin^2\theta \dot{\theta}^2 \; = \; \tfrac{1}{16}L^2(1 + 15\sin^2\theta)\dot{\theta}^2 \; = \; \frac{gL(1 - \cos\theta)(1 + 15\sin^2\theta)}{2(4 + 7\cos^2\theta)} The question should say: "what is the speed of AA when the rods make an angle θ\theta with the horizontal", that that will create the necessary swap between sinθ\sin\theta and cosθ\cos\theta.

@Karan Chatrath @Lil Doug

Mark Hennings - 1 week, 1 day ago

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@Mark Hennings Thanks but sir which rod?

Lil Doug - 1 week, 1 day ago

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Both. They have to make the same angle with the ground, since they have the same length, and so triangle ABCABC is isosceles!

Mark Hennings - 1 week, 1 day ago

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@Mark Hennings @Mark Hennings But sir at right side mass is 2 times of left side. Still they will form an isosceles triangle

Lil Doug - 1 week, 1 day ago

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@Lil Doug The masses at the bottom do not affect the length of the rods! The uneven masses mean that xx does not stay constant. If the masses at the bottom were equal, then xx would remain zero throughout.

Mark Hennings - 1 week, 1 day ago

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@Mark Hennings @Mark Hennings Thank you so much, the problem is corrected now.

Lil Doug - 1 week, 1 day ago

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@Lil Doug Is it? The correct formula gives 1.3114664081.311466408 when θ=30\theta = 30^\circ, which is not the official answer. When changing the angle from against the vertical to against the horizontal, don't you have to change your value of θ\theta?

Mark Hennings - 1 week, 1 day ago

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@Mark Hennings @Mark Hennings So 1.311...1.311... and 30°30° are correct set.

Lil Doug - 1 week, 1 day ago

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@Lil Doug Yes, but 1.311... is not the official answer. Should it be θ=60\theta = 60^\circ and v=5.160468...v = 5.160468...? Or was the old idea that θ\theta was the angle between the rods, in which case we now want θ=75\theta = 75^\circ and v=7.8851...v = 7.8851...?

Mark Hennings - 1 week, 1 day ago

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@Mark Hennings @Mark Hennings Thanks I have corrected it.

Lil Doug - 1 week, 1 day ago

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@Lil Doug I'm getting 1.311 1.311 for the speed of the hinge mass when the angle between the rod and the horizontal is 60 60 degrees. But it's rejecting that.

Steven Chase - 1 week, 1 day ago

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@Steven Chase @Steven Chase Sir go and rock now.

Lil Doug - 1 week, 1 day ago

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@Lil Doug Thanks. I'll post a solution some time today

Steven Chase - 1 week, 1 day ago

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@Steven Chase @Steven Chase You are Welcome\Huge Welcome

Lil Doug - 1 week, 1 day ago

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@Steven Chase @Steven Chase Sir please check your last 1 hour notification

Lil Doug - 1 week, 1 day ago

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@Mark Hennings So as I understand, θ\theta is to the horizontal in the problem statement. The expression derived is by taking the angle definition as the half-angle to the vertical, so in the expression that I obtained, ϕ=θ\phi=\theta should be 60 degrees instead of 30.

Karan Chatrath - 1 week, 1 day ago

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Thank you for posting this. This clarifies my concern about the answer.

Karan Chatrath - 1 week, 1 day ago

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@Karan Chatrath The problem is corrected now.

Lil Doug - 1 week, 1 day ago

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@Krishna Karthik

Lil Doug - 1 week, 1 day ago

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@Karan Chatrath Thanks for the note and attempt

Lil Doug - 1 week, 1 day ago

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@Karan Chatrath sorry, yes but that type of sign are generally used in Lagrange.

Lil Doug - 1 week, 1 day ago

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The answer which is given behind the book is gl2(1sinθ)(1+15cos2θ)(4+7sin2θ)\sqrt{\frac{gl}{2}(1-\sin \theta)\frac{(1+15 \cos^{2} \theta )}{(4+7 \sin^{2} \theta)}}

Lil Doug - 1 week, 1 day ago

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I have added the expression that I have found in the note.

Karan Chatrath - 1 week, 1 day ago

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@Karan Chatrath How do you make that beautiful figure in your attempt?

Lil Doug - 1 week, 1 day ago

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@Karan Chatrath Substituting the values in your expression gives 1.31146641\approx 1.31146641

Lil Doug - 1 week, 1 day ago

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In my expression, ϕ=π/12\phi = \pi/12 and not π/6\pi/6

Karan Chatrath - 1 week, 1 day ago

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@Karan Chatrath oh yes, after substituting the correct Values I am getting 7.8857.885

Lil Doug - 1 week, 1 day ago

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@Lil Doug I still do not know what is my mistake. Maybe you can spot it.

Karan Chatrath - 1 week, 1 day ago

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@Karan Chatrath @Karan Chatrath at that time when you reported the problem you were saying, the answer is 0.40.4

By the way, i have deleted that problem and again , I have reuploaded , 2 hours ago.
Did you noticed that?

Lil Doug - 1 week, 1 day ago

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@Lil Doug Yes, that is the answer I get with the expression in the note. No, I did not notice

Karan Chatrath - 1 week, 1 day ago

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@Karan Chatrath @Karan Chatrath if you don't mind But now putting values in your expression gives 1.414\approx 1.414

Lil Doug - 1 week, 1 day ago

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@Lil Doug Check again

Karan Chatrath - 1 week, 1 day ago

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@Karan Chatrath @Karan Chatrath Now I am. Getting 0.175\approx 0.175
As per again and again I put values, everytime I get a new answer.

Lil Doug - 1 week, 1 day ago

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@Lil Doug Did you solve it at all? Made any headway? Really tough problem. I used Lagrangian mech. and got 90% close to the answer.

Krishna Karthik - 1 week, 1 day ago

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@Krishna Karthik what is tough in this. It is a medium level problem

Lil Doug - 1 week, 1 day ago

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Medium level? Not in most of the stuff in the community... nope. It's kind of tough. I would rate it medium-hard.

Krishna Karthik - 1 week, 1 day ago

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@Krishna Karthik @Krishna Karthik Okk okay

Lil Doug - 1 week, 1 day ago

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@Lil Doug I'm gonna do a time-domain version of this problem. Wish me luck :)

Krishna Karthik - 1 week, 1 day ago

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@Krishna Karthik @Krishna Karthik ok. Can we have 1 chess match

Lil Doug - 1 week, 1 day ago

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@Lil Doug Sorry; not now. I'm working on physics. Btw my school term has ended! YAYYYYYYY

Yeah, so it's now the holidays. More math and physics :)

Krishna Karthik - 1 week, 1 day ago

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@Lil Doug Hi Lil Doug, please refrain from using swear words on Brilliant. We do not want to encourage such behaviors.

I've edited your comment.

Brilliant Mathematics Staff - 6 days, 3 hours ago

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