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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

I don't think that I am at that level to check your solution. But I have tried to solve the question and I got the same value of \(v_{1}\). Here is my solution.

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?

Log in to reply

your answer as far as i can comprehend seems absolutely correct

Log 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.

Log in to reply

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

Log in to reply

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 !!

Log in to reply

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. \( f.R=I\alpha \\ mg-Ncos\theta -fsin\theta =ma_{ 1 }\\ Nsin\theta -fcos\theta =ma_{ 2 }\\ \\ Equating\quad the\quad acceleration\quad of\quad point\quad of\quad contact\quad on\quad both\quad cylinders:\\ -a_{ 1 }\hat { j } +\quad R\alpha cos\theta \hat { i } +\quad R\alpha sin\theta \hat { j } =\quad a_{ 2 }\hat { i } -\quad R\alpha cos\theta \hat { i } -\quad R\alpha sin\theta \hat { j } \\ \therefore \quad a_{ 2 }=2R\alpha cos\theta \quad And\quad a_{ 1 }=2R\alpha sin\theta \quad \\ On\quad solving\quad a_{ 2 }=\frac { 2gsin\theta cos\theta }{ 3 } \quad ...(1)\\ X\quad co-ordinate\quad of\quad Centre\quad of\quad Bottom\quad cylinder,\quad x\quad =\quad R+2Rsin\theta \\ \therefore \quad dx=2Rcos\theta \quad \dot { \theta } \\ Writing\quad a_{ 2 }\quad as\quad v\quad dv/dx\quad and\quad substituting\quad with\quad dx\quad in\quad (1)\\ \\ v\quad dv\quad =\quad \frac { 4Rgsin\theta \cos ^{ 2 }{ \theta } }{ 3 } d\theta \\ Integration\quad gives\quad us\quad { v }^{ 2 }=\frac { 8gR }{ 9 } (1-\cos ^{ 3 }{ \theta } )\\ Which\quad gives\quad me\quad a\quad different\quad result\quad for\quad maximum\quad velocity.\quad Is\quad this\quad wrong?\quad Why\quad so?\\ Thanks\quad a\quad lot. \)Log 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

Log in to reply

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. :)

Log in to reply

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. \( f.R=I\alpha \\ mg-Ncos\theta -fsin\theta =ma_{ 1 }\\ Nsin\theta -fcos\theta =ma_{ 2 }\\ \\ Equating\quad the\quad acceleration\quad of\quad point\quad of\quad contact\quad on\quad both\quad cylinders:\\ -a_{ 1 }\hat { j } +\quad R\alpha cos\theta \hat { i } +\quad R\alpha sin\theta \hat { j } =\quad a_{ 2 }\hat { i } -\quad R\alpha cos\theta \hat { i } -\quad R\alpha sin\theta \hat { j } \\ \therefore \quad a_{ 2 }=2R\alpha cos\theta \quad And\quad a_{ 1 }=2R\alpha sin\theta \quad \\ On\quad solving\quad a_{ 2 }=\frac { 2gsin\theta cos\theta }{ 3 } \quad ...(1)\\ X\quad co-ordinate\quad of\quad Centre\quad of\quad Bottom\quad cylinder,\quad x\quad =\quad R+2Rsin\theta \\ \therefore \quad dx=2Rcos\theta \quad \dot { \theta } \\ Writing\quad a_{ 2 }\quad as\quad v\quad dv/dx\quad and\quad substituting\quad with\quad dx\quad in\quad (1)\\ \\ v\quad dv\quad =\quad \frac { 4Rgsin\theta \cos ^{ 2 }{ \theta } }{ 3 } d\theta \\ Integration\quad gives\quad us\quad { v }^{ 2 }=\frac { 8gR }{ 9 } (1-\cos ^{ 3 }{ \theta } )\\ Which\quad gives\quad me\quad a\quad different\quad result\quad for\quad maximum\quad velocity.\quad Is\quad this\quad wrong?\quad Why\quad so?\\ Thanks\quad a\quad lot. \)Log 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

Log in to reply

Log in to reply

Log in to reply

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 instant

Log in to reply

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

Log in to reply

Log in to reply

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 *Log in to reply

You are done Excellent work!! Hats off!!

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 ??

Log in to reply

Probably because initially the upper cylinder presses the lower cylinder more on the vertical direction and less on the horizontal direction initially ,, so the rate of change of velocity of lower cylinder gradually increases along with velocity,, but slowly as the relative velocity of seperation of the cylinders increase,,, they touch less and less harder with each other till a point where departure happens and further motion cant be treated as a single rigid body system,, funny thing though,, ( i have updated the calculations)

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)Log in to reply

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?

Log in to reply

Pandey jiwith love)I think that will be more interesting and difficult...!! I will surly give a try to it. !!

Log in to reply

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?

Log in to reply

Comment deleted Oct 31, 2014

Log in to reply

Log in to reply

EDITthat is correct and awesome

Log in to reply

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. :(

Log in to reply

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

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,Log in to reply

Log in to reply

how do I prepare for IPHO

Log in to reply

HOW do I prepare for ipho? please tell

Log in to reply