As shown in figure there is Two

IdenticalCylinders each of mass "M" and Radius "R". Now If Lower cylinder isdisplace slightlyto the rightward direction then, Find Themaximum Velocityof each Cylinder ?\(\bullet \) All Surfaces are

smooth.\(\bullet \) Gravity is in downward direction as usual.

Please Post Solution Instead of posting answers.

**Update**:

Repeat this question if there is friction between the cylinders.

## And Then Try question based on it Click here

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TopNewestHere is my method when

frictionis present between only cylinders. By using concept of Instantaneous centre of rotation [ ICR ]Because This system is an equivalent Rigid Body. Since There is no relative motion between point of contact of cylinder.Figure

The whole system is assumed to be purely rotated about "O"

Then

so velocity of any point P of this combined system in ICR ( O ) frame is given By

\[{ V }_{ p/o }\quad =\quad { r }_{ p/0 }\quad \times \quad { \omega }_{ 0 }\].

\[{ V }_{ 1 }\quad =\quad (2R\cos { \theta } ){ \omega }_{ 0 }\\ \\ { V }_{ 2 }\quad =\quad (2R\sin { \theta } ){ \omega }_{ 0 }\\ \\ \Rightarrow \quad { \omega }_{ 0 }\quad =\quad \cfrac { \sqrt { { V }_{ 1 }^{ 2 }\quad +\quad { V }_{ 2 }^{ 2 } } }{ 2R } \].

Now using energy conservation of system in ICR frame we get

\(2gR(1-\cos { \theta } )\quad =\quad \frac { 1 }{ 2 } { I }_{ o }{ { \omega }_{ 0 } }^{ 2 }\quad \quad \quad \longrightarrow (1)\\ \\ { I }_{ o }\quad =\quad { mR }^{ 2 }\quad +{ \quad mR }^{ 2 }\quad +\quad { 4mR }^{ 2 }\cos ^{ 2 }{ \theta } \quad +{ \quad 4mR }^{ 2 }\sin ^{ 2 }{ \theta } \\ \\ { I }_{ o }\quad =\quad 6{ mR }^{ 2 }\\ \\ \\ \Rightarrow \quad { \omega }_{ 0 }\quad =\quad \sqrt { \cfrac { 2g }{ 3R } (1-\cos { \theta } ) } \quad \longrightarrow \quad (2)\\ \\ \\ { V }_{ 1 }^{ 2 }\quad =\quad \cfrac { 8gR }{ 3 } (1-\cos { \theta } )\cos ^{ 2 }{ \theta } \quad \longrightarrow \quad (3)\\ \\ \\ { V }_{ 2 }^{ 2 }\quad =\quad \cfrac { 8gR }{ 3 } (1-\cos { \theta } )\sin ^{ 2 }{ \theta } \quad \longrightarrow \quad (4)\).

Now using maxima concept on velocity of cylinder - 1 ( Lower Cylinder )

\[\cfrac { d({ V }_{ 1 }^{ 2 }) }{ d\theta } \quad =\quad 0\\ \\ \boxed { \theta \quad =\quad \cos ^{ -1 }{ \cfrac { 2 }{ 3 } } } \\ \\ Or\\ \quad \quad \\ \quad \quad \quad \quad \theta \quad =\quad 0\quad \quad \quad \times \quad (rejeccted\quad )\\ \quad \quad \quad \quad \quad \\ Or\\ \quad \quad \quad \quad \theta \quad =\quad \cfrac { \pi }{ 2 } \quad \quad \times \quad (\quad rejected\quad )\].

So again cylinder will loose contact at the same hight as that of in 1st Part of this question ( when no friction ) I rejected other values of angle because velocity of lower cylinder never be zero. which we have discuss earlier. And Now Further Motion is independent

So we get maximum velocity of lower block

\[\boxed { { V }_{ 1,max }\quad =\quad \sqrt { \cfrac { 32gR }{ 81 } } } \].

Now velocity of 2nd cylinder can be easily calculated. – Deepanshu Gupta · 2 years, 6 months ago

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As \(r{ \omega }_{ 2 }-{ v }_{ 2 }sin\theta =r{ \omega }_{ 1 }+{ v }_{ 1 }cos\theta \)

and \({ \omega }_{ 2 }={ -\omega }_{ 1 }\).

So \({ \omega }_{ 1 }=\frac { { v }_{ 1 } }{ 2rcos\theta } \)

and \({ \omega }_{ 2 }=-\frac { { v }_{ 1 } }{ 2rcos\theta } \)

By energy conservation

\(mg2r(1-cos\theta )=\frac { 1 }{ 2 } m{ v }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { v }_{ 2 }^{ 2 }+\frac { 1 }{ 2 } { I\omega }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { I\omega }_{ 2 }^{ 2 }\)

On putting the values

\(g2r(1-cos\theta )=\frac { { v }_{ 1 }^{ 2 } }{ 2{ cos }^{ 2 }\theta } +\frac { { v }_{ 1 }^{ 2 } }{ 4{ cos }^{ 2 }\theta } \)

and \({ v }_{ 1 }^{ 2 }=\frac { 8gr(1-cos\theta ){ cos }^{ 2 }\theta }{ 3 } \).

I got the same expression of \(v_{1}\).

Deepanshu and Mvs ,what you guys think about my method. Is this correct? – Satvik Pandey · 2 years, 6 months ago

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– Mvs Saketh · 2 years, 6 months ago

your answer as far as i can comprehend seems absolutely correctLog in to reply

I have tried

as \({ \alpha }_{ 1 }=-{ \alpha }_{ 2 }\)

so \(\frac { d{ \omega }_{ 1 } }{ dt } =-\frac { d{ \omega }_{ 2 } }{ dt } \)

or \(\int { d{ \omega }_{ 1 } } =-\int { d{ \omega }_{ 2 } }\)

or \({ \omega }_{ 1 }=-{ \omega }_{ 2 }+C\)

At \(t=0\) \({ \omega }_{ 1 }={ \omega }_{ 2 }=0\) So \(C=0\)

So \({ \omega }_{ 1 }=-{ \omega }_{ 2 }\)

Is that how you proved \({ \omega }_{ 1 }=-{ \omega }_{ 2 }\)?

@Mvs Saketh and @DEEPANSHU GUPTA please reply. – Satvik Pandey · 2 years, 6 months ago

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heres what we have

w1=w2

also that \(r{ w }_{ 1 }-{ v }_{ 2 }sin\theta \quad =\quad r{ w }_{ 1 }+{ v }_{ 1 }cos\theta \)

as w1=w2 so rw1=rw2, thus cancelling like terms

\(-{ v }_{ 2 }sin\theta \quad =\quad { +v }_{ 1 }cos\theta \)...(1)

now already through translatory constraint

we have \({ v }_{ 2 }cos\theta \quad =\quad { +v }_{ 1 }sin\theta \)...(2)

dividing one and 2 , we get

\(tan\theta \quad =\quad -cot\theta \) which is not true for all angles,, clearly we are wrong,, i do not know where,, i do now know how,, but i would say we are wrong,, this is giving contradictory results,, either this means that such a situation is not possible or that we are missing something

@DEEPANSHU GUPTA @satvik pandey – Mvs Saketh · 2 years, 6 months ago

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And Satvik your solution is Partially incorrect. !!

See carefully What is I'am Trying to Saying ...!!

Fig-1

Fig-1

Now Since Cylinder are Rolling without slipping so There is no relative motion along common Tangent.

\({ \omega }_{ 2 }R\quad -\quad { V }_{ 2 }\sin { \theta } \quad =\quad { V }_{ 1 }\cos { \theta } -\quad \quad { \omega }_{ 1 }R\quad \quad ...........\quad (1)\).

Now Since Rod is Rigid So there is no relative motion between cylinders along Common Normal .

\({ V }_{ 2 }\cos { \theta } \quad =\quad { V }_{ 1 }\sin { \theta } \quad \quad .................(2)\).

Now By writing Torque equation we get :

\({ \alpha }_{ 1 }\quad =\quad \frac { fR }{ I } \quad \quad (\quad Anti\quad clock\quad wise\quad )\\ \\ { \alpha }_{ 2 }\quad =\quad \frac { fR }{ I } \quad (\quad Anti\quad Clock\quad wise\quad )\\ \\ { \alpha }_{ 1 }\quad =\quad { \alpha }_{ 2 }\).

Now after integrating it and put limits we get

\({ \omega }_{ 1 }\quad =\quad { \omega }_{ 2 }\quad ........................\quad (3)\).

So @satvik pandey Your Method is incorrect . But Your Final expression for \({ V }_{ 1 }\). as a function of theta is correct .

Because in energy equation :

\(\boxed { \\ { { (-\omega }_{ 1 }) }^{ 2 }{ { \quad =\quad (\omega }_{ 1 }) }^{ 2 } } \).

And If You want more Rigorous Proof of the fact that

\(\boxed { { \omega }_{ 1 }\quad =\quad { \omega }_{ 2 } } \).

Then I would say that In My solution I used ICR .. and calculate \({ \omega }_{ ICR }\quad =\quad { \omega }_{ 0 }\).

Note That this system (

both cylinder) is considers asrigid Body( not discrete because there is no relative motion between them ) So according to fundamental rule ofRotational Dynamicsthat allRotational ParametersLike as \(\\ { \alpha }\quad ,{ \quad \omega }\quad etc.\). wouldremain sameabout all Point's onRigid body system.So In Fact\(\boxed { { \omega }_{ 1 }\quad =\quad { \omega }_{ 2 }\quad ={ \quad \omega }_{ ICR }\quad =\quad { \omega }_{ 0 }\quad =\quad ...........(Any\quad Point)\quad \quad } \).

Hope This might be Helps You !! – Deepanshu Gupta · 2 years, 6 months ago

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Here a

{1} : Acceleration of upper cylinder in downward dir. And a{2} : Acceleration of lower cylinder in horizontal dir. N , f : Normal reaction and friction(As in your diagram above) at point of contact. Theta: Angle made by line joining centres with the vertical. \( f.R=I\alpha \\ mg-Ncos\theta -fsin\theta =ma_{ 1 }\\ Nsin\theta -fcos\theta =ma_{ 2 }\\ \\ Equating\quad the\quad acceleration\quad of\quad point\quad of\quad contact\quad on\quad both\quad cylinders:\\ -a_{ 1 }\hat { j } +\quad R\alpha cos\theta \hat { i } +\quad R\alpha sin\theta \hat { j } =\quad a_{ 2 }\hat { i } -\quad R\alpha cos\theta \hat { i } -\quad R\alpha sin\theta \hat { j } \\ \therefore \quad a_{ 2 }=2R\alpha cos\theta \quad And\quad a_{ 1 }=2R\alpha sin\theta \quad \\ On\quad solving\quad a_{ 2 }=\frac { 2gsin\theta cos\theta }{ 3 } \quad ...(1)\\ X\quad co-ordinate\quad of\quad Centre\quad of\quad Bottom\quad cylinder,\quad x\quad =\quad R+2Rsin\theta \\ \therefore \quad dx=2Rcos\theta \quad \dot { \theta } \\ Writing\quad a_{ 2 }\quad as\quad v\quad dv/dx\quad and\quad substituting\quad with\quad dx\quad in\quad (1)\\ \\ v\quad dv\quad =\quad \frac { 4Rgsin\theta \cos ^{ 2 }{ \theta } }{ 3 } d\theta \\ Integration\quad gives\quad us\quad { v }^{ 2 }=\frac { 8gR }{ 9 } (1-\cos ^{ 3 }{ \theta } )\\ Which\quad gives\quad me\quad a\quad different\quad result\quad for\quad maximum\quad velocity.\quad Is\quad this\quad wrong?\quad Why\quad so?\\ Thanks\quad a\quad lot. \) – Jatin Sharma · 1 year, 3 months agoLog in to reply

@Deepanshu Gupta sir when the radius of two cylinders are different then w1,w2,w(icr) are not same eventhough they form a rigid body – Sashank Bonda · 5 months, 4 weeks ago

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– Sashank Bonda · 5 months, 4 weeks ago

will this result hold when radius of two cylinder are different?Log in to reply

But this shows that rw2 can never be greater than v2(sin(x)) in the process,,, otherwise we get the contradiction. :) – Mvs Saketh · 2 years, 6 months ago

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Here a

{1} : Acceleration of upper cylinder in downward dir. And a{2} : Acceleration of lower cylinder in horizontal dir. N , f : Normal reaction and friction(As in your diagram above) at point of contact. Theta: Angle made by line joining centres with the vertical. \( f.R=I\alpha \\ mg-Ncos\theta -fsin\theta =ma_{ 1 }\\ Nsin\theta -fcos\theta =ma_{ 2 }\\ \\ Equating\quad the\quad acceleration\quad of\quad point\quad of\quad contact\quad on\quad both\quad cylinders:\\ -a_{ 1 }\hat { j } +\quad R\alpha cos\theta \hat { i } +\quad R\alpha sin\theta \hat { j } =\quad a_{ 2 }\hat { i } -\quad R\alpha cos\theta \hat { i } -\quad R\alpha sin\theta \hat { j } \\ \therefore \quad a_{ 2 }=2R\alpha cos\theta \quad And\quad a_{ 1 }=2R\alpha sin\theta \quad \\ On\quad solving\quad a_{ 2 }=\frac { 2gsin\theta cos\theta }{ 3 } \quad ...(1)\\ X\quad co-ordinate\quad of\quad Centre\quad of\quad Bottom\quad cylinder,\quad x\quad =\quad R+2Rsin\theta \\ \therefore \quad dx=2Rcos\theta \quad \dot { \theta } \\ Writing\quad a_{ 2 }\quad as\quad v\quad dv/dx\quad and\quad substituting\quad with\quad dx\quad in\quad (1)\\ \\ v\quad dv\quad =\quad \frac { 4Rgsin\theta \cos ^{ 2 }{ \theta } }{ 3 } d\theta \\ Integration\quad gives\quad us\quad { v }^{ 2 }=\frac { 8gR }{ 9 } (1-\cos ^{ 3 }{ \theta } )\\ Which\quad gives\quad me\quad a\quad different\quad result\quad for\quad maximum\quad velocity.\quad Is\quad this\quad wrong?\quad Why\quad so?\\ Thanks\quad a\quad lot. \) – Jatin Sharma · 1 year, 3 months agoLog in to reply

fig

I also want to show how I got w1=-w2

as \({ \alpha }_{ 1 }={ \alpha }_{ 2 }\)

so \(-\frac { d{ \omega }_{ 1 } }{ dt } =\frac { d{ \omega }_{ 2 } }{ dt } \) (I have added -ve sign because direction of a1 and w1 are opposite to each other)

or \(-\int { d{ \omega }_{ 1 } } =\int { d{ \omega }_{ 2 } }\)

or \({- \omega }_{ 1 }={ \omega }_{ 2 }+C\)

At \(t=0\) \({ \omega }_{ 1 }={ \omega }_{ 2 }=0\) So \(C=0\)

So \({ \omega }_{ 1 }=-{ \omega }_{ 2 }\)

What you guys think?

@DEEPANSHU GUPTA @Mvs Saketh – Satvik Pandey · 2 years, 6 months ago

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– Satvik Pandey · 2 years, 6 months ago

The equation is \(r{ \omega }_{ 2 }-{ v }_{ 2 }sin\theta =r{ \omega }_{ 1 }+{ v }_{ 1 }sin\theta \) (I think you typed it wrong there. You wrote w1 instead of w2) and also \( \omega_{1}=- \omega_{2}\) (you missed that minus sign). If I am not wrong I think you are getting this result due to wrong signs.Log in to reply

– Mvs Saketh · 2 years, 6 months ago

no w1=w2 is okay i am comparing magnitudes,, but yes wrong signs actually i do not know yet whether rw2 greater or v2 sin(x) so my contradiction arises because i am assuming rw2 is greater and the contradiction shows that such a case is never possibleLog in to reply

– Mvs Saketh · 2 years, 6 months ago

I think it is correct, absolutely,, correct,, except dont reject theta=0 , at that instant the lower cylinder is actually at rest, only that we already know it was at rest at that instantLog in to reply

– Deepanshu Gupta · 2 years, 6 months ago

why I do not reject \(\theta \) = 0 ?? I dont think that it is useful since it gives intial condition in that case it is minimum not maximum so it is useless so I reject it..! Sorry But I can't understand what are you trying to say ? Isn't I make any mistake ?Log in to reply

– Mvs Saketh · 2 years, 6 months ago

No its not a mistake its not useful surely,, i meant that you rejected pi/2 because it cant exist,, but rejecting 0 along with that makes it appear as if it cant exist either,, :P not that you are wrong or it is usefulLog in to reply

– Deepanshu Gupta · 2 years, 6 months ago

Okay :)Log in to reply

– Satvik Pandey · 2 years, 6 months ago

Thank you Deepanshu and Saketh for discussing this question. This discussion is really very useful. Once again thank you guys. :)Log in to reply

Note-

EDITEDdue to some errorsThe situation

Ok so heres the thing, some quick conclusions to draw from observation--1) The only energy supplier in the system is gravity,

2) That as long as the balls remain in contact , their velocities are constrained, by the relation, \({ v }_{ 1 }tan\theta \quad =\quad { v }_{ 2 }\) where the angle is between the line joining their radii and the normal vertical( v1 for lower ball, v2 for upper ball)

3) For now i am going to assume that the balls remain in contact till the upper ball reaches the floor, we will soon see how this is not true and they depart before that

4)The upper ball only falls vertically because the lower ball only presses it against the wall which offers an equivalent normal reaction for that to happen,

5) Consequently i can say the lower ball only moves horizontally,

Now applying energy conservation ( with potential energy taken 0 at the height of the

comof lower sphere) and substituting constraint relation i can say\(mg(2R(1-cos\theta ))\quad =\quad \frac { m{ { v }_{ 1 } }^{ 2 } }{ 2 } +\frac { m{ v }_{ 2 }^{ 2 } }{ 2 } =\frac { m{ { v }_{ 1 } }^{ 2 } }{ 2 } (sec^{ 2 }\theta )\)

now using calculus i can show that maxima occurs at

theta= arc cos(2/3)degrees (provided i did the calculations correct)which is

* \( \sqrt {16 gR /27} \) *HOWEVERthis is where theTWISToccurs,, because if my equations were correct then according to it the speed of the lower ball becomes 0 at angle 90 degrees,, but how is that possible when there is no force acting oppsite to it,, so its velocity can never decrease, so at its maxima or at anglearc cos (2/3)degrees,, the spheres depart and further motion is independent,,Thus at that angle the two balls get departed,, now you can easily find subsequent velocity of the upper ball as it is free falling under gravity

Now velocity of upper ball at instant of departure is \( \sqrt {20 gR /27} \) now individually applying energy conservation for upper ball find velocity max which occurs just before touching ground,,,

*Am i correct @DEEPANSHU GUPTA *– Mvs Saketh · 2 years, 7 months agoLog in to reply

But I think final answer should be \[{ V }_{ 1,max }=\quad \sqrt { \frac { 16gR }{ 27 } } \\ \\ { V }_{ 2,max }=\quad \sqrt { \frac { 216gR }{ 45 } } \].

And According to you

\(\frac { d{ V }_{ 1 }^{ 2 } }{ d\theta } \quad =\quad 0\\ \\ \quad \theta \quad =\quad 0\quad \quad \times \quad (\quad initial\quad condition\quad )\\ \\ \quad or\quad \\ \quad \quad \theta \quad =\quad \frac { \pi }{ 2 } \quad \quad (\quad rejected\quad )\\ \quad or\quad \\ \boxed { \theta \quad =\quad \cos ^{ -1 }{ \frac { 2 }{ 3 } } } \).

I'am agreeing with you that the condition

\[\theta \quad =\quad \frac { \pi }{ 2 } \]. is rejected because \[{ V }_{ 1 }\quad =\quad 0\quad \quad \quad (\because \quad { V }_{ 2 }\quad \neq \quad 0\quad )\quad \].

which is not possible since lower cylinder never get's stop.

But I have doubt that why does this condition is arouse after differentiation ? Is it coincident or some logic behind it ?? – Deepanshu Gupta · 2 years, 7 months ago

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Also @DEEPANSHU GUPTA it is not a coincidence,,, when the derivative of the velocity of lower cylinder becomes 0 , it means its acceleration(except here rate of change with theta and not time, both implie its independence for all future instants) becomes 0 and when does it become 0,, when no force acting on it anymore and when does that happen, when it loses contact with the upper cylinder,,

(note acceleration is rate of change with time and not theta obviously,but if velocity becomes constant,, then its derivative with everything becomes 0, which is what i have shown, ofcourse this should happen just once and we should see it as a maxima,,, however through contradiction i have already shown that the cylinders must depart after this instant)– Mvs Saketh · 2 years, 6 months agoLog in to reply

– Deepanshu Gupta · 2 years, 6 months ago

yeah...!! Thanks a lot Saketh !!Log in to reply

@Mvs Saketh . The way you gave the mathematical explanation of stuff that why the cylinders loose contact is awesome. Thank you Deepanshu for sharing this problem.

Nice solutionIf there would have been friction between the contact contact surface then the cylinders would roll over each other. How should this question be solved? – Satvik Pandey · 2 years, 6 months ago

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Pandey jiwith love)I think that will be more interesting and difficult...!! I will surly give a try to it. !! – Deepanshu Gupta · 2 years, 6 months ago

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fig

Now if we apply conservation of energy we have to consider rotational energies also.

I tried to find relation between \( \omega\) and \(v\)

I came up with this equation

\({ v }_{ 2 }cos\theta \hat { i } +(r{ \omega }_{ 2 }-{ v }_{ 2 }sin\theta )\hat { j } ={ v }_{ 1 }sin\theta \hat { i } +(r{ \omega }_{ 1 }+{ v }_{ 1 }cos\theta )\hat { j } \)

On equating i cap I got \({ v }_{ 2 }={ v }_{ 1 }tan\theta \)

and on equating j cap I got \(r{ \omega }_{ 2 }-{ v }_{ 2 }sin\theta =r{ \omega }_{ 1 }+{ v }_{ 1 }cos\theta \)

or \(r{ \omega }_{ 2 }-r{ \omega }_{ 1 }=\frac { { v }_{ 1 } }{ cos\theta } \)

Also I think \({ \omega }_{ 2 }={ -\omega }_{ 1 }\).

@DEEPANSHU GUPTA and @Mvs Saketh , what are you guys getting? – Satvik Pandey · 2 years, 6 months ago

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– Satvik Pandey · 2 years, 6 months ago

Oh! yes w2 can not be equal to w1 but \( \alpha_{1}=-\alpha_{2}\). Can this help in any way? I don't think so. :DLog in to reply

EDITthat is correct and awesome – Mvs Saketh · 2 years, 6 months ago

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fig

The cylinder 1 tends to slip downwards w.r.t to cylinder 2 if there is no friction. So frictional force acts in upward direction.Right? So friction force act in downward direction on cylinder 2.Right?

So the magnitude of torque acting on the cylinders are equal but they are opposite in direction. Also the moment of inertia of the cylinders are equal.

So \(I \alpha_{1}=-I \alpha_{2}\). This shows that \( \alpha_{1}=- \alpha_{2}\). This is not correct. But where am I going wrong?

I have neglected torque due to normal force because they are passing through c.o.m and I have found the torques about the c.o.m.

Sorry if I am missing something very obvious. :( – Satvik Pandey · 2 years, 6 months ago

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\(r{ w }_{ 2 }-r{ w }_{ 1 }=\frac { { v }_{ 1 } }{ cos\theta } \\ \)

differentiating w.r.t time

\({ \alpha }_{ 2 }-r\alpha _{ 1 }=\frac { d }{ dt } \frac { { v }_{ 1 } }{ cos\theta } =0\\ \) from force analsys as you have shown

so in the process\( \frac { { v }_{ 1 } }{ cos\theta } \quad \) remains constant which is truely remarkable

And we can see that this only holds iff the cylinders have equal radii

Now this is equal to \({ v }_{ 1 }cos\theta \quad +\quad { v }_{ 2 }sin\theta \) which is nothing but the relative velocities of cylinders

comsperpendicular to the line joining their radii and you have shown that it remains constant,,, great, – Mvs Saketh · 2 years, 6 months agoLog in to reply

– Satvik Pandey · 2 years, 6 months ago

Most of my friends also call me by this name. :DLog in to reply

how do I prepare for IPHO – Avieroop Ghosh · 1 year, 7 months ago

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HOW do I prepare for ipho? please tell – Avieroop Ghosh · 1 year, 7 months ago

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